有没有办法在 Matplotlib 中制作不连续的轴？

问题描述：

我正在尝试使用 pyplot 创建一个具有不连续 x 轴的图。通常绘制的方式是轴将具有如下内容：

（值）----//----（后续值）

其中 // 表示您跳过 (值) 和 (后面的值) 之间的所有内容。

我找不到任何这方面的例子，所以我想知道这是否可行。我知道您可以在不连续性上连接数据，例如财务数据，但我想让轴上的跳跃更加明确。目前我只是使用子图，但我真的很想让所有内容最终都出现在同一张图上。

解决方案 1：

保罗的回答是一种非常好的方法。

但是，如果您不想进行自定义转换，则可以使用两个子图来创建相同的效果。

无需从头开始编写示例，Paul Ivanov 在 matplotlib 示例中编写了一个很好的示例（它仅出现在当前的 git 提示中，因为它是几个月前才提交的。它尚未出现在网页上。）。

这只是对此示例的简单修改，使其具有不连续的 x 轴而不是 y 轴。（这就是我将这篇文章设为 CW 的原因）

基本上，你只需做这样的事情：

import matplotlib.pylab as plt
import numpy as np

# If you're not familiar with np.r_, don't worry too much about this. It's just 
# a series with points from 0 to 1 spaced at 0.1, and 9 to 10 with the same spacing.
x = np.r_[0:1:0.1, 9:10:0.1]
y = np.sin(x)

fig,(ax,ax2) = plt.subplots(1, 2, sharey=True)

# plot the same data on both axes
ax.plot(x, y, 'bo')
ax2.plot(x, y, 'bo')

# zoom-in / limit the view to different portions of the data
ax.set_xlim(0,1) # most of the data
ax2.set_xlim(9,10) # outliers only

# hide the spines between ax and ax2
ax.spines['right'].set_visible(False)
ax2.spines['left'].set_visible(False)
ax.yaxis.tick_left()
ax.tick_params(labeltop='off') # don't put tick labels at the top
ax2.yaxis.tick_right()

# Make the spacing between the two axes a bit smaller
plt.subplots_adjust(wspace=0.15)

plt.show()

要添加断轴线//效果，我们可以这样做（再次根据 Paul Ivanov 的示例进行修改）：

import matplotlib.pylab as plt
import numpy as np

# If you're not familiar with np.r_, don't worry too much about this. It's just 
# a series with points from 0 to 1 spaced at 0.1, and 9 to 10 with the same spacing.
x = np.r_[0:1:0.1, 9:10:0.1]
y = np.sin(x)

fig,(ax,ax2) = plt.subplots(1, 2, sharey=True)

# plot the same data on both axes
ax.plot(x, y, 'bo')
ax2.plot(x, y, 'bo')

# zoom-in / limit the view to different portions of the data
ax.set_xlim(0,1) # most of the data
ax2.set_xlim(9,10) # outliers only

# hide the spines between ax and ax2
ax.spines['right'].set_visible(False)
ax2.spines['left'].set_visible(False)
ax.yaxis.tick_left()
ax.tick_params(labeltop='off') # don't put tick labels at the top
ax2.yaxis.tick_right()

# Make the spacing between the two axes a bit smaller
plt.subplots_adjust(wspace=0.15)

# This looks pretty good, and was fairly painless, but you can get that
# cut-out diagonal lines look with just a bit more work. The important
# thing to know here is that in axes coordinates, which are always
# between 0-1, spine endpoints are at these locations (0,0), (0,1),
# (1,0), and (1,1). Thus, we just need to put the diagonals in the
# appropriate corners of each of our axes, and so long as we use the
# right transform and disable clipping.

d = .015 # how big to make the diagonal lines in axes coordinates
# arguments to pass plot, just so we don't keep repeating them
kwargs = dict(transform=ax.transAxes, color='k', clip_on=False)
ax.plot((1-d,1+d),(-d,+d), **kwargs) # top-left diagonal
ax.plot((1-d,1+d),(1-d,1+d), **kwargs) # bottom-left diagonal

kwargs.update(transform=ax2.transAxes) # switch to the bottom axes
ax2.plot((-d,d),(-d,+d), **kwargs) # top-right diagonal
ax2.plot((-d,d),(1-d,1+d), **kwargs) # bottom-right diagonal

