Is there a parameter to force horizontal labels in an mplstyle file? and/or using rcParams?
I'm currently using ax.xaxis.set_tick_params(rotation=0) at plot construction. I'd like a permanent style or setting. Thanks!
Default look (with x_compat=True in a pandas dataframes):
Desired look:
import pandas as pd
import matplotlib.pyplot as plt
df = pd.DataFrame({'Date': {0: '1950-01-01', 1: '1960-01-02', 2: '1970-01-03', 3: '1980-01-04', 4: '1990-01-05'}, 'Value': {0 : 0, 1: 1, 2: 0, 3: 1, 4: 0}})
df['Date'] = pd.to_datetime(df['Date'], format='%Y-%m-%d')
df = df.set_index('Date', drop=False)
f, ax = plt.subplots()
df.plot(ax=ax, x='Date', x_compat=True)
#ax.xaxis.set_tick_params(rotation=0)
plt.show()
I looked in there, but may have missed it:
customizing-with-matplotlibrc-files
matplotlib_configuration_api.html
Use parameter rot from df.plot
df.plot(ax=ax, x='Date', x_compat=True, rot=0)
I'll answer my own question to put the matter to rest.
No, there isn't.
[as of January 2022] There is no way to control tick label rotation via a style. This is because the pandas plot wrapper resets the rotation parameter. To quote from pandas/doc/source/user_guide/visualization.rst,
pandas includes automatic tick resolution adjustment for regular
frequency time-series data. For limited cases where pandas cannot
infer the frequency information (e.g., in an externally created
twinx), you can choose to suppress this behavior for alignment
purposes.
[...]
Using the x_compat parameter, you can suppress this behavior
Despite the wording here --- namely "alignment purposes" ---, setting x_compat=True does not reset the rotation parameter back to its matplotlib default of 0, as I'd incorrectly expected.
There seem to be mainly two ways around this:
Use matplotlib directly without pandas.
Reset the rotation inside the pandas plot call. This may be done
the pandas way [See Vishnudev's answer] with df.plot(... rot=0...) or the matplotlib way [See my OP] with an axis object
setting ax.xaxis.set_tick_params(rotation=0).
Source and Thanks to: Jody Klymak in comments and Marco Gorelli at Github.
Related
lately people have been telling me that seaborn is the best data visualization package in Python, so I decided to try it out. However, my plots look exactly identical to my matplotlib plots..
I am using PyCharm, Python 3.6.
Here is my super simple code to test the two:
x = [1, 2, 3, 4, 5]
y = [11, 12, 13, 14, 15]
plt.plot(x, y)
plt.show()
sns.lineplot(x, y)
plt.show()
and they both look like:
Normally, seaborn plots should have at leasthave a gridded blue background. Why is mines not working?
Seaborn is an extention to matplotlib, which simplifies certain tasks. I.e. often creating a plot with seaborn takes only 30 to 50% of the number of code lines it would take to plot with matplotlib directly. But every seaborn plot necessarily is a matplotlib plot.
Concerning the style, seaborn has some shortcuts on setting style parameters. Those are explained in detail in the Aestetics tutorial.
In short, you can use
seaborn.set()
to get the "darkgrid" theme; you can use
seaborn.reset_defaults()
to reset the parameters back to the matplotlib defaults.
Essentially[*] the same can be achieved with matplotlib via
plt.style.use("seaborn-darkgrid")
and
plt.style.use("default")
For this read the matplotlib customizing tutorial.
[*] There are small differences, because seaborn.set also sets other parameters like defaults of figure size etc.
maybe I am a little too late for your answer but try using:
seaborn.set()
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 would like to set different levels of transparency (= alpha) for the edge line and fill of a distribution plot that I created in matplotlib/seaborn. For example:
ax1 = sns.distplot(BSRDI_DF, label="BsrDI", bins=newBins, kde=False,
hist_kws={"edgecolor": (1,0,0,1), "color":(1,0,0,0.25)})
The above approach does not work, unfortunately. Does anybody have any idea how I could accomplish this?
The problem seems to be that seaborn sets an alpha parameter for the histogram. While alpha defaults to None for a usual histogram, such that something like
plt.hist(x, lw=3, edgecolor=(1,0,0,0.75), color=(1,0,0,0.25))
works as expected, seaborn sets this alpha to some given value. This overwrites the alpha that is set in the RGBA tuples.
The solution is to set alpha explicitely to None:
ax = sns.distplot(x, kde=False, hist_kws={"lw":3, "edgecolor": (1,0,0,0.75),
"color":(1,0,0,0.25),"alpha":None})
A complete example:
import seaborn as sns
import matplotlib.pyplot as plt
import numpy as np
x = np.random.randn(60)
ax = sns.distplot(x, label="BsrDI", bins=np.linspace(-3,3,10), kde=False,
hist_kws={"lw":3, "edgecolor": (1,0,0,0.75),
"color":(1,0,0,0.25),"alpha":None})
plt.show()
EDIT Nevermind, I thought using color instead of facecolor was causing the problem but it seems the output that I got only looked right because the patches were overlapping, giving seemingly darker edges.
After investigating the issue further, it looks like seaborn is hard-setting the alpha level at 0.4, which supersedes the arguments passed to hist_kws=
sns.distplot(x, kde=False, hist_kws={"edgecolor": (1,0,0,1), "lw":5, "facecolor":(0,1,0,0.1), "rwidth":0.8})
While using the same parameters to plt.hist() gives:
plt.hist(x, edgecolor=(1,0,0,1), lw=5, facecolor=(0,1,0,0.1), rwidth=0.8)
Conclusion: if you want different alpha levels for edges and face colors, you'll have to use matplotlib directly, and not seaborn.
I have been trying to plot some time series graphs using the pandas dataframe plot function. I was trying to add markers at some arbitrary points on the plot to show anomalous points. The code I used :
df1 = pd.DataFrame({'Entropy Values' : MeanValues}, index=DateRange)
df1.plot(linestyle = '-')
I have a list of Dates on which I need to add markers.Such as:
Dates = ['15:45:00', '15:50:00', '15:55:00', '16:00:00']
I had a look at this link matplotlib: Set markers for individual points on a line. Does DF.plot have a similar functionality?
I really appreciate the help. Thanks!
DataFrame.plot passes all keyword arguments it does not recognize to the matplotlib plotting method. To put markers at a few points in the plot you can use the markevery argument. Here is an example:
import pandas as pd
df = pd.DataFrame({'A': range(10), 'B': range(10)}).set_index('A')
df.plot(linestyle='-', markevery=[1, 5, 7, 8], marker='o', markerfacecolor='r')
In your case, you would have to do something like
df1.plot(linestyle='-', markevery=Dates, marker='o', markerfacecolor='r')
I am using pandas for graphing data for a cluster of nodes. I find that pandas is repeating color values for the different series, which makes them indistinguishable.
I tried giving custom color values like this and passed the my_colors to the colors field in plot:
my_colors = []
for node in nodes_list:
my_colors.append(rand_color())
rand_color() is defined as follows:
def rand_color():
from random import randrange
return "#%s" % "".join([hex(randrange(16, 255))[2:] for i in range(3)])
But here also I need to avoid color values that are too close to distinguish. I sometimes have as many as 60 nodes (series). Most probably a hard-coded list of color values would be best option?
You can get a list of colors from any colormap defined in Matplotlib, and even custom colormaps, by:
>>> import matplotlib.pyplot as plt
>>> colors = plt.cm.Paired(np.linspace(0,1,60))
Plotting an example with these colors:
>>> plt.scatter( range(60), [0]*60, color=colors )
<matplotlib.collections.PathCollection object at 0x04ED2830>
>>> plt.axis("off")
(-10.0, 70.0, -0.0015, 0.0015)
>>> plt.show()
I found the "Paired" colormap to be especially useful for this kind of things, but you can use any other available or custom colormap.