I would like to have different font weights for each of my colorbar labels.
I have tried to let LaTeX format the labels in the following way:
import numpy as np
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
im = ax.imshow(np.random.rand(50, 50)/20)
cbar = ax.figure.colorbar(im, ticks=np.arange(0, 0.05, 0.01))
cbar.ax.set_yticklabels([r'{\fontsize{50pt}{3em}\selectfont{}{0}}',
r'{\fontsize{40pt}{3em}\selectfont{}{0.01}}',
r'{\fontsize{30pt}{3em}\selectfont{}{0.03}}',
r'{\fontsize{20pt}{3em}\selectfont{}{0.03}}',
r'{\fontsize{10pt}{3em}\selectfont{}{0.04}}',
r'{\fontsize{1pt}{3em}\selectfont{}{0.05}}', ])
but this only updates the text of the labels to the whole string (e.g., {\fontsize{50pt}{3em}\selectfont{}{0}}). The pyplot TeX demo works for me. Even if this solution would work it would not be ideal as I would probably need to specify everything manually.
Much more convenient would be something like in this question. There, I learned that the font size of single labels of the regular x and y axis can be updated by calling
label = axes.yaxis.get_major_ticks()[2].label
label.set_fontsize(size)
replacing set_fontsize by set_fontweight correctly updates the weight of the selected label.
Unfortunately I could not find the equivalent of axes.yaxis.get_major_ticks()[2].label for the colorbar.
Is it possible to change the font weight of individual labels of the colorbar directly? With directly I mean without using a workaround like plotting some new text above existing labels.
If this is not possible a solution plotting text above existing labels which automatically uses the position and content the previous labels and only adjusts the font weight would also be appreciated.
Thanks!
As pointed out by #ImportanceOfBingErnest , set_fontweight works for setting the weight of single colorbar labels too.
I had to try a couple of things to find which call would give me the text objects defining the colorbar labels. They are accessible in cbar.ax.get_yticklabels().
The code snippet below now properly changes the weight of the second colorbar label:
import numpy as np
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
im = ax.imshow(np.random.rand(50, 50)/20)
cbar = ax.figure.colorbar(im, ticks=np.arange(0, 0.05, 0.01))
cbar.ax.get_yticklabels()[1].set_fontweight(1000)
plt.show()
Output of code (not enough reputation for inline images)
Related
I've been working for a while with the matplotlib package in Python, and I know that you can do 2D graphs (usually involving two "dimensions", x and y) or 3D graphs (with functions like plot3D). However, I am unable to find documentation about giving a '3D aesthetic' to a 2D plot.
That is, giving the plot a bit of volume, some shadows, etc.
To give an example, let's say I wanted to create a donut chart in matplotlib. A first draft could be something like this:
import matplotlib.pyplot as plt
#Given an array of values 'values' and,
#optionally, an array of colors 'colors'
#and an array of labels 'labels':
ax = plt.subplot()
ax.pie(
x = values,
labels = labels,
colors = colors
)
center_circle = plt.Circle((0,0), radius = 0.5, fc = "white")
ax.add_artist(center_circle)
plt.show()
However, a quick graph with Excel can give a much more appealing result:
Looking at the documentation of plt.pie, I was not able to find anything significant, apart from the parameter shadow, which when set to True, gives an underwhelming result:
Also, I would like to add effect such as the use of bevel (like the 3d-look of the borders of each wedge of the pie) and more style things. How could I improve the look of my graph with matplotlib? Is it even possible to accomplish it with this library?
One solution might be using a different library. I am not familiar with seaborn, but I know it is also a powerful visualisation library. The same with plotly. Does any one of these libraries allow for these kind of customisations?
There are a whole bunch of options on the matplotlib website for pie charts here: https://matplotlib.org/stable/gallery/pie_and_polar_charts/index.html
Matplotlib does not have a built-in option to add a bevel to a 2D pie chart or any other types of charts directly.
