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Changing the "tick frequency" on x or y axis in matplotlib

I am trying to fix how python plots my data. Say:

x = [0,5,9,10,15]
y = [0,1,2,3,4]

matplotlib.pyplot.plot(x,y)
matplotlib.pyplot.show()

The x axis' ticks are plotted in intervals of 5. Is there a way to make it show intervals of 1?

Though ticks is the appropriate word here, change ticks to step size will definitely guide more newbies to this question.
Closely related question: stackoverflow.com/questions/6682784/… and a great solution: pyplot.locator_params(nbins=4)
I've rolled back the edit because there's no need to drastically change a canonical question. The question had already been found 1.3 million times in its previous form.
@tdy thanks, I almost did the same thing. (the last thing we need is assertions that examples aren't necessary)

u
unutbu

You could explicitly set where you want to tick marks with plt.xticks:

plt.xticks(np.arange(min(x), max(x)+1, 1.0))

For example,

import numpy as np
import matplotlib.pyplot as plt

x = [0,5,9,10,15]
y = [0,1,2,3,4]
plt.plot(x,y)
plt.xticks(np.arange(min(x), max(x)+1, 1.0))
plt.show()

(np.arange was used rather than Python's range function just in case min(x) and max(x) are floats instead of ints.)

The plt.plot (or ax.plot) function will automatically set default x and y limits. If you wish to keep those limits, and just change the stepsize of the tick marks, then you could use ax.get_xlim() to discover what limits Matplotlib has already set.

start, end = ax.get_xlim()
ax.xaxis.set_ticks(np.arange(start, end, stepsize))

The default tick formatter should do a decent job rounding the tick values to a sensible number of significant digits. However, if you wish to have more control over the format, you can define your own formatter. For example,

ax.xaxis.set_major_formatter(ticker.FormatStrFormatter('%0.1f'))

Here's a runnable example:

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker

x = [0,5,9,10,15]
y = [0,1,2,3,4]
fig, ax = plt.subplots()
ax.plot(x,y)
start, end = ax.get_xlim()
ax.xaxis.set_ticks(np.arange(start, end, 0.712123))
ax.xaxis.set_major_formatter(ticker.FormatStrFormatter('%0.1f'))
plt.show()

Is there no way to get it to still decide it's own limits, but just change the step size? This method is not very good if the min is something like 3523.232512!
@Corone, It has been a while since you asked, but I have posted an answer below that allows for easy control of step size while still using automatic bounds determination.
Note that the +1 in plt.xticks(np.arange(min(x), max(x)+1, 1.0)) is required to show the last tick mark.
Yes, np.arange(start, stop) generates values in the half-open interval [start, stop), including start but excluding stop. So I used max(x)+1 to ensure that max(x) is included.
is there an equivalent for datetime e.g. plt.xticks(np.arange(min(dates), max(dates)+0.1,0.1) ? it seems to only plots the year
r
robochat

Another approach is to set the axis locator:

import matplotlib.ticker as plticker

loc = plticker.MultipleLocator(base=1.0) # this locator puts ticks at regular intervals
ax.xaxis.set_major_locator(loc)

There are several different types of locator depending upon your needs.

Here is a full example:

import matplotlib.pyplot as plt
import matplotlib.ticker as plticker

x = [0,5,9,10,15]
y = [0,1,2,3,4]
fig, ax = plt.subplots()
ax.plot(x,y)
loc = plticker.MultipleLocator(base=1.0) # this locator puts ticks at regular intervals
ax.xaxis.set_major_locator(loc)
plt.show()

This does not work as expected. Specifically, when using dates, it does not use the appropriate dates.
When using dates, you should use the methods in the matplotlib.dates module. For example matplotlib.dates.AutoDateLocator()
It worked as expected for me, with dates. This solution is much easier than the accepted one.
What does base=1.0 actually mean/do ?
base=1.0 means that there will a locator for every whole number. The documentation says that MultipleLocator "Set[s] a tick on each integer multiple of a base within the view interval.". So if base=2 then there'll be a tick for even numbers and I think that you could event put base=2.5.
j
jthomas

I like this solution (from the Matplotlib Plotting Cookbook):

import matplotlib.pyplot as plt
import matplotlib.ticker as ticker

x = [0,5,9,10,15]
y = [0,1,2,3,4]

tick_spacing = 1

fig, ax = plt.subplots(1,1)
ax.plot(x,y)
ax.xaxis.set_major_locator(ticker.MultipleLocator(tick_spacing))
plt.show()

This solution give you explicit control of the tick spacing via the number given to ticker.MultipleLocater(), allows automatic limit determination, and is easy to read later.


