Question

我试图在matplotlib的情节之外放置一个相当广泛的传奇。这个图例有很多条目，每个条目都可能很长（但我不知道确切的长度）。

显然，使用

非常容易

legendHandle = plt.legend(loc = "center left", bbox_to_anchor = (1, 0.5))

但问题是图例被窗口边缘切断了。我花了很长时间寻找解决方案。到目前为止我能找到的最好的事情是：

box = ax.get_position()
ax.set_position([box.x0, box.y0, box.width * 0.8, box.height])
plt.legend(loc = "center left", bbox_to_anchor = (1, 0.5))

不幸的是，这并没有真正解决我的问题。由于应用于框宽的显式因子0.8，这仅适用于图形和图例宽度的一个特定组合。如果我调整图形窗口的大小，或者我的图例条目具有不同的长度，则它不起作用。

我只是不明白如何在图中放置一个传奇是如此困难。我已经习惯了Matlab，它就像

一样简单

legend('Location', 'eastoutside')

Python中是否存在类似的东西？

Answer 1

在尝试了很多之后，这是我能想到的最好的：

from matplotlib.lines import Line2D
from matplotlib.gridspec import GridSpec
from enum import Enum

class Location(Enum):
    EastOutside = 1
    WestOutside = 2
    NorthOutside = 3
    SouthOutside = 4

class Legend:
    def __init__(self, figure, plotAxes, location: Location):
        self.figure = figure
        self.plotAxes = plotAxes
        self.location = location

        # Create a separate subplot for the legend. Actual location doesn't matter - will be modified anyway.
        self.legendAxes = figure.add_subplot(1, 2, 1)
        self.legendAxes.clear() # remove old lines
        self.legendAxes.set_axis_off()

        # Add all lines from the plot to the legend subplot
        for line in plotAxes.get_lines():
            legendLine = Line2D([], [])
            legendLine.update_from(line)
            self.legendAxes.add_line(legendLine)

        if self.location == Location.EastOutside:
            self.legend = self.legendAxes.legend(loc = "center left")
        elif self.location == Location.WestOutside:
            self.legend = self.legendAxes.legend(loc = "center right")
        elif self.location == Location.NorthOutside:
            self.legend = self.legendAxes.legend(loc = "lower center")
        elif self.location == Location.SouthOutside:
            self.legend = self.legendAxes.legend(loc = "upper center")
        else:
            raise Exception("Unknown legend location.")

        self.UpdateSize()

        # Recalculate legend size if the size changes
        figure.canvas.mpl_connect('resize_event', lambda event: self.UpdateSize())

    def UpdateSize(self):
        self.figure.canvas.draw() # draw everything once in order to get correct legend size

        # Extract legend size in percentage of the figure width
        legendSize = self.legend.get_window_extent().inverse_transformed(self.figure.transFigure)
        legendWidth = legendSize.width
        legendHeight = legendSize.height

        # Update subplot such that it is only as large as the legend
        if self.location == Location.EastOutside:
            gridspec = GridSpec(1, 2, width_ratios = [1 - legendWidth, legendWidth])
            legendLocation = 1
            plotLocation = 0
        elif self.location == Location.WestOutside:
            gridspec = GridSpec(1, 2, width_ratios = [legendWidth, 1 - legendWidth])
            legendLocation = 0
            plotLocation = 1
        elif self.location == Location.NorthOutside:
            gridspec = GridSpec(2, 1, height_ratios = [legendHeight, 1 - legendHeight])
            