But, you could do this (raised shaddow) for a 3d effect:
import matplotlib.pyplot as plt
# Pie chart, where the slices will be ordered and plotted counter-clockwise:
labels = 'Frogs', 'Hogs', 'Dogs', 'Logs'
sizes = [15, 30, 45, 10]
explode = (0, 0.1, 0, 0) # only "explode" the 2nd slice (i.e. 'Hogs')
fig1, ax1 = plt.subplots()
ax1.pie(sizes, explode=explode, labels=labels, autopct='%1.1f%%',
shadow=True, startangle=90)
ax1.axis('equal') # Equal aspect ratio ensures that pie is drawn as a circle.
plt.show()
which give this:
I can't figure out how to rotate the text on the X Axis. Its a time stamp, so as the number of samples increase, they get closer and closer until they overlap. I'd like to rotate the text 90 degrees so as the samples get closer together, they aren't overlapping.
Below is what I have, it works fine with the exception that I can't figure out how to rotate the X axis text.
import sys
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
import datetime
font = {'family' : 'normal',
'weight' : 'bold',
'size' : 8}
matplotlib.rc('font', **font)
values = open('stats.csv', 'r').readlines()
time = [datetime.datetime.fromtimestamp(float(i.split(',')[0].strip())) for i in values[1:]]
delay = [float(i.split(',')[1].strip()) for i in values[1:]]
plt.plot(time, delay)
plt.grid(b='on')
plt.savefig('test.png')
This works for me:
plt.xticks(rotation=90)
Many "correct" answers here but I'll add one more since I think some details are left out of several. The OP asked for 90 degree rotation but I'll change to 45 degrees because when you use an angle that isn't zero or 90, you should change the horizontal alignment as well; otherwise your labels will be off-center and a bit misleading (and I'm guessing many people who come here want to rotate axes to something other than 90).
Easiest / Least Code
Option 1
plt.xticks(rotation=45, ha='right')
As mentioned previously, that may not be desirable if you'd rather take the Object Oriented approach.
Option 2
Another fast way (it's intended for date objects but seems to work on any label; doubt this is recommended though):
fig.autofmt_xdate(rotation=45)
fig you would usually get from:
fig = plt.gcf()
fig = plt.figure()
fig, ax = plt.subplots()
fig = ax.figure
Object-Oriented / Dealing directly with ax
Option 3a
If you have the list of labels:
labels = ['One', 'Two', 'Three']
ax.set_xticks([1, 2, 3])
ax.set_xticklabels(labels, rotation=45, ha='right')
In later versions of Matplotlib (3.5+), you can just use set_xticks alone:
ax.set_xticks([1, 2, 3], labels, rotation=45, ha='right')
Option 3b
If you want to get the list of labels from the current plot:
# Unfortunately you need to draw your figure first to assign the labels,
# otherwise get_xticklabels() will return empty strings.
plt.draw()
ax.set_xticks(ax.get_xticks())
ax.set_xticklabels(ax.get_xticklabels(), rotation=45, ha='right')
As above, in later versions of Matplotlib (3.5+), you can just use set_xticks alone:
ax.set_xticks(ax.get_xticks(), ax.get_xticklabels(), rotation=45, ha='right')
Option 4
Similar to above, but loop through manually instead.
for label in ax.get_xticklabels():
label.set_rotation(45)
label.set_ha('right')
Option 5
We still use pyplot (as plt) here but it's object-oriented because we're changing the property of a specific ax object.
plt.setp(ax.get_xticklabels(), rotation=45, ha='right')
Option 6
This option is simple, but AFAIK you can't set label horizontal align this way so another option might be better if your angle is not 90.
ax.tick_params(axis='x', labelrotation=45)
Edit:
There's discussion of this exact "bug" but a fix hasn't been released (as of 3.4.0):
https://github.com/matplotlib/matplotlib/issues/13774
Easy way
As described here, there is an existing method in the matplotlib.pyplot figure class that automatically rotates dates appropriately for you figure.
You can call it after you plot your data (i.e.ax.plot(dates,ydata) :
fig.autofmt_xdate()
If you need to format the labels further, checkout the above link.
Non-datetime objects
As per languitar's comment, the method I suggested for non-datetime xticks would not update correctly when zooming, etc. If it's not a datetime object used as your x-axis data, you should follow Tommy's answer:
for tick in ax.get_xticklabels():
tick.set_rotation(45)
Try pyplot.setp. I think you could do something like this:
x = range(len(time))
plt.xticks(x, time)
locs, labels = plt.xticks()
plt.setp(labels, rotation=90)
plt.plot(x, delay)
Appart from
plt.xticks(rotation=90)
this is also possible:
plt.xticks(rotation='vertical')
I came up with a similar example. Again, the rotation keyword is.. well, it's key.
from pylab import *
fig = figure()
ax = fig.add_subplot(111)
ax.bar( [0,1,2], [1,3,5] )
ax.set_xticks( [ 0.5, 1.5, 2.5 ] )
ax.set_xticklabels( ['tom','dick','harry'], rotation=45 ) ;
If you want to apply rotation on the axes object, the easiest way is using tick_params. For example.
ax.tick_params(axis='x', labelrotation=90)
Matplotlib documentation reference here.
This is useful when you have an array of axes as returned by plt.subplots, and it is more convenient than using set_xticks because in that case you need to also set the tick labels, and also more convenient that those that iterate over the ticks (for obvious reasons)
If using plt:
plt.xticks(rotation=90)
In case of using pandas or seaborn to plot, assuming ax as axes for the plot:
ax.set_xticklabels(ax.get_xticklabels(), rotation=90)
Another way of doing the above:
for tick in ax.get_xticklabels():
tick.set_rotation(45)
My answer is inspired by cjohnson318's answer, but I didn't want to supply a hardcoded list of labels; I wanted to rotate the existing labels:
for tick in ax.get_xticklabels():
tick.set_rotation(45)
The simplest solution is to use:
plt.xticks(rotation=XX)
but also
# Tweak spacing to prevent clipping of tick-labels
plt.subplots_adjust(bottom=X.XX)
e.g for dates I used rotation=45 and bottom=0.20 but you can do some test for your data
import pylab as pl
pl.xticks(rotation = 90)
To rotate the x-axis label to 90 degrees
for tick in ax.get_xticklabels():
tick.set_rotation(45)
It will depend on what are you plotting.
import matplotlib.pyplot as plt
x=['long_text_for_a_label_a',
'long_text_for_a_label_b',
'long_text_for_a_label_c']
y=[1,2,3]
myplot = plt.plot(x,y)
for item in myplot.axes.get_xticklabels():
item.set_rotation(90)
For pandas and seaborn that give you an Axes object:
df = pd.DataFrame(x,y)
#pandas
myplot = df.plot.bar()
#seaborn
myplotsns =sns.barplot(y='0', x=df.index, data=df)
# you can get xticklabels without .axes cause the object are already a
# isntance of it
for item in myplot.get_xticklabels():
item.set_rotation(90)
If you need to rotate labels you may need change the font size too, you can use font_scale=1.0 to do that.
I created a scatter plot in seaborn using seaborn.relplot, but am having trouble putting the legend all in one graph.
When I do this simple way, everything works fine:
import pandas as pd
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
import seaborn as sns
df2 = df[df.ln_amt_000s < 700]
sns.relplot(x='ln_amt_000s', y='hud_med_fm_inc', hue='outcome', size='outcome', legend='brief', ax=ax, data=df2)
The result is a scatter plot as desired, with the legend on the right hand side.
However, when I try to generate a matplotlib figure and axes objects ahead of time to specify the figure dimensions I run into problems:
a4_dims = (10, 10) # generating a matplotlib figure and axes objects ahead of time to specify figure dimensions
df2 = df[df.ln_amt_000s < 700]
fig, ax = plt.subplots(figsize = a4_dims)
sns.relplot(x='ln_amt_000s', y='hud_med_fm_inc', hue='outcome', size='outcome', legend='brief', ax=ax, data=df2)
The result is two graphs -- one that has the scatter plots as expected but missing the legend, and another one below it that is all blank except for the legend on the right hand side.
How do I fix this such? My desired result is one graph where I can specify the figure dimensions and have the legend at the bottom in two rows, below the x-axis (if that is too difficult, or not supported, then the default legend position to the right on the same graph would work too)? I know the problem lies with "ax=ax", and in the way I am specifying the dimensions as matplotlib figure, but I'd like to know specifically why this causes a problem so I can learn from this.
Thank you for your time.
