reducing number of plot ticks

Question

I have too many ticks on my graph and they are running into each other.

How can I reduce the number of ticks?

For example, I have ticks:

1E-6, 1E-5, 1E-4, ... 1E6, 1E7

And I only want:

1E-5, 1E-3, ... 1E5, 1E7

I've tried playing with the LogLocator, but I haven't been able to figure this out.

Answer 1

Alternatively, if you want to simply set the number of ticks while allowing matplotlib to position them (currently only with MaxNLocator), there is pyplot.locator_params,

pyplot.locator_params(nbins=4)

You can specify specific axis in this method as mentioned below, default is both:

# To specify the number of ticks on both or any single axes
pyplot.locator_params(axis='y', nbins=6)
pyplot.locator_params(axis='x', nbins=10)

• benjaminmgross: This was a great suggestion, also being able to specify pyplot.locator_params(axis = 'x', nbins = 4) (or axis = 'y') made the process really straightforward. Thanks @bgamari!
• meduz: With log scale, this worked with numticks instead of nbins
• Vim: This doesn't seem to place the labels where they should be. For example, if the original tick labels are [0, 1, ..., 99] and now one sets nticks=10, then the new sparse labels will be placed ten times as long apart along the axis, i.e. now 1 will sit where 9 was, 2 where 19 was... and 9 where 99 was.
• David J.: Check your results before trusting this method. @Vim is correct. The tick values will be placed incorrectly.
• ldmtwo: I got incorrect labeling with this method, but @Katerina's worked as expected.

Answer 2

To solve the issue of customisation and appearance of the ticks, see the Tick Locators guide on the matplotlib website

ax.xaxis.set_major_locator(plt.MaxNLocator(3))

would set the total number of ticks in the x-axis to 3, and evenly distribute them across the axis.

There is also a nice tutorial about this

• Mithril: Only select first 3 datetime index. when get ax from pandas.DataFrame.plot ax = df.plot()
• Prageeth Jayathissa: @Mithril sorry I don't quite understand your comment. Could you please elaborate?
• Mithril: If I have a df ( pandas.DataFrame ) with datetime index [2019年01月01日, ...2019年11月01日], call ax = df.plot() , return a figure object . call ax.xaxis.set_major_locator(plt.MaxNLocator(3)) only show first 3 index [2019年01月01日, 2019年01月02日, 2019年01月03日] .
• Prageeth Jayathissa: @Mithril, df.plot() often displays the minor_locator, so you might want to try ax1.xaxis.set_minor_locator(plt.MaxNLocator(3)). Also remember to substitute the 3 for the number of ticks that you want to display. For pandas timeseries I recommend import matplotlib.dates as mdates and run ax.xaxis.set_minor_locator(mdates.MonthLocator(interval = 1)) with ax.xaxis.set_minor_formatter(mdates.DateFormatter('%m-%Y'))
• Sparkofska: As I had a lot of datapoints to plot, pyplot was printing a lot (>500) ticks per axis. This heavily slowed down the plotting process. By reducing the number of thicks with set_major_locator(5) the plotting process is much faster.

Answer 3

If somebody still gets this page in search results:

fig, ax = plt.subplots()
plt.plot(...)
every_nth = 4
for n, label in enumerate(ax.xaxis.get_ticklabels()):
 if n % every_nth != 0:
 label.set_visible(False)

• Joonho Park: In my case with pandas and numerically spaced, this is working as is expected while locator_params(nbins) and plt.MaxNLocator give a strange behavior.
• jgholder: This works for non-numerical x axis labels, which is helpful

Answer 4

There's a set_ticks() function for axis objects.

• jlconlin: This would work if I knew beforehand what ticks I wanted. The example I gave above was only an example. I don't know what the ticks are, I just know I want fewer of them, i.e., every other one.
• user812786: You could call get_xticks() or get_yticks() first for the axes object, edit as needed, and then pass the list back to set_ticks().
• Gauthier: I don't have set_ticks(), but I do have set_xticks() and set_yticks(). These are attributes of axes objects, not axis. Maybe this has changed during the last couple of years.
• Gauthier: I am not sure if I should, some people have found your answer useful as is, and just because it is different for me does not mean it is for everybody.
• Richard: An example would go a long way towards making this answer useful.

