Given the following code:
import matplotlib.pyplot as plt
import numpy as np
x = [1.0, 1.1, 2.0, 5.7]
y = np.arange(len(x))
fsize=(2,2)
fig, ax = plt.subplots(1,1,figsize=fsize)
ax.set_yticklabels(['a','b','c','d'])
ax.barh(y,x,align='center',color='grey')
plt.show()
Why are the labels not showing as expected ('a' does not show up and everything is shifted down by 1 place)?
The locator is generating an extra tick on each side (which are not being shown because they is outside the plotted data). Try the following:
>>> ax.get_yticks()
array([-1., 0., 1., 2., 3., 4.])
You have a couple of options. You can either hard-code your tick labels to include the extra ticks (which I think is a bad idea):
ax.set_yticklabels(list(' abcd')) # You don't really need 'e'
Or, you can set the ticks to where you want them to be along with the labels:
ax.set_yticks(y)
ax.set_yticklabels(list('abcd'))
A more formal solution to the tick problem would be to set a Locator object on the y-axis. The tick label problem is formally solved by setting the Formatter for the y-axis. That is essentially what is happening under the hood when you call set_yticks and set_yticklabels anyway, but this way you have full control:
from matplotlib.ticker import FixedLocator, FixedFormatter
...
ax.yaxis.set_major_locator(FixedLocator(y))
ax.yaxis.set_major_formatter(FixedFormatter(list('abcd')))
No, none of that works for me on Windows, Python 3.7.
Just print it as text with the x-axis location slightly to the left of the smallest x value.
import matplotlib.pyplot as plt
import numpy as np
ax.set_yticklabels("")
x = [1.0, 1.1, 2.0, 5.7]
y = np.arange(len(x))
fsize=(2,2)
fig, ax = plt.subplots(1,1,figsize=fsize)
names = ['a','b','c','d']
for i in np.arange(len(x)):
plt.text(-0.05,len(x)-i-1,names[i])
plt.show()
Related
I'm trying to transform the scales on y-axis to the log values. For example, if one of the numbers on y is 0.01, I want to get -2 (which is log(0.01)). How should I do this in matplotlib (or any other library)?!
Thanks,
Without plt.yscale('log') there will be few y-ticks visible that have a nice number as log. You can change the "formatter" to a function that only shows the exponent. Also note that in the latest seaborn version distplot has been replaced by histplot(..., kde=True) or kdeplot(...).
Here is an example:
import matplotlib.pyplot as plt
from matplotlib.ticker import LogFormatterExponent
import numpy as np
import seaborn as sns
x = np.random.randn(10, 1000).cumsum(axis=1).ravel()
ax = sns.histplot(x, kde=True, stat='density', color='purple')
ax.set_yscale('log')
ax.yaxis.set_major_formatter(LogFormatterExponent(base=10.0, labelOnlyBase=True))
ax.set_ylabel(ax.get_ylabel() + ' (exponent)')
ax.margins(x=0)
plt.show()
I have an existing plot that was created with pandas like this:
df['myvar'].plot(kind='bar')
The y axis is format as float and I want to change the y axis to percentages. All of the solutions I found use ax.xyz syntax and I can only place code below the line above that creates the plot (I cannot add ax=ax to the line above.)
How can I format the y axis as percentages without changing the line above?
Here is the solution I found but requires that I redefine the plot:
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.ticker as mtick
data = [8,12,15,17,18,18.5]
perc = np.linspace(0,100,len(data))
fig = plt.figure(1, (7,4))
ax = fig.add_subplot(1,1,1)
ax.plot(perc, data)
fmt = '%.0f%%' # Format you want the ticks, e.g. '40%'
xticks = mtick.FormatStrFormatter(fmt)
ax.xaxis.set_major_formatter(xticks)
plt.show()
Link to the above solution: Pyplot: using percentage on x axis
This is a few months late, but I have created PR#6251 with matplotlib to add a new PercentFormatter class. With this class you just need one line to reformat your axis (two if you count the import of matplotlib.ticker):
import ...
import matplotlib.ticker as mtick
ax = df['myvar'].plot(kind='bar')
ax.yaxis.set_major_formatter(mtick.PercentFormatter())
PercentFormatter() accepts three arguments, xmax, decimals, symbol. xmax allows you to set the value that corresponds to 100% on the axis. This is nice if you have data from 0.0 to 1.0 and you want to display it from 0% to 100%. Just do PercentFormatter(1.0).
The other two parameters allow you to set the number of digits after the decimal point and the symbol. They default to None and '%', respectively. decimals=None will automatically set the number of decimal points based on how much of the axes you are showing.
