使用matplotlib / python的平方根刻度
我想使用Python绘制平方根比例的图:
但是,我不知道该怎么做。Matplotlib允许进行对数刻度,但是在这种情况下,我需要像幂函数刻度之类的东西。
回答:
您可以创建自己的ScaleBase
课程来做。我已根据您的目的从此处修改了示例(该示例制作了正方形比例,而不是平方根比例)。另外,请参阅此处的文档。
请注意,要正确执行此操作,您可能还应该创建自己的自定义刻度定位器。我在这里还没有做到;我只是使用手动设置主要和次要刻度线ax.set_yticks()
。
import matplotlib.scale as mscaleimport matplotlib.pyplot as plt
import matplotlib.transforms as mtransforms
import matplotlib.ticker as ticker
import numpy as np
class SquareRootScale(mscale.ScaleBase):
"""
ScaleBase class for generating square root scale.
"""
name = 'squareroot'
def __init__(self, axis, **kwargs):
# note in older versions of matplotlib (<3.1), this worked fine.
# mscale.ScaleBase.__init__(self)
# In newer versions (>=3.1), you also need to pass in `axis` as an arg
mscale.ScaleBase.__init__(self, axis)
def set_default_locators_and_formatters(self, axis):
axis.set_major_locator(ticker.AutoLocator())
axis.set_major_formatter(ticker.ScalarFormatter())
axis.set_minor_locator(ticker.NullLocator())
axis.set_minor_formatter(ticker.NullFormatter())
def limit_range_for_scale(self, vmin, vmax, minpos):
return max(0., vmin), vmax
class SquareRootTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = True
def transform_non_affine(self, a):
return np.array(a)**0.5
def inverted(self):
return SquareRootScale.InvertedSquareRootTransform()
class InvertedSquareRootTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = True
def transform(self, a):
return np.array(a)**2
def inverted(self):
return SquareRootScale.SquareRootTransform()
def get_transform(self):
return self.SquareRootTransform()
mscale.register_scale(SquareRootScale)
fig, ax = plt.subplots(1)
ax.plot(np.arange(0, 9)**2, label='$y=x^2$')
ax.legend()
ax.set_yscale('squareroot')
ax.set_yticks(np.arange(0,9,2)**2)
ax.set_yticks(np.arange(0,8.5,0.5)**2, minor=True)
plt.show()
以上是 使用matplotlib / python的平方根刻度 的全部内容, 来源链接: utcz.com/qa/405305.html