Answer 5

in case somebody still needs it, and since nothing here really worked for me, i came up with a very simple way that keeps the appearance of the generated plot "as is" while fixing the number of ticks to exactly N:

import numpy as np
import matplotlib.pyplot as plt
f, ax = plt.subplots()
ax.plot(range(100))
ymin, ymax = ax.get_ylim()
ax.set_yticks(np.round(np.linspace(ymin, ymax, N), 2))

• Nick Settje: I had to modify the last line slightly in order to get it to return the values as int instead of float: ax.set_yticks(np.linspace(int(ymin), int(ymax), N), 2)
• Mohd: @NickSettje still floats with me!

Answer 6

The solution @raphael gave is straightforward and quite helpful.

Still, the displayed tick labels will not be values sampled from the original distribution but from the indices of the array returned by np.linspace(ymin, ymax, N).

To display N values evenly spaced from your original tick labels, use the set_yticklabels() method. Here is a snippet for the y axis, with integer labels:

import numpy as np
import matplotlib.pyplot as plt
ax = plt.gca()
ymin, ymax = ax.get_ylim()
custom_ticks = np.linspace(ymin, ymax, N, dtype=int)
ax.set_yticks(custom_ticks)
ax.set_yticklabels(custom_ticks)

Answer 7

If you need one tick every N=3 ticks :

N = 3 # 1 tick every 3
xticks_pos, xticks_labels = plt.xticks() # get all axis ticks
myticks = [j for i,j in enumerate(xticks_pos) if not i%N] # index of selected ticks
newlabels = [label for i,label in enumerate(xticks_labels) if not i%N]

or with fig,ax = plt.subplots() :

N = 3 # 1 tick every 3
xticks_pos = ax.get_xticks()
xticks_labels = ax.get_xticklabels()
myticks = [j for i,j in enumerate(xticks_pos) if not i%N] # index of selected ticks
newlabels = [label for i,label in enumerate(xticks_labels) if not i%N]

(obviously you can adjust the offset with (i+offset)%N).

Note that you can get uneven ticks if you wish, e.g. myticks = [1, 3, 8].

Then you can use

plt.gca().set_xticks(myticks) # set new X axis ticks

or if you want to replace labels as well

plt.xticks(myticks, newlabels) # set new X axis ticks and labels

Beware that axis limits must be set after the axis ticks.

Finally, you may wish to draw only an arbitrary set of ticks :

mylabels = ['03/2018', '09/2019', '10/2020']
plt.draw() # needed to populate xticks with actual labels
xticks_pos, xticks_labels = plt.xticks() # get all axis ticks
myticks = [i for i,j in enumerate(xticks_labels) if j.get_text() in mylabels]
plt.xticks(myticks, mylabels)

(assuming mylabels is ordered ; if it is not, then sort myticks and reorder it).

• gtristan: there is a little error in the last code block. One of the lines should be written as: myticks = [i for i,j in enumerate(xticks_labels) if j.get_text() in mylabels]
• Skippy le Grand Gourou: @gtristan Right, fixed, thanks !

Answer 8

xticks function auto iterates with range function

start_number = 0

end_number = len(data you have)

step_number = how many skips to make from strat to end

rotation = 90 degrees tilt will help with long ticks

plt.xticks(range(start_number,end_number,step_number),rotation=90)

• Joe Still: I came to contribute a range based solution. Your example is much more concise. But to add to it, you can use the shape to get what you want eg plt.xticks(range(0, df.shape[0], 12),rotation=90)

Answer 9

if you want 10 ticks:

for y axis: ax.set_yticks(ax.get_yticks()[::len(ax.get_yticks())//10])

for x axis: ax.set_xticks(ax.get_xticks()[::len(ax.get_xticks())//10])

this simply gets your ticks and chooses every 10th of the list and sets it back to your ticks. you can change the number of ticks as you wish.

Answer 10

When a log scale is used the number of major ticks can be fixed with the following command

import matplotlib.pyplot as plt
....
plt.locator_params(numticks=12)
plt.show()

The value set to numticks determines the number of axis ticks to be displayed.

Credits to @bgamari's post for introducing the locator_params() function, but the nticks parameter throws an error when a log scale is used.

• Dawid Laszuk: The question and answer are for previous matplotlib, i.e. 1, and you're referring to 2.
• troymyname00: I do not think an argument called 'numticks' exists.
• Ébe Isaac: Kindly refer the matplotlib doc for numticks @troymyname00