Vertical ranges of Column Plots in 12 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?CustomTicks and small rangesMultiple ranges in one plotFilling between quantile rangesCombine 3D plots with different ranges into oneMake a column of two plotsTransferFunctionModel and BodePlot shows different result between Mathematica and MatlabHow do I control the ranges in plots made with LogLogPlotMultiple plots sharing a single vertical axisVertical and Horizontal Shifts of PlotsConfiguration of `PlotLayout` as “Row”/“Column”
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Vertical ranges of Column Plots in 12
Vertical ranges of Column Plots in 12
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?CustomTicks and small rangesMultiple ranges in one plotFilling between quantile rangesCombine 3D plots with different ranges into oneMake a column of two plotsTransferFunctionModel and BodePlot shows different result between Mathematica and MatlabHow do I control the ranges in plots made with LogLogPlotMultiple plots sharing a single vertical axisVertical and Horizontal Shifts of PlotsConfiguration of `PlotLayout` as “Row”/“Column”
$begingroup$
Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.
It appears that the new capability always sets the PlotRange of all the data sets to the same value:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic]

I have tried what seems reasonable to set the ranges separately:
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, Automatic, Automatic, Automatic]
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]
But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?
EDIT:
Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)
d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
GraphicsColumn[
ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
FrameTicks -> Automatic, None, None, None,
FrameLabel -> None, "Linear"],
ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
ScalingFunctions -> "Log"]
]

plotting version-12
$endgroup$
add a comment |
$begingroup$
Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.
It appears that the new capability always sets the PlotRange of all the data sets to the same value:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic]

I have tried what seems reasonable to set the ranges separately:
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, Automatic, Automatic, Automatic]
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]
But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?
EDIT:
Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)
d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
GraphicsColumn[
ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
FrameTicks -> Automatic, None, None, None,
FrameLabel -> None, "Linear"],
ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
ScalingFunctions -> "Log"]
]

plotting version-12
$endgroup$
add a comment |
$begingroup$
Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.
It appears that the new capability always sets the PlotRange of all the data sets to the same value:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic]

I have tried what seems reasonable to set the ranges separately:
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, Automatic, Automatic, Automatic]
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]
But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?
EDIT:
Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)
d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
GraphicsColumn[
ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
FrameTicks -> Automatic, None, None, None,
FrameLabel -> None, "Linear"],
ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
ScalingFunctions -> "Log"]
]

plotting version-12
$endgroup$
Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.
It appears that the new capability always sets the PlotRange of all the data sets to the same value:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic]

I have tried what seems reasonable to set the ranges separately:
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, Automatic, Automatic, Automatic]
ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]
But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?
EDIT:
Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)
d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
GraphicsColumn[
ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
FrameTicks -> Automatic, None, None, None,
FrameLabel -> None, "Linear"],
ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
ScalingFunctions -> "Log"]
]

plotting version-12
plotting version-12
edited 2 hours ago
David Keith
asked 3 hours ago
David KeithDavid Keith
1,6901314
1,6901314
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is another new feature in M12.0: GraphicsColumn will now align the frames.
You can use this:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
GraphicsColumn[
ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
]

While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.
Here's what the same looks like in M11.3:

The grids and frames are not quite aligned.
$endgroup$
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
There is another new feature in M12.0: GraphicsColumn will now align the frames.
You can use this:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
GraphicsColumn[
ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
]

While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.
Here's what the same looks like in M11.3:

The grids and frames are not quite aligned.
$endgroup$
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
add a comment |
$begingroup$
There is another new feature in M12.0: GraphicsColumn will now align the frames.
You can use this:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
GraphicsColumn[
ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
]

While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.
Here's what the same looks like in M11.3:

The grids and frames are not quite aligned.
$endgroup$
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
add a comment |
$begingroup$
There is another new feature in M12.0: GraphicsColumn will now align the frames.
You can use this:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
GraphicsColumn[
ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
]

While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.
Here's what the same looks like in M11.3:

The grids and frames are not quite aligned.
$endgroup$
There is another new feature in M12.0: GraphicsColumn will now align the frames.
You can use this:
d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];
GraphicsColumn[
ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
]

While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.
Here's what the same looks like in M11.3:

The grids and frames are not quite aligned.
answered 3 hours ago
SzabolcsSzabolcs
165k14450954
165k14450954
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
add a comment |
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
$begingroup$
Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
$endgroup$
– David Keith
2 hours ago
add a comment |
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