A way to do this without calculating the ticks explicitly!
This is the same answer as this one. It does not make sense to add an identical answer two years later.
Good catch. I did not recognize them as the same when I posted the answer. Still, I think this presentation is a little easier to understand.
The book reference in this answer also provide a helpful source for more information.
This is the same answer as that of robochat, which came three years earlier.
g
glopes

In case anyone is interested in a general one-liner, simply get the current ticks and use it to set the new ticks by sampling every other tick.

ax.set_xticks(ax.get_xticks()[::2])

This is the only generalisable answer for different tick types (str, float, datetime)
Remove non-integer ticks: ax.set_xticks([tick for tick in ax.get_xticks() if tick % 1 == 0])
Lots of detailed solutions above but I agree this is the most concise. You could even extract the length of ax.get_xticks() and set the slicing frequency by this length divided by the number of required ticks.
I think this is the best answer. Most other answers are too complicated and hard to apply/generalize. Thank you!
It can only reduce number of sticks, whereas in the question (and mine goal how I found it) was to increase it.
C
Community

This is a bit hacky, but by far the cleanest/easiest to understand example that I've found to do this. It's from an answer on SO here:

Cleanest way to hide every nth tick label in matplotlib colorbar?

for label in ax.get_xticklabels()[::2]:
    label.set_visible(False)

Then you can loop over the labels setting them to visible or not depending on the density you want.

edit: note that sometimes matplotlib sets labels == '', so it might look like a label is not present, when in fact it is and just isn't displaying anything. To make sure you're looping through actual visible labels, you could try:

visible_labels = [lab for lab in ax.get_xticklabels() if lab.get_visible() is True and lab.get_text() != '']
plt.setp(visible_labels[::2], visible=False)

This is the most simple and generic solution. A tiny adjustment: usually ax.get_xticklabels()[1::2] are the labels to be hidden.
This doesn't work with matplotlib.finance.candlestick2
@BCR it could be that some of the xticklabels are just set to '' so that when you loop through them, you're making xticklabels that are empty invisible (which would have no effect on the visualization, but might mean that you aren't pulling the correct labels). You could try: vis_labels = [label for label in ax.get_xticklabels() if label.get_visible() is True]; plt.setp(vis_labels[::2], visible==False)
G
Gary Steele

if you just want to set the spacing a simple one liner with minimal boilerplate:

plt.gca().xaxis.set_major_locator(plt.MultipleLocator(1))

also works easily for minor ticks:

plt.gca().xaxis.set_minor_locator(plt.MultipleLocator(1))

a bit of a mouthfull, but pretty compact


Worth mentioning the argument plt.MultipleLocator(arg) is the tick interval. So, if you want your ticks to be 5 units away from each other, just use plt.MultipleLocator(5). Otherwise like this solution the most. Thank you!
T
Tompa

This is an old topic, but I stumble over this every now and then and made this function. It's very convenient:

import matplotlib.pyplot as pp
import numpy as np

def resadjust(ax, xres=None, yres=None):
    """
    Send in an axis and I fix the resolution as desired.
    """

    if xres:
        start, stop = ax.get_xlim()
        ticks = np.arange(start, stop + xres, xres)
        ax.set_xticks(ticks)
    if yres:
        start, stop = ax.get_ylim()
        ticks = np.arange(start, stop + yres, yres)
        ax.set_yticks(ticks)

One caveat of controlling the ticks like this is that one does no longer enjoy the interactive automagic updating of max scale after an added line. Then do

gca().set_ylim(top=new_top) # for example

and run the resadjust function again.


D
Deninhos

I developed an inelegant solution. Consider that we have the X axis and also a list of labels for each point in X.

import matplotlib.pyplot as plt

x = [0,1,2,3,4,5]
y = [10,20,15,18,7,19]
xlabels = ['jan','feb','mar','apr','may','jun']
xlabelsnew = []
for i in xlabels:
    if i not in ['feb','jun']:
        i = ' '
        xlabelsnew.append(i)
    else:
        xlabelsnew.append(i)
plt.plot(x,y)
plt.xticks(range(0,len(x)),xlabels,rotation=45)
plt.show()
plt.plot(x,y)
plt.xticks(range(0,len(x)),xlabelsnew,rotation=45)
plt.show()

G
Greenstick

Pure Python Implementation

Below's a pure python implementation of the desired functionality that handles any numeric series (int or float) with positive, negative, or mixed values and allows for the user to specify the desired step size:

import math

def computeTicks (x, step = 5):
    """
    Computes domain with given step encompassing series x
    @ params
    x    - Required - A list-like object of integers or floats
    step - Optional - Tick frequency
    """
    xMax, xMin = math.ceil(max(x)), math.floor(min(x))
    dMax, dMin = xMax + abs((xMax % step) - step) + (step if (xMax % step != 0) else 0), xMin - abs((xMin % step))
    return range(dMin, dMax, step)

Sample Output

# Negative to Positive
series = [-2, 18, 24, 29, 43]
print(list(computeTicks(series)))

[-5, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45]

# Negative to 0
series = [-30, -14, -10, -9, -3, 0]
print(list(computeTicks(series)))

[-30, -25, -20, -15, -10, -5, 0]

# 0 to Positive
series = [19, 23, 24, 27]
print(list(computeTicks(series)))

[15, 20, 25, 30]

# Floats
series = [1.8, 12.0, 21.2]
print(list(computeTicks(series)))

[0, 5, 10, 15, 20, 25]

# Step – 100
series = [118.3, 293.2, 768.1]
print(list(computeTicks(series, step = 100)))

[100, 200, 300, 400, 500, 600, 700, 800]

Sample Usage

import matplotlib.pyplot as plt

x = [0,5,9,10,15]
y = [0,1,2,3,4]
plt.plot(x,y)
plt.xticks(computeTicks(x))
plt.show()

https://i.stack.imgur.com/imxIG.png

Notice the x-axis has integer values all evenly spaced by 5, whereas the y-axis has a different interval (the matplotlib default behavior, because the ticks weren't specified).


J
Johnny V

Generalisable one liner, with only Numpy imported:

ax.set_xticks(np.arange(min(x),max(x),1))

Set in the context of the question:

import numpy as np
import matplotlib.pyplot as plt 
fig, ax = plt.subplots()
x = [0,5,9,10,15]
y = [0,1,2,3,4]
ax.plot(x,y)
ax.set_xticks(np.arange(min(x),max(x),1))
plt.show()

How it works:

fig, ax = plt.subplots() gives the ax object which contains the axes. np.arange(min(x),max(x),1) gives an array of interval 1 from the min of x to the max of x. This is the new x ticks that we want. ax.set_xticks() changes the ticks on the ax object.


so, can you explain the code?
I've rewritten it to make it clearer. Hope that helps
B
BartoszKP
xmarks=[i for i in range(1,length+1,1)]

plt.xticks(xmarks)

This worked for me

if you want ticks between [1,5] (1 and 5 inclusive) then replace

length = 5

fyi, you could simply write xmarks = range(1, length+1, 1). pretty sure the list comprehension is redundant.
k
kmario23

Since None of the above solutions worked for my usecase, here I provide a solution using None (pun!) which can be adapted to a wide variety of scenarios.

Here is a sample piece of code that produces cluttered ticks on both X and Y axes.

# Note the super cluttered ticks on both X and Y axis.

# inputs
x = np.arange(1, 101)
y = x * np.log(x) 

fig = plt.figure()     # create figure
ax = fig.add_subplot(111)
ax.plot(x, y)
ax.set_xticks(x)        # set xtick values
ax.set_yticks(y)        # set ytick values

plt.show()

Now, we clean up the clutter with a new plot that shows only a sparse set of values on both x and y axes as ticks.

# inputs
x = np.arange(1, 101)
y = x * np.log(x)

fig = plt.figure()       # create figure
ax = fig.add_subplot(111)
ax.plot(x, y)

ax.set_xticks(x)
ax.set_yticks(y)

# which values need to be shown?
# here, we show every third value from `x` and `y`
show_every = 3

sparse_xticks = [None] * x.shape[0]
sparse_xticks[::show_every] = x[::show_every]

sparse_yticks = [None] * y.shape[0]
sparse_yticks[::show_every] = y[::show_every]

ax.set_xticklabels(sparse_xticks, fontsize=6)   # set sparse xtick values
ax.set_yticklabels(sparse_yticks, fontsize=6)   # set sparse ytick values

plt.show()

Depending on the usecase, one can adapt the above code simply by changing show_every and using that for sampling tick values for X or Y or both the axes.

If this stepsize based solution doesn't fit, then one can also populate the values of sparse_xticks or sparse_yticks at irregular intervals, if that is what is desired.


A
Ahmad

You can loop through labels and show or hide those you want:

   for i, label in enumerate(ax.get_xticklabels()):
        if i % interval != 0:
            label.set_visible(False)