matplotlib绘制了太多的滴答声 [英] matplotlib plots too many ticks

问题描述

我有一个大代码，可以输出多个图像.有时，情节看起来像这样(请参见下面的MWE):

I have a large code which outputs several images. At times, a plot will look like this (see MWE below):

请注意，x轴上的刻度线几乎彼此重叠，这看起来很糟糕.

Notice how the ticks in the x axis are almost overlapping each other, which looks pretty bad.

我的问题是:为什么matplotlib无法实现这一点，而只是绘制较少的滴答声(通常不会出现问题)?我该如何强制matplotlib以 general 的方式绘制较少的刻度线(即:我不是只针对该特定情节应用 的解决方案，这只是一个示例)?

My question is: why doesn't matplotlib realize this and simply plots fewer ticks (as it usually does without issues)? How can I force matplotlib to draw fewer ticks in a general way (ie: I'm not after a solution that applies only for this particular plot, this is just an example)?

MWE(如果代码看起来不必要地复杂，那是因为它只是更大代码的一小部分)

MWE (if the code looks unnecessarily complicated, it's because it's a small portion of a much larger code)

import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
import random


def scatter_plot(x, y):

    ax = plt.subplot(gs[0:2, 0:2])
    plt.xlim(min(x), 0.00158)
    plt.xlabel('$x$', fontsize=16)
    plt.ylabel('$y$', fontsize=16)
    ax.minorticks_on()

    plt.scatter(x, y)


# Generate random data.
x = [random.random() / 100. for i in xrange(10000)]
y = [random.random() for i in xrange(10000)]
# Define size of output figure.
fig = plt.figure(figsize=(30, 25))  # create the top-level container
gs = gridspec.GridSpec(10, 12)      # create a GridSpec object
# Create plot.
scatter_plot(x, y)
# Save plot to file.
fig.tight_layout()
plt.savefig('out.png', dpi=150)

推荐答案

首先 import numpy as np 然后调用 ax.xaxis.set_ticks(np.arange(xmin, xmax, stepsize)))，其中 xmin 是最小刻度值， xmax 是最大刻度值，而 stepsize 是连续刻度之间的差滴答声.使用 stepsize =(xmax-xmin)/tick_count 可能是有意义的，其中 tick_count 是您想要的刻度数.

First import numpy as np and then call ax.xaxis.set_ticks(np.arange(xmin, xmax, stepsize)), where xmin is the minimum tick value, xmax is the maximum tick value, and stepsize is the difference between consecutive ticks. It probably makes sense to use stepsize = (xmax - xmin) / tick_count, where tick_count is the number of ticks you want.

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