Update
PercentFormatter was introduced into Matplotlib proper in version 2.1.0.
pandas dataframe plot will return the ax for you, And then you can start to manipulate the axes whatever you want.
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(100,5))
# you get ax from here
ax = df.plot()
type(ax) # matplotlib.axes._subplots.AxesSubplot
# manipulate
vals = ax.get_yticks()
ax.set_yticklabels(['{:,.2%}'.format(x) for x in vals])
Jianxun's solution did the job for me but broke the y value indicator at the bottom left of the window.
I ended up using FuncFormatterinstead (and also stripped the uneccessary trailing zeroes as suggested here):
import pandas as pd
import numpy as np
from matplotlib.ticker import FuncFormatter
df = pd.DataFrame(np.random.randn(100,5))
ax = df.plot()
ax.yaxis.set_major_formatter(FuncFormatter(lambda y, _: '{:.0%}'.format(y)))
Generally speaking I'd recommend using FuncFormatter for label formatting: it's reliable, and versatile.
For those who are looking for the quick one-liner:
plt.gca().set_yticklabels([f'{x:.0%}' for x in plt.gca().get_yticks()])
this assumes
import: from matplotlib import pyplot as plt
Python >=3.6 for f-String formatting. For older versions, replace f'{x:.0%}' with '{:.0%}'.format(x)
I'm late to the game but I just realize this: ax can be replaced with plt.gca() for those who are not using axes and just subplots.
Echoing #Mad Physicist answer, using the package PercentFormatter it would be:
import matplotlib.ticker as mtick
plt.gca().yaxis.set_major_formatter(mtick.PercentFormatter(1))
#if you already have ticks in the 0 to 1 range. Otherwise see their answer
I propose an alternative method using seaborn
Working code:
import pandas as pd
import seaborn as sns
data=np.random.rand(10,2)*100
df = pd.DataFrame(data, columns=['A', 'B'])
ax= sns.lineplot(data=df, markers= True)
ax.set(xlabel='xlabel', ylabel='ylabel', title='title')
#changing ylables ticks
y_value=['{:,.2f}'.format(x) + '%' for x in ax.get_yticks()]
ax.set_yticklabels(y_value)
You can do this in one line without importing anything:
plt.gca().yaxis.set_major_formatter(plt.FuncFormatter('{}%'.format))
If you want integer percentages, you can do:
plt.gca().yaxis.set_major_formatter(plt.FuncFormatter('{:.0f}%'.format))
You can use either ax.yaxis or plt.gca().yaxis. FuncFormatter is still part of matplotlib.ticker, but you can also do plt.FuncFormatter as a shortcut.
Based on the answer of #erwanp, you can use the formatted string literals of Python 3,
x = '2'
percentage = f'{x}%' # 2%
inside the FuncFormatter() and combined with a lambda expression.
All wrapped:
ax.yaxis.set_major_formatter(FuncFormatter(lambda y, _: f'{y}%'))
Another one line solution if the yticks are between 0 and 1:
plt.yticks(plt.yticks()[0], ['{:,.0%}'.format(x) for x in plt.yticks()[0]])
add a line of code
ax.yaxis.set_major_formatter(ticker.PercentFormatter())
I seem to have got stuck at a relatively simple problem but couldn't fix it after searching for last hour and after lot of experimenting.
I have two numpy arrays x and y and I am using seaborn's jointplot to plot them:
sns.jointplot(x, y)
Now I want to label the xaxis and yaxis as "X-axis label" and "Y-axis label" respectively. If I use plt.xlabel, the labels goes to the marginal distribution. How can I make them appear on the joint axes?
sns.jointplot returns a JointGrid object, which gives you access to the matplotlib axes and you can then manipulate from there.
import seaborn as sns
import numpy as np
# example data
X = np.random.randn(1000,)
Y = 0.2 * np.random.randn(1000) + 0.5
h = sns.jointplot(X, Y)
# JointGrid has a convenience function
h.set_axis_labels('x', 'y', fontsize=16)
# or set labels via the axes objects
h.ax_joint.set_xlabel('new x label', fontweight='bold')
# also possible to manipulate the histogram plots this way, e.g.
h.ax_marg_y.grid('on') # with ugly consequences...
# labels appear outside of plot area, so auto-adjust
h.figure.tight_layout()
(The problem with your attempt is that functions such as plt.xlabel("text") operate on the current axis, which is not the central one in sns.jointplot; but the object-oriented interface is more specific as to what it will operate on).
