以科学风格设置自定义刻度的精度 [英] Set precision on custom ticks with scientific style

问题描述

我已生成此代码以生成以下图表:

 将matplotlib.pyplot导入为plt将numpy导入为npx = np.linspace(1/10000, 1/2000, 13)y = x ** 2plt.plot(x，y，'ro')plt.ticklabel_format(style='sci', axis='x', scilimits=(0,0), useMathText=True)

我想在数据的位置设置xticks.如果再执行 plt.xticks(x，rotation = 45)，我会在所需的位置获得刻度，但是小数点太多(请参见下图).如何在指定位置获得刻度，但精度可控?

解决方案

为了获得刻度标签的预定义格式同时保持科学乘数，您可以使用 的简化版本我在

I have produced this code to generate the following graph:

import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(1/10000, 1/2000, 13)
y = x**2
plt.plot(x, y, 'ro')
plt.ticklabel_format(style='sci', axis='x', scilimits=(0,0), useMathText=True)

I would like to set the xticks at the position of the data. If I then do plt.xticks(x, rotation=45) I get the ticks at the desired locations but with too many decimal places (see next picture). How do I get the ticks at the specified locations but with a controllable precision?

解决方案

In order to get a predefined format for the ticklabels while maintaining the scientific multiplier you can use a simplified version of the OOMFormatter from my answer here.

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

class FFormatter(matplotlib.ticker.ScalarFormatter):
    def __init__(self, fformat="%1.1f", offset=True, mathText=True):
        self.fformat = fformat
        matplotlib.ticker.ScalarFormatter.__init__(self,useOffset=offset,useMathText=mathText)
    def _set_format(self, vmin, vmax):
        self.format = self.fformat
        if self._useMathText:
            self.format = '$%s$' % matplotlib.ticker._mathdefault(self.format)

x = np.linspace(1/10000, 1/2000, 13)
y = x**2
plt.plot(x, y, 'ro')
plt.xticks(x, rotation=45)

fmt = plt.gca().xaxis.set_major_formatter(FFormatter(fformat="%1.1f"))
plt.ticklabel_format(style='sci', axis='x', scilimits=(0,0), useMathText=True)

plt.show()

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