Question

我想用Python创建一个平方根比例图：

但是，我不知道如何制作它。 Matplotlib允许制作对数刻度，但在这种情况下我需要像功率函数量表这样的东西。

Answer 1

您可以创建自己的ScaleBase课程来执行此操作。我已经修改了here（为了您的目的制作了一个方形比例，而不是平方根比例）的示例。另请参阅文档here。

请注意，要正确执行此操作，您可能还应创建自己的自定义刻度定位器;我虽然没有这样做过;我只需使用ax.set_yticks()手动设置主要和次要刻度。

import matplotlib.scale as mscale
import 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):
        mscale.ScaleBase.__init__(self)

    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()

Answer 2

这是绘制 $sqrt(x)$ 的简单方法

import numpy as np
from matplotlib import pyplot as plt

plt.rcParams["figure.dpi"] = 140

fig, ax = plt.subplots()
ax.spines["left"].set_position("zero")
ax.spines["bottom"].set_position("zero")
ax.spines["right"].set_color("none")
ax.spines["top"].set_color("none")
ax.xaxis.set_ticks_position("bottom")
ax.yaxis.set_ticks_position("left")

origin = [0, 0]

# 45
plt.plot(
    np.linspace(0, 1, 1000),
    np.sqrt(np.linspace(0, 1, 1000)),
    color="k",
)

ax.set_aspect("equal")
plt.xlim(-0.25, 1)
plt.ylim(0, 1)
plt.yticks(ticks=np.linspace(0, 1, 6))
plt.show()

Answer 3

我喜欢lolopop的评论和汤姆的答案，更快速和肮脏的解决方案将使用set_yticks and set_yticklabels，如下所示：

x = np.arange(2, 15, 2)
y = x * x

fig = plt.figure()
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)

ax1.plot(x,y)

ax2.plot(x, np.sqrt(y))
ax2.set_yticks([2,4,6,8,10,12,14])
ax2.set_yticklabels(['4','16','36','64','100','144','196'])

Answer 4

Matplotlib现在提供powlaw标准。因此，将功率设置为0.5应该可以解决问题！

C.f。 Matplotlib Powerlaw norm

他们的榜样：

"""
Demonstration of using norm to map colormaps onto data in non-linear ways.
"""

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.colors as colors
from matplotlib.mlab import bivariate_normal

N = 100
X, Y = np.mgrid[-3:3:complex(0, N), -2:2:complex(0, N)]

'''
PowerNorm: Here a power-law trend in X partially obscures a rectified
sine wave in Y. We can remove gamma to 0.5 should do the trick using  PowerNorm.
'''
X, Y = np.mgrid[0:3:complex(0, N), 0:2:complex(0, N)]
Z1 = (1 + np.sin(Y * 10.)) * X**(2.)

fig, ax = plt.subplots(2, 1)

pcm = ax[0].pcolormesh(X, Y, Z1, norm=colors.PowerNorm(gamma=1./2.),
                       cmap='PuBu_r')
fig.colorbar(pcm, ax=ax[0], extend='max')

pcm = ax[1].pcolormesh(X, Y, Z1, cmap='PuBu_r')
fig.colorbar(pcm, ax=ax[1], extend='max')
fig.show()

使用matplotlib / python的平方根比例

4 个答案: