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How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?

7

I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.

Ex :

#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4] 
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
 asd = i[0] + i[1]
 asd2= i[1] + i[2]

 list2 = [asd, asd2]
 if all(elem in list2 for elem in list1):
 print (i,list2)

It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.

python python-3.x

share|improve this question

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.
  
  – gilch
  4 hours ago
  

  
  
  
  

add a comment | 

7

I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.

Ex :

#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4] 
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
 asd = i[0] + i[1]
 asd2= i[1] + i[2]

 list2 = [asd, asd2]
 if all(elem in list2 for elem in list1):
 print (i,list2)

It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.

python python-3.x

share|improve this question

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.
  
  – gilch
  4 hours ago
  

  
  
  
  

add a comment | 

I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.

Ex :

#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4] 
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
 asd = i[0] + i[1]
 asd2= i[1] + i[2]

 list2 = [asd, asd2]
 if all(elem in list2 for elem in list1):
 print (i,list2)

It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.

python python-3.x

share|improve this question

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4] 
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.

list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
 asd = i[0] + i[1]
 asd2= i[1] + i[2]

 list2 = [asd, asd2]
 if all(elem in list2 for elem in list1):
 print (i,list2)

share|improve this question

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

share|improve this question

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

edited 4 hours ago

petezurich

3,76581936

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

asked 5 hours ago

Vitor OliveiraVitor Oliveira

525

2 Answers
2

active

oldest

votes

8

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
 """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
 return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

share|improve this answer

edited 4 hours ago

answered 5 hours ago

gilchgilch

4,3201716

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  that's slick @gilch
  
  – modesitt
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Does this work for something like a_all_in_b([1], [1, 1])?
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".
  
  – gilch
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Also it should be <=, not <.
  
  – user2357112
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

1

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 

list1 = [2,2,2,6] 
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
 c1 = Counter(a)
 c2 = Counter(b)

 for key,value in c1.items():
 if c2[key] != value:
 return False
 return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

Output:

True
False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

share|improve this answer

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works too. Thanks !
  
  – Vitor Oliveira
  3 hours ago
  

  
  
  
  

add a comment | 

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8

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
 """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
 return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

share|improve this answer

edited 4 hours ago

answered 5 hours ago

gilchgilch

4,3201716

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  that's slick @gilch
  
  – modesitt
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Does this work for something like a_all_in_b([1], [1, 1])?
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".
  
  – gilch
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Also it should be <=, not <.
  
  – user2357112
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

8

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
 """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
 return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

share|improve this answer

edited 4 hours ago

answered 5 hours ago

gilchgilch

4,3201716

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  that's slick @gilch
  
  – modesitt
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Does this work for something like a_all_in_b([1], [1, 1])?
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".
  
  – gilch
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
  
  – Tomothy32
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Also it should be <=, not <.
  
  – user2357112
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
 """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
 return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

share|improve this answer

edited 4 hours ago

answered 5 hours ago

gilchgilch

4,3201716

from collections import Counter

def a_all_in_b(a, b):
 """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
 return Counter(a).items() <= Counter(b).items()

share|improve this answer

edited 4 hours ago

answered 5 hours ago

gilchgilch

4,3201716

answered 5 hours ago

gilchgilch

4,3201716

answered 5 hours ago

gilchgilch

4,3201716

1

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 

list1 = [2,2,2,6] 
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
 c1 = Counter(a)
 c2 = Counter(b)

 for key,value in c1.items():
 if c2[key] != value:
 return False
 return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

Output:

True
False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

share|improve this answer

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works too. Thanks !
  
  – Vitor Oliveira
  3 hours ago
  

  
  
  
  

add a comment | 

1

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 

list1 = [2,2,2,6] 
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
 c1 = Counter(a)
 c2 = Counter(b)

 for key,value in c1.items():
 if c2[key] != value:
 return False
 return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

Output:

True
False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

share|improve this answer

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works too. Thanks !
  
  – Vitor Oliveira
  3 hours ago
  

  
  
  
  

add a comment | 

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 

list1 = [2,2,2,6] 
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
 c1 = Counter(a)
 c2 = Counter(b)

 for key,value in c1.items():
 if c2[key] != value:
 return False
 return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

Output:

True
False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

share|improve this answer

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

from collections import Counter 

list1 = [2,2,2,6] 
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
 c1 = Counter(a)
 c2 = Counter(b)

 for key,value in c1.items():
 if c2[key] != value:
 return False
 return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

True
False

share|improve this answer

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544

answered 4 hours ago

Patrick ArtnerPatrick Artner

26k62544