Develop Reference

Setting the same x-scale but different x-limits for adjacent subplots matplotlib

I am trying to create a figure with three bar plots side by side. These bar plots have different yscales, but the data is fundamentally similar so I'd like all the bars to have the same width.
The only way I was able to get the bars to have the exact same width was by using sharex when creating the subplots, in order to keep the same x scale.
import matplotlib.pyplot as plt
BigData = [[100,300],[400,200]]
MediumData = [[40, 30],[50,20],[60,50],[30,30]]
SmallData = [[3,2],[11,3],[7,5]]
data = [BigData, MediumData, SmallData]
colors = ['#FC766A','#5B84B1']
fig, axs = plt.subplots(1, 3, figsize=(30,5), sharex=True)
subplot = 0
for scale in data:
for type in range(2):
bar_x = [x + type*0.2 for x in range(len(scale))]
bar_y = [d[type] for d in scale]
axs[subplot].bar(bar_x,bar_y, width = 0.2, color = colors[type])
subplot += 1
plt.show()
This creates this figure:
The problem with this is that the x-limits of the plot are also shared, leading to unwanted whitespace. I've tried setting the x-bounds after the fact, but it doesn't seem to override sharex. Is there a way to make the bars have the same width, without each subplot also being the same width?
Additionally, is there a way to create such a plot (one with different y scales to depending on the size of the data) without having to sort the data manually beforehand, like shown in my code?
Thanks!

Thanks to Jody Klymak for help finding this solution! I thought I should document it for future users.
We can make use of the 'width_ratios' GridSpec parameter. Unfortunately there's no way to specify these ratios after we've already drawn a graph, so the best way I found to implement this is to write a function that creates a dummy graph, and measures the x-limits from that graph:
def getXRatios(data, size):
phig, aks = plt.subplots(1, 3, figsize=size)
subplot = 0
for scale in data:
for type in range(2):
bar_x = [x + type*0.2 for x in range(len(scale))]
bar_y = [d[type] for d in scale]
aks[subplot].bar(bar_x,bar_y, width = 0.2)
subplot += 1
ratios = [aks[i].get_xlim()[1] for i in range(3)]
plt.close(phig)
return ratios
This is essentially identical to the code that creates the actual figure, with the cosmetic aspects removed, as all we want from this dummy figure is the x-limits of the graph (something we can't get from our actual figure as we need to define those limits before we start in order to solve the problem).
Now all you need to do is call this function when you're creating your subplots:
fig, axs = plt.subplots(1, 3, figsize=(40,5), gridspec_kw = {'width_ratios':getXRatios(data,(40,5))})
As long as your XRatio function creates your graph in the same way your actual graph does, everything should work! Here's my output using this solution.
To save space you could re-purpose the getXRatios function to also construct your final graph, by calling itself in the arguments and giving an option to return either the ratios or the final figure. I couldn't be bothered.

How to use a colored shape as yticks in matplotlib or seaborn?

I am working on a task called knowledge tracing which estimates the student mastery level over time. I would like to plot a similar figure as below using the Matplotlib or Seaborn.
It uses different colors to represent a knowledge concept, instead of a text. However, I have googled and found there is no article is talking about how we can do this.
I tried the following
# simulate a record of student mastery level
student_mastery = np.random.rand(5, 30)
df = pd.DataFrame(student_mastery)
# plot the heatmap using seaborn
marker = matplotlib.markers.MarkerStyle(marker='o', fillstyle='full')
sns_plot = sns.heatmap(df, cmap="RdYlGn", vmin=0.0, vmax=1.0)
y_limit = 5
y_labels = [marker for i in range(y_limit)]
plt.yticks(range(y_limit), y_labels)
Yet it simply returns the __repr__ of the marker, e.g., <matplotlib.markers.MarkerStyle at 0x1c5bb07860> on the yticks.
Thanks in advance!