# What's cool about this is that now if we vary the distance between
# ax and ax2 via f.subplots_adjust(hspace=...) or plt.subplot_tool(),
# the diagonal lines will move accordingly, and stay right at the tips
# of the spines they are 'breaking'

plt.show()

解决方案 2：

我看到很多关于此功能的建议，但没有迹象表明它已经实现。这是暂时可行的解决方案。它将阶跃函数变换应用于 x 轴。它有很多代码，但它相当简单，因为其中大部分是样板自定义比例的东西。我没有添加任何图形来指示中断的位置，因为这是一个风格问题。祝你好运完成工作。

from matplotlib import pyplot as plt
from matplotlib import scale as mscale
from matplotlib import transforms as mtransforms
import numpy as np

def CustomScaleFactory(l, u):
    class CustomScale(mscale.ScaleBase):
        name = 'custom'

        def __init__(self, axis, **kwargs):
            mscale.ScaleBase.__init__(self)
            self.thresh = None #thresh

        def get_transform(self):
            return self.CustomTransform(self.thresh)

        def set_default_locators_and_formatters(self, axis):
            pass

        class CustomTransform(mtransforms.Transform):
            input_dims = 1
            output_dims = 1
            is_separable = True
            lower = l
            upper = u
            def __init__(self, thresh):
                mtransforms.Transform.__init__(self)
                self.thresh = thresh

            def transform(self, a):
                aa = a.copy()
                aa[a>self.lower] = a[a>self.lower]-(self.upper-self.lower)
                aa[(a>self.lower)&(a<self.upper)] = self.lower
                return aa

            def inverted(self):
                return CustomScale.InvertedCustomTransform(self.thresh)

        class InvertedCustomTransform(mtransforms.Transform):
            input_dims = 1
            output_dims = 1
            is_separable = True
            lower = l
            upper = u

            def __init__(self, thresh):
                mtransforms.Transform.__init__(self)
                self.thresh = thresh

            def transform(self, a):
                aa = a.copy()
                aa[a>self.lower] = a[a>self.lower]+(self.upper-self.lower)
                return aa

            def inverted(self):
                return CustomScale.CustomTransform(self.thresh)

    return CustomScale

mscale.register_scale(CustomScaleFactory(1.12, 8.88))

x = np.concatenate((np.linspace(0,1,10), np.linspace(9,10,10)))
xticks = np.concatenate((np.linspace(0,1,6), np.linspace(9,10,6)))
y = np.sin(x)
plt.plot(x, y, '.')
ax = plt.gca()
ax.set_xscale('custom')
ax.set_xticks(xticks)
plt.show()

解决方案 3：

检查brokenaxes包：

import matplotlib.pyplot as plt
from brokenaxes import brokenaxes
import numpy as np

fig = plt.figure(figsize=(5,2))
bax = brokenaxes(
    xlims=((0, .1), (.4, .7)),
    ylims=((-1, .7), (.79, 1)),
    hspace=.05
)
x = np.linspace(0, 1, 100)
bax.plot(x, np.sin(10 * x), label='sin')
bax.plot(x, np.cos(10 * x), label='cos')
bax.legend(loc=3)
bax.set_xlabel('time')
bax.set_ylabel('value')

解决方案 4：

一个非常简单的技巧就是

1. 在轴的脊上散点图矩形，以及

2. 在该位置绘制“//”作为文本。

对我来说非常有效：

# FAKE BROKEN AXES
# plot a white rectangle on the x-axis-spine to "break" it
xpos = 10 # x position of the "break"
ypos = plt.gca().get_ylim()[0] # y position of the "break"
plt.scatter(xpos, ypos, color='white', marker='s', s=80, clip_on=False, zorder=100)
# draw "//" on the same place as text
plt.text(xpos, ymin-0.125, r'//', fontsize=label_size, zorder=101, horizontalalignment='center', verticalalignment='center')

示例图：

解决方案 5：

对于那些感兴趣的人，我扩展了@Paul的答案并将其添加到matplotlib包装器proplot中。 它可以执行轴“跳跃”，“加速”和“减速”。

目前没有办法像 Joe 的回答中那样添加表示离散跳跃的“十字”，但我计划在将来添加它。我还计划添加一个默认的“刻度定位器”，根据CutoffScale参数设置合理的默认刻度位置。

解决方案 6：

解决Frederick Nord 的问题：当使用比例不等于 1:1 的网格规范时，如何使对角“断裂”线平行定向，基于 Paul Ivanov 和 Joe Kingtons 的建议进行的以下更改可能会有所帮助。可以使用变量 n 和 m 改变宽度比。

import matplotlib.pylab as plt
import numpy as np
import matplotlib.gridspec as gridspec

x = np.r_[0:1:0.1, 9:10:0.1]
y = np.sin(x)

n = 5; m = 1;
gs = gridspec.GridSpec(1,2, width_ratios = [n,m])

plt.figure(figsize=(10,8))

ax = plt.subplot(gs[0,0])
ax2 = plt.subplot(gs[0,1], sharey = ax)
plt.setp(ax2.get_yticklabels(), visible=False)
plt.subplots_adjust(wspace = 0.1)

ax.plot(x, y, 'bo')
ax2.plot(x, y, 'bo')

ax.set_xlim(0,1)
ax2.set_xlim(10,8)

# hide the spines between ax and ax2
ax.spines['right'].set_visible(False)
ax2.spines['left'].set_visible(False)
ax.yaxis.tick_left()
ax.tick_params(labeltop='off') # don't put tick labels at the top
ax2.yaxis.tick_right()

d = .015 # how big to make the diagonal lines in axes coordinates
# arguments to pass plot, just so we don't keep repeating them
kwargs = dict(transform=ax.transAxes, color='k', clip_on=False)

on = (n+m)/n; om = (n+m)/m;
ax.plot((1-d*on,1+d*on),(-d,d), **kwargs) # bottom-left diagonal
ax.plot((1-d*on,1+d*on),(1-d,1+d), **kwargs) # top-left diagonal
kwargs.update(transform=ax2.transAxes) # switch to the bottom axes
ax2.plot((-d*om,d*om),(-d,d), **kwargs) # bottom-right diagonal
ax2.plot((-d*om,d*om),(1-d,1+d), **kwargs) # top-right diagonal

plt.show()

解决方案 7：

对于 x 轴断裂，这是一个比较粗糙但很好的解决方案。

该解决方案基于https://matplotlib.org/stable/gallery/subplots_axes_and_figures/broken_axis.html，它解决了将断点定位在脊柱上方的问题，通过如何使用matplotlib绘制点使它们出现在脊柱顶部来解决？

from matplotlib.patches import Rectangle
import matplotlib.pyplot as plt

def axis_break(axis, xpos=[0.1, 0.125], slant=1.5):
    d = slant  # proportion of vertical to horizontal extent of the slanted line
    anchor = (xpos[0], -1)
    w = xpos[1] - xpos[0]
    h = 1

    kwargs = dict(marker=[(-1, -d), (1, d)], markersize=12, zorder=3,
                linestyle="none", color='k', mec='k', mew=1, clip_on=False)
    axis.add_patch(Rectangle(
        anchor, w, h, fill=True, color="white",
        transform=axis.transAxes, clip_on=False, zorder=3)
    )
    axis.plot(xpos, [0, 0], transform=axis.transAxes, **kwargs)

fig, ax = plt.subplots(1,1)
plt.plot(np.arange(10))
axis_break(ax, xpos=[0.1, 0.12], slant=1.5)
axis_break(ax, xpos=[0.3, 0.31], slant=-10)

如果你想替换轴标签，这样做就可以了：

from matplotlib import ticker

def replace_pos_with_label(fig, pos, label, axis):
    fig.canvas.draw()  # this is needed to set up the x-ticks
    labs = axis.get_xticklabels()
    labels = []
    locs = []
    for text in labs:
        x = text._x
        lab = text._text

        if x == pos:
            lab = label

        labels.append(lab)
        locs.append(x)
        
    axis.xaxis.set_major_locator(ticker.FixedLocator(locs))
    axis.set_xticklabels(labels)

fig, ax = plt.subplots(1,1)
plt.plot(np.arange(10))
replace_pos_with_label(fig, 0, "-10", axis=ax)
replace_pos_with_label(fig, 6, "$10^{4}$", axis=ax)
axis_break(ax, xpos=[0.1, 0.12], slant=2)