Note that the last command uses the figure attribute of the JointGrid. The initial version of this answer used the simpler - but not object-oriented - approach via the matplotlib.pyplot interface.
To use the pyplot interface:
import matplotlib.pyplot as plt
plt.tight_layout()
Alternatively, you can specify the axes labels in a pandas DataFrame in the call to jointplot.
import pandas as pd
import seaborn as sns
x = ...
y = ...
data = pd.DataFrame({
'X-axis label': x,
'Y-axis label': y,
})
sns.jointplot(x='X-axis label', y='Y-axis label', data=data)
I have following simple plot, and I would like to display the origin axis (x, y). I already have grid, but I need the x, y axis to be emphasized.
this is my code:
x = linspace(0.2,10,100)
plot(x, 1/x)
plot(x, log(x))
axis('equal')
grid()
I have seen this question. The accepted answer suggests to use "Axis spine" and just links to some example. The example is however too complicated, using subplots. I am unable to figure out, how to use "Axis spine" in my simple example.
Using subplots is not too complicated, the spines might be.
Dumb, simple way:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
And I get:
(you can't see the vertical axis since the lower x-limit is zero.)
Alternative using simple spines
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')
# set the x-spine (see below for more info on `set_position`)
ax.spines['left'].set_position('zero')
# turn off the right spine/ticks
ax.spines['right'].set_color('none')
ax.yaxis.tick_left()
# set the y-spine
ax.spines['bottom'].set_position('zero')
# turn off the top spine/ticks
ax.spines['top'].set_color('none')
ax.xaxis.tick_bottom()
Alternative using seaborn (my favorite)
import numpy as np
import matplotlib.pyplot as plt
import seaborn
seaborn.set(style='ticks')
x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')
seaborn.despine(ax=ax, offset=0) # the important part here
Using the set_position method of a spine
Here are the docs for a the set_position method of spines:
Spine position is specified by a 2 tuple of (position type, amount).
The position types are:
'outward' : place the spine out from the data area by the specified number of points. (Negative values specify placing the
spine inward.)
'axes' : place the spine at the specified Axes coordinate (from
0.0-1.0).
'data' : place the spine at the specified data coordinate.
Additionally, shorthand notations define a special positions:
'center' -> ('axes',0.5)
'zero' -> ('data', 0.0)
So you can place, say the left spine anywhere with:
ax.spines['left'].set_position((system, poisition))
where system is 'outward', 'axes', or 'data' and position in the place in that coordinate system.
Some time has passed since this question was asked. With Matplotlib 3.6.2 it looks like this works:
plt.axhline(0, color='black', linewidth=.5)
plt.axvline(0, color='black', linewidth=.5)
and there are other options.
Let me answer to this (rather old) question for those who will search for it as I just did. Although it suggested working solutions, I consider the (only) provided answer as way too complex, when it comes to such a simple situation like that described in the question (note: this method requires you to specify all axes endpoints).
I found a simple working solution in one of the first tutorials on matplotlib's pyplot. It is sufficient to add the following line after the creation of the plot
plt.axis([xmin, xmax, ymin, ymax])
as in the following example:
from matplotlib import pyplot as plt
xs = [1,2,3,4,5]
ys = [3,5,1,2,4]
plt.scatter(xs, ys)
plt.axis([0,6,0,6]) #this line does the job
plt.show()
which produces the following result:
Is there a way to automatically not display tick mark labels if they would protrude past the axis itself? For example, consider the following code
#!/usr/bin/python
import pylab as P, numpy as N, math as M
xvals=N.arange(-10,10,0.1)
yvals=[ M.sin(x) for x in xvals ]
P.plot( xvals, yvals )
P.show()
See how the -10 and 10 labels on the x-axis poke out to the left and right of the plot? And similar for the -1.0 and 1.0 labels on the y-axis. Can I automatically suppress plotting these but retain the ones that do not go outside the plot limits?
I think you could just format the axis ticks yourself and then prune the ones
that are hanging over. The recommended way to deal with setting up the axis is
to use the ticker API. So for example
from matplotlib.ticker import MaxNLocator
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111)
xvals=np.arange(-10,10,0.1)
yvals=[ np.sin(x) for x in xvals ]
ax.plot( xvals, yvals )
ax.xaxis.set_major_locator(MaxNLocator(prune='both'))
plt.show()
Here we are creating a figure and axes, plotting the data, and then setting the xaxis
major ticks. The formatter MaxNLocator is given the
argument prune='both' which is described in the docs here.
This is not exactly what you were asking for, but maybe it will solve your problem.