While How can I make the xtick labels of a plot be simple drawings using matplotlib? gives you a general solution for arbitrary shapes, for the shapes shown here, it may make sense to use unicode symbols as text and colorize them according to your needs.
import matplotlib.pyplot as plt
import numpy as np; np.random.seed(42)
fig, ax = plt.subplots()
ax.imshow(np.random.rand(3,10), cmap="Greys")
symbolsx = ["⚪", "⚪", "⚫", "⚫", "⚪", "⚫","⚪", "⚫", "⚫","⚪"]
colorsx = np.random.choice(["#3ba1ab", "#b43232", "#8ecc3a", "#893bab"], 10)
ax.set_xticks(range(len(symbolsx)))
ax.set_xticklabels(symbolsx, size=40)
for tick, color in zip(ax.get_xticklabels(), colorsx):
tick.set_color(color)
symbolsy = ["◾", "◾", "◾"]
ax.set_yticks(range(len(symbolsy)))
ax.set_yticklabels(symbolsy, size=40)
for tick, color in zip(ax.get_yticklabels(), ["crimson", "gold", "indigo"]):
tick.set_color(color)
plt.show()

Matplotlib: Adjust legend location/position

I'm creating a figure with multiple subplots. One of these subplots is giving me some trouble, as none of the axes corners or centers are free (or can be freed up) for placing the legend. What I'd like to do is to have the legend placed somewhere in between the 'upper left' and 'center left' locations, while keeping the padding between it and the y-axis equal to the legends in the other subplots (that are placed using one of the predefined legend location keywords).
I know I can specify a custom position by using loc=(x,y), but then I can't figure out how to get the padding between the legend and the y-axis to be equal to that used by the other legends. Would it be possible to somehow use the borderaxespad property of the first legend? Though I'm not succeeding at getting that to work.
Any suggestions would be most welcome!
Edit: Here is a (very simplified) illustration of the problem:
import matplotlib.pyplot as plt
fig, ax = plt.subplots(1, 2, sharex=False, sharey=False)
ax[0].axhline(y=1, label='one')
ax[0].axhline(y=2, label='two')
ax[0].set_ylim([0.8,3.2])
ax[0].legend(loc=2)
ax[1].axhline(y=1, label='one')
ax[1].axhline(y=2, label='two')
ax[1].axhline(y=3, label='three')
ax[1].set_ylim([0.8,3.2])
ax[1].legend(loc=2)
plt.show()
What I'd like is that the legend in the right plot is moved down somewhat so it no longer overlaps with the line.
As a last resort I could change the axis limits, but I would very much like to avoid that.

I saw the answer you posted and tried it out. The problem however is that it is also depended on the figure size.
Here's a new try:
import numpy
import matplotlib.pyplot as plt
x = numpy.linspace(0, 10, 10000)
y = numpy.cos(x) + 2.
x_value = .014 #Offset by eye
y_value = .55
fig, ax = plt.subplots(1, 2, sharex = False, sharey = False)
fig.set_size_inches(50,30)
ax[0].plot(x, y, label = "cos")
ax[0].set_ylim([0.8,3.2])
ax[0].legend(loc=2)
line1 ,= ax[1].plot(x,y)
ax[1].set_ylim([0.8,3.2])
axbox = ax[1].get_position()
fig.legend([line1], ["cos"], loc = (axbox.x0 + x_value, axbox.y0 + y_value))
plt.show()
So what I am now doing is basically getting the coordinates from the subplot. I then create the legend based on the dimensions of the entire figure. Hence, the figure size does not change anything to the legend positioning anymore.
With the values for x_value and y_value the legend can be positioned in the subplot. x_value has been eyeballed for a good correspondence with the "normal" legend. This value can be changed at your desire. y_value determines the height of the legend.
Good luck!

After spending way too much time on this, I've come up with the following satisfactory solution (the Transformations Tutorial definitely helped):
bapad = plt.rcParams['legend.borderaxespad']
fontsize = plt.rcParams['font.size']
axline = plt.rcParams['axes.linewidth'] #need this, otherwise the result will be off by a few pixels
pad_points = bapad*fontsize + axline #padding is defined in relative to font size
pad_inches = pad_points/72.0 #convert from points to inches
pad_pixels = pad_inches*fig.dpi #convert from inches to pixels using the figure's dpi
Then, I found that both of the following work and give the same value for the padding:
# Define inverse transform, transforms display coordinates (pixels) to axes coordinates
inv = ax[1].transAxes.inverted()
# Inverse transform two points on the display and find the relative distance
pad_axes = inv.transform((pad_pixels, 0)) - inv.transform((0,0))
pad_xaxis = pad_axes[0]
or
# Find how may pixels there are on the x-axis
x_pixels = ax[1].transAxes.transform((1,0)) - ax[1].transAxes.transform((0,0))
# Compute the ratio between the pixel offset and the total amount of pixels
pad_xaxis = pad_pixels/x_pixels[0]
And then set the legend with:
ax[1].legend(loc=(pad_xaxis,0.6))
Plot:

Moving matplotlib legend outside of the axis makes it cutoff by the figure box

I'm familiar with the following questions:
Matplotlib savefig with a legend outside the plot
How to put the legend out of the plot
It seems that the answers in these questions have the luxury of being able to fiddle with the exact shrinking of the axis so that the legend fits.
Shrinking the axes, however, is not an ideal solution because it makes the data smaller making it actually more difficult to interpret; particularly when its complex and there are lots of things going on ... hence needing a large legend
The example of a complex legend in the documentation demonstrates the need for this because the legend in their plot actually completely obscures multiple data points.
http://matplotlib.sourceforge.net/users/legend_guide.html#legend-of-complex-plots
What I would like to be able to do is dynamically expand the size of the figure box to accommodate the expanding figure legend.
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(-2*np.pi, 2*np.pi, 0.1)
fig = plt.figure(1)
ax = fig.add_subplot(111)
ax.plot(x, np.sin(x), label='Sine')
ax.plot(x, np.cos(x), label='Cosine')
ax.plot(x, np.arctan(x), label='Inverse tan')
lgd = ax.legend(loc=9, bbox_to_anchor=(0.5,0))
ax.grid('on')
Notice how the final label 'Inverse tan' is actually outside the figure box (and looks badly cutoff - not publication quality!)
Finally, I've been told that this is normal behaviour in R and LaTeX, so I'm a little confused why this is so difficult in python... Is there a historical reason? Is Matlab equally poor on this matter?
I have the (only slightly) longer version of this code on pastebin http://pastebin.com/grVjc007

Sorry EMS, but I actually just got another response from the matplotlib mailling list (Thanks goes out to Benjamin Root).
The code I am looking for is adjusting the savefig call to:
fig.savefig('samplefigure', bbox_extra_artists=(lgd,), bbox_inches='tight')
#Note that the bbox_extra_artists must be an iterable
This is apparently similar to calling tight_layout, but instead you allow savefig to consider extra artists in the calculation. This did in fact resize the figure box as desired.
import matplotlib.pyplot as plt
import numpy as np
plt.gcf().clear()
x = np.arange(-2*np.pi, 2*np.pi, 0.1)
fig = plt.figure(1)
ax = fig.add_subplot(111)
ax.plot(x, np.sin(x), label='Sine')
ax.plot(x, np.cos(x), label='Cosine')
ax.plot(x, np.arctan(x), label='Inverse tan')
handles, labels = ax.get_legend_handles_labels()
lgd = ax.legend(handles, labels, loc='upper center', bbox_to_anchor=(0.5,-0.1))
text = ax.text(-0.2,1.05, "Aribitrary text", transform=ax.transAxes)
ax.set_title("Trigonometry")
ax.grid('on')
fig.savefig('samplefigure', bbox_extra_artists=(lgd,text), bbox_inches='tight')
This produces:
[edit] The intent of this question was to completely avoid the use of arbitrary coordinate placements of arbitrary text as was the traditional solution to these problems. Despite this, numerous edits recently have insisted on putting these in, often in ways that led to the code raising an error. I have now fixed the issues and tidied the arbitrary text to show how these are also considered within the bbox_extra_artists algorithm.

Added: I found something that should do the trick right away, but the rest of the code below also offers an alternative.
Use the subplots_adjust() function to move the bottom of the subplot up:
fig.subplots_adjust(bottom=0.2) # <-- Change the 0.02 to work for your plot.
Then play with the offset in the legend bbox_to_anchor part of the legend command, to get the legend box where you want it. Some combination of setting the figsize and using the subplots_adjust(bottom=...) should produce a quality plot for you.
Alternative:
I simply changed the line:
fig = plt.figure(1)
to:
fig = plt.figure(num=1, figsize=(13, 13), dpi=80, facecolor='w', edgecolor='k')
and changed
lgd = ax.legend(loc=9, bbox_to_anchor=(0.5,0))
to
lgd = ax.legend(loc=9, bbox_to_anchor=(0.5,-0.02))
and it shows up fine on my screen (a 24-inch CRT monitor).
Here figsize=(M,N) sets the figure window to be M inches by N inches. Just play with this until it looks right for you. Convert it to a more scalable image format and use GIMP to edit if necessary, or just crop with the LaTeX viewport option when including graphics.

Here is another, very manual solution. You can define the size of the axis and paddings are considered accordingly (including legend and tickmarks). Hope it is of use to somebody.
Example (axes size are the same!):
Code:
#==================================================
# Plot table
colmap = [(0,0,1) #blue
,(1,0,0) #red
,(0,1,0) #green
,(1,1,0) #yellow
,(1,0,1) #magenta
,(1,0.5,0.5) #pink
,(0.5,0.5,0.5) #gray
,(0.5,0,0) #brown
,(1,0.5,0) #orange
]
import matplotlib.pyplot as plt
import numpy as np
import collections
df = collections.OrderedDict()
df['labels'] = ['GWP100a\n[kgCO2eq]\n\nasedf\nasdf\nadfs','human\n[pts]','ressource\n[pts]']
df['all-petroleum long name'] = [3,5,2]
df['all-electric'] = [5.5, 1, 3]
df['HEV'] = [3.5, 2, 1]
df['PHEV'] = [3.5, 2, 1]
numLabels = len(df.values()[0])
numItems = len(df)-1
posX = np.arange(numLabels)+1
width = 1.0/(numItems+1)
fig = plt.figure(figsize=(2,2))
ax = fig.add_subplot(111)
for iiItem in range(1,numItems+1):
ax.bar(posX+(iiItem-1)*width, df.values()[iiItem], width, color=colmap[iiItem-1], label=df.keys()[iiItem])
ax.set(xticks=posX+width*(0.5*numItems), xticklabels=df['labels'])
#--------------------------------------------------
# Change padding and margins, insert legend
fig.tight_layout() #tight margins
leg = ax.legend(loc='upper left', bbox_to_anchor=(1.02, 1), borderaxespad=0)
plt.draw() #to know size of legend
padLeft = ax.get_position().x0 * fig.get_size_inches()[0]
padBottom = ax.get_position().y0 * fig.get_size_inches()[1]
padTop = ( 1 - ax.get_position().y0 - ax.get_position().height ) * fig.get_size_inches()[1]
padRight = ( 1 - ax.get_position().x0 - ax.get_position().width ) * fig.get_size_inches()[0]
dpi = fig.get_dpi()
padLegend = ax.get_legend().get_frame().get_width() / dpi
widthAx = 3 #inches
heightAx = 3 #inches
widthTot = widthAx+padLeft+padRight+padLegend
heightTot = heightAx+padTop+padBottom
# resize ipython window (optional)
posScreenX = 1366/2-10 #pixel
posScreenY = 0 #pixel
canvasPadding = 6 #pixel
canvasBottom = 40 #pixel
ipythonWindowSize = '{0}x{1}+{2}+{3}'.format(int(round(widthTot*dpi))+2*canvasPadding
,int(round(heightTot*dpi))+2*canvasPadding+canvasBottom
,posScreenX,posScreenY)
fig.canvas._tkcanvas.master.geometry(ipythonWindowSize)
plt.draw() #to resize ipython window. Has to be done BEFORE figure resizing!
# set figure size and ax position
fig.set_size_inches(widthTot,heightTot)
ax.set_position([padLeft/widthTot, padBottom/heightTot, widthAx/widthTot, heightAx/heightTot])
plt.draw()
plt.show()
#--------------------------------------------------
#==================================================

I tried a very simple way, just make the figure a bit wider:
fig, ax = plt.subplots(1, 1, figsize=(a, b))
adjust a and b to a proper value such that the legend is included in the figure

How can I set the aspect ratio in matplotlib?

I'm trying to make a square plot (using imshow), i.e. aspect ratio of 1:1, but I can't. None of these work:
import matplotlib.pyplot as plt
ax = fig.add_subplot(111,aspect='equal')
ax = fig.add_subplot(111,aspect=1.0)
ax.set_aspect('equal')
plt.axes().set_aspect('equal')
It seems like the calls are just being ignored (a problem I often seem to have with matplotlib).

Third times the charm. My guess is that this is a bug and Zhenya's answer suggests it's fixed in the latest version. I have version 0.99.1.1 and I've created the following solution:
import matplotlib.pyplot as plt
import numpy as np
def forceAspect(ax,aspect=1):
im = ax.get_images()
extent = im[0].get_extent()
ax.set_aspect(abs((extent[1]-extent[0])/(extent[3]-extent[2]))/aspect)
data = np.random.rand(10,20)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.imshow(data)
ax.set_xlabel('xlabel')
ax.set_aspect(2)
fig.savefig('equal.png')
ax.set_aspect('auto')
fig.savefig('auto.png')
forceAspect(ax,aspect=1)
fig.savefig('force.png')
This is 'force.png':
Below are my unsuccessful, yet hopefully informative attempts.
Second Answer:
My 'original answer' below is overkill, as it does something similar to axes.set_aspect(). I think you want to use axes.set_aspect('auto'). I don't understand why this is the case, but it produces a square image plot for me, for example this script:
import matplotlib.pyplot as plt
import numpy as np
data = np.random.rand(10,20)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.imshow(data)
ax.set_aspect('equal')
fig.savefig('equal.png')
ax.set_aspect('auto')
fig.savefig('auto.png')
Produces an image plot with 'equal' aspect ratio:
and one with 'auto' aspect ratio:
The code provided below in the 'original answer' provides a starting off point for an explicitly controlled aspect ratio, but it seems to be ignored once an imshow is called.
Original Answer:
Here's an example of a routine that will adjust the subplot parameters so that you get the desired aspect ratio:
import matplotlib.pyplot as plt
def adjustFigAspect(fig,aspect=1):
'''
Adjust the subplot parameters so that the figure has the correct
aspect ratio.
'''
xsize,ysize = fig.get_size_inches()
minsize = min(xsize,ysize)
xlim = .4*minsize/xsize
ylim = .4*minsize/ysize
if aspect < 1:
xlim *= aspect
else:
ylim /= aspect
fig.subplots_adjust(left=.5-xlim,
right=.5+xlim,
bottom=.5-ylim,
top=.5+ylim)
fig = plt.figure()
adjustFigAspect(fig,aspect=.5)
ax = fig.add_subplot(111)
ax.plot(range(10),range(10))
fig.savefig('axAspect.png')
This produces a figure like so:
I can imagine if your having multiple subplots within the figure, you would want to include the number of y and x subplots as keyword parameters (defaulting to 1 each) to the routine provided. Then using those numbers and the hspace and wspace keywords, you can make all the subplots have the correct aspect ratio.

A simple option using plt.gca() to get current axes and set aspect
plt.gca().set_aspect('equal')
in place of your last line

What is the matplotlib version you are running? I have recently had to upgrade to 1.1.0, and with it, add_subplot(111,aspect='equal') works for me.

After many years of success with the answers above, I have found this not to work again - but I did find a working solution for subplots at
https://jdhao.github.io/2017/06/03/change-aspect-ratio-in-mpl
With full credit of course to the author above (who can perhaps rather post here), the relevant lines are:
ratio = 1.0
xleft, xright = ax.get_xlim()
ybottom, ytop = ax.get_ylim()
ax.set_aspect(abs((xright-xleft)/(ybottom-ytop))*ratio)
The link also has a crystal clear explanation of the different coordinate systems used by matplotlib.
Thanks for all great answers received - especially #Yann's which will remain the winner.

you should try with figaspect. It works for me. From the docs:
Create a figure with specified aspect ratio. If arg is a number, use that aspect ratio. > If arg is an array, figaspect will
determine the width and height for a figure that would fit array
preserving aspect ratio. The figure width, height in inches are
returned. Be sure to create an axes with equal with and height, eg
Example usage:
# make a figure twice as tall as it is wide
w, h = figaspect(2.)
fig = Figure(figsize=(w,h))
ax = fig.add_axes([0.1, 0.1, 0.8, 0.8])
ax.imshow(A, **kwargs)
# make a figure with the proper aspect for an array
A = rand(5,3)
w, h = figaspect(A)
fig = Figure(figsize=(w,h))
ax = fig.add_axes([0.1, 0.1, 0.8, 0.8])
ax.imshow(A, **kwargs)
Edit: I am not sure of what you are looking for. The above code changes the canvas (the plot size). If you want to change the size of the matplotlib window, of the figure, then use:
In [68]: f = figure(figsize=(5,1))
this does produce a window of 5x1 (wxh).

This answer is based on Yann's answer. It will set the aspect ratio for linear or log-log plots. I've used additional information from https://stackoverflow.com/a/16290035/2966723 to test if the axes are log-scale.
def forceAspect(ax,aspect=1):
#aspect is width/height
scale_str = ax.get_yaxis().get_scale()
xmin,xmax = ax.get_xlim()
ymin,ymax = ax.get_ylim()
if scale_str=='linear':
asp = abs((xmax-xmin)/(ymax-ymin))/aspect
elif scale_str=='log':
asp = abs((scipy.log(xmax)-scipy.log(xmin))/(scipy.log(ymax)-scipy.log(ymin)))/aspect
ax.set_aspect(asp)
Obviously you can use any version of log you want, I've used scipy, but numpy or math should be fine.

In my case, the following setting works best:
plt.figure(figsize=(16,9))
where (16,9) is your plot aspect ratio.