The issue is that sns.relplot is a "Figure-level interface for drawing relational plots onto a FacetGrid" (see the API page). With a simple sns.scatterplot (the default type of plot used by sns.relplot), your code works (changed to use reproducible data):
df = pd.read_csv("https://vincentarelbundock.github.io/Rdatasets/csv/datasets/iris.csv", index_col=0)
fig, ax = plt.subplots(figsize = (5,5))
sns.scatterplot(x = 'Sepal.Length', y = 'Sepal.Width',
hue = 'Species', legend = 'brief',
ax=ax, data = df)
plt.show()
Further edits to legend
Seaborn's legends are a bit finicky. Some tweaks you may want to employ:
Remove the default seaborn title, which is actually a legend entry, by getting and slicing the handles and labels
Set a new title that is actually a title
Move the location and make use of bbox_to_anchor to move outside the plot area (note that the bbox parameters need some tweaking depending on your plot size)
Specify the number of columns
fig, ax = plt.subplots(figsize = (5,5))
sns.scatterplot(x = 'Sepal.Length', y = 'Sepal.Width',
hue = 'Species', legend = 'brief',
ax=ax, data = df)
handles, labels = ax.get_legend_handles_labels()
ax.legend(handles=handles[1:], labels=labels[1:], loc=8,
ncol=2, bbox_to_anchor=[0.5,-.3,0,0])
plt.show()
I am using python-3.x and I would like to plot several boxplots in one figure, all the data from one numpy array where the shape of this array is (100, 301)
If I use the code below it will plot them all (I will have 301 boxplots in one figure which is too much)
fig, ax = plt.subplots()
ax.boxplot(my_data)
plt.show()
I don't want to plot all the data, I just want to plot 10, 15 or 20 (variable number) of the data by using for loop or any method that work best.
for example, I want to plot boxplots every 50 number of data that mean I will have around 6 boxplots from 301 in my figure, I tried to use for loop but no luck
Any advice would be much appreciated
You can just use indexing to plot every 50th data points using a variable step. To have separate box plots and avoid overlapping, you can specify the positions of individual box plot using the positions parameter. my_data[:, ::step] gives you the desired data to plot. Below is an example using some random data.
import numpy as np
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
my_data = np.random.randint(0, 20, (100, 301))
step = 50
posit = range(my_data[:, ::step].shape[1])
ax.boxplot(my_data[:, ::step], positions=posit)
plt.show()
I have the results of a (H,ranges) = numpy.histogram2d() computation and I'm trying to plot it.
Given H I can easily put it into plt.imshow(H) to get the corresponding image. (see http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.imshow )
My problem is that the axis of the produced image are the "cell counting" of H and are completely unrelated to the values of ranges.
I know I can use the keyword extent (as pointed in: Change values on matplotlib imshow() graph axis ). But this solution does not work for me: my values on range are not growing linearly (actually they are going exponentially)
My question is: How can I put the value of range in plt.imshow()? Or at least, or can I manually set the label values of the plt.imshow resulting object?
Editing the extent is not a good solution.
You can just change the tick labels to something more appropriate for your data.
For example, here we'll set every 5th pixel to an exponential function:
import numpy as np
import matplotlib.pyplot as plt
im = np.random.rand(21,21)
fig,(ax1,ax2) = plt.subplots(1,2)
ax1.imshow(im)
ax2.imshow(im)
# Where we want the ticks, in pixel locations
ticks = np.linspace(0,20,5)
# What those pixel locations correspond to in data coordinates.
# Also set the float format here
ticklabels = ["{:6.2f}".format(i) for i in np.exp(ticks/5)]
ax2.set_xticks(ticks)
ax2.set_xticklabels(ticklabels)
ax2.set_yticks(ticks)
ax2.set_yticklabels(ticklabels)
plt.show()
Expanding a bit on #thomas answer
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.image as mi
im = np.random.rand(20, 20)
ticks = np.exp(np.linspace(0, 10, 20))
fig, ax = plt.subplots()
ax.pcolor(ticks, ticks, im, cmap='viridis')
ax.set_yscale('log')
ax.set_xscale('log')
ax.set_xlim([1, np.exp(10)])
ax.set_ylim([1, np.exp(10)])
By letting mpl take care of the non-linear mapping you can now accurately over-plot other artists. There is a performance hit for this (as pcolor is more expensive to draw than AxesImage), but getting accurate ticks is worth it.
imshow is for displaying images, so it does not support x and y bins.
You could either use pcolor instead,
H,xedges,yedges = np.histogram2d()
plt.pcolor(xedges,yedges,H)
or use plt.hist2d which directly plots your histogram.