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Why can I use a list index as an indexing variable in a for loop?



The 2019 Stack Overflow Developer Survey Results Are InHow do I check if a list is empty?Finding the index of an item given a list containing it in PythonAccessing the index in 'for' loops?Understanding Python super() with __init__() methodsHow do I sort a dictionary by value?How to make a flat list out of list of lists?How do I pass a variable by reference?Loop through an array in JavaScriptHow do I list all files of a directory?Why is printing “B” dramatically slower than printing “#”?



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12















I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    42 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    21 mins ago

















12















I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    42 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    21 mins ago













12












12








12


7






I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?







python for-loop






share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 34 mins ago









Amir A. Shabani

457616




457616






New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Kundan VermaKundan Verma

712




712




New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    42 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    21 mins ago












  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    42 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    21 mins ago







7




7





Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

– gameon67
1 hour ago





Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

– gameon67
1 hour ago




4




4





I don't know why you would ever want to do this, but now I know you can

– Nathan
1 hour ago





I don't know why you would ever want to do this, but now I know you can

– Nathan
1 hour ago




8




8





This would be a great question for an awful coding interview

– Nathan
1 hour ago





This would be a great question for an awful coding interview

– Nathan
1 hour ago




1




1





OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

– Christian Dean
42 mins ago





OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

– Christian Dean
42 mins ago




1




1





This question first got two down votes :) I guess you shouldn't panic by down votes!

– Amir A. Shabani
21 mins ago





This question first got two down votes :) I guess you shouldn't panic by down votes!

– Amir A. Shabani
21 mins ago












5 Answers
5






active

oldest

votes


















10














To bring a language-lawyer aspect to the question, let's look at documentation:




for_stmt ::= "for" target_list "in" expression_list ":" suite


The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



Let's extend the print statement:



a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])


This gives the following output:



[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)


As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)





share|improve this answer




















  • 2





    Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

    – Mateen Ulhaq
    59 mins ago







  • 2





    Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

    – Christian Dean
    47 mins ago


















2














it is an interesting question, you can understand it by that:



for v in a:
a[-1] = v
print(a[-1])

print(a)


actually a becomes: [0, 1, 2, 2] after loop



output:



0
1
2
2
[0, 1, 2, 2]


Hope that helps you, and comment if you have further questions. : )






share|improve this answer






























    2














    (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



    If you had



    x = 0
    l = [1, 2, 3]
    for x in l:
    print(x)


    you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



    When we do



    a = [0, 1, 2, 3]

    for a[-1] in a:
    print(a[-1])


    even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






    share|improve this answer

























    • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

      – Amir A. Shabani
      50 mins ago


















    1














    The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



    for n in a:
    print(n)


    is just a fancy way of doing:



    for i in range(len(a)):
    n = a[i]
    print(n)


    Likewise,



    for a[-1] in a:
    print(a[-1])


    is just a fancy way of doing:



    for i in range(len(a)):
    a[-1] = a[i]
    print(a[-1])


    where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






    share|improve this answer






























      1














      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






      share|improve this answer

























      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

        – Amir A. Shabani
        1 hour ago






      • 1





        Agree, I don't really understand by reading this answer

        – gameon67
        1 hour ago











      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

        – Nathan
        1 hour ago











      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

        – Tom Karzes
        1 hour ago






      • 1





        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

        – Christian Dean
        39 mins ago












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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10














      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer




















      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        59 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        47 mins ago















      10














      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer




















      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        59 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        47 mins ago













      10












      10








      10







      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer















      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 4 mins ago

























      answered 1 hour ago









      TrebledJTrebledJ

      3,76421329




      3,76421329







      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        59 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        47 mins ago












      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        59 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        47 mins ago







      2




      2





      Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

      – Mateen Ulhaq
      59 mins ago






      Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

      – Mateen Ulhaq
      59 mins ago





      2




      2





      Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

      – Christian Dean
      47 mins ago





      Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

      – Christian Dean
      47 mins ago













      2














      it is an interesting question, you can understand it by that:



      for v in a:
      a[-1] = v
      print(a[-1])

      print(a)


      actually a becomes: [0, 1, 2, 2] after loop



      output:



      0
      1
      2
      2
      [0, 1, 2, 2]


      Hope that helps you, and comment if you have further questions. : )






      share|improve this answer



























        2














        it is an interesting question, you can understand it by that:



        for v in a:
        a[-1] = v
        print(a[-1])

        print(a)


        actually a becomes: [0, 1, 2, 2] after loop



        output:



        0
        1
        2
        2
        [0, 1, 2, 2]


        Hope that helps you, and comment if you have further questions. : )






        share|improve this answer

























          2












          2








          2







          it is an interesting question, you can understand it by that:



          for v in a:
          a[-1] = v
          print(a[-1])

          print(a)


          actually a becomes: [0, 1, 2, 2] after loop



          output:



          0
          1
          2
          2
          [0, 1, 2, 2]


          Hope that helps you, and comment if you have further questions. : )






          share|improve this answer













          it is an interesting question, you can understand it by that:



          for v in a:
          a[-1] = v
          print(a[-1])

          print(a)


          actually a becomes: [0, 1, 2, 2] after loop



          output:



          0
          1
          2
          2
          [0, 1, 2, 2]


          Hope that helps you, and comment if you have further questions. : )







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          recnacrecnac

          1,218123




          1,218123





















              2














              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer

























              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                50 mins ago















              2














              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer

























              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                50 mins ago













              2












              2








              2







              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer















              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 47 mins ago









              Amir A. Shabani

              457616




              457616










              answered 54 mins ago









              NathanNathan

              2,02511326




              2,02511326












              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                50 mins ago

















              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                50 mins ago
















              Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

              – Amir A. Shabani
              50 mins ago





              Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

              – Amir A. Shabani
              50 mins ago











              1














              The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



              for n in a:
              print(n)


              is just a fancy way of doing:



              for i in range(len(a)):
              n = a[i]
              print(n)


              Likewise,



              for a[-1] in a:
              print(a[-1])


              is just a fancy way of doing:



              for i in range(len(a)):
              a[-1] = a[i]
              print(a[-1])


              where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






              share|improve this answer



























                1














                The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                for n in a:
                print(n)


                is just a fancy way of doing:



                for i in range(len(a)):
                n = a[i]
                print(n)


                Likewise,



                for a[-1] in a:
                print(a[-1])


                is just a fancy way of doing:



                for i in range(len(a)):
                a[-1] = a[i]
                print(a[-1])


                where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






                share|improve this answer

























                  1












                  1








                  1







                  The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                  for n in a:
                  print(n)


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  n = a[i]
                  print(n)


                  Likewise,



                  for a[-1] in a:
                  print(a[-1])


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  a[-1] = a[i]
                  print(a[-1])


                  where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






                  share|improve this answer













                  The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                  for n in a:
                  print(n)


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  n = a[i]
                  print(n)


                  Likewise,



                  for a[-1] in a:
                  print(a[-1])


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  a[-1] = a[i]
                  print(a[-1])


                  where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 59 mins ago









                  blhsingblhsing

                  43.5k41744




                  43.5k41744





















                      1














                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer

























                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        39 mins ago
















                      1














                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer

























                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        39 mins ago














                      1












                      1








                      1







                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer















                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 20 mins ago

























                      answered 1 hour ago









                      Tom KarzesTom Karzes

                      11.2k1926




                      11.2k1926












                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        39 mins ago


















                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        39 mins ago

















                      How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                      – Amir A. Shabani
                      1 hour ago





                      How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                      – Amir A. Shabani
                      1 hour ago




                      1




                      1





                      Agree, I don't really understand by reading this answer

                      – gameon67
                      1 hour ago





                      Agree, I don't really understand by reading this answer

                      – gameon67
                      1 hour ago













                      I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                      – Nathan
                      1 hour ago





                      I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                      – Nathan
                      1 hour ago













                      It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                      – Tom Karzes
                      1 hour ago





                      It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                      – Tom Karzes
                      1 hour ago




                      1




                      1





                      "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                      – Christian Dean
                      39 mins ago






                      "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                      – Christian Dean
                      39 mins ago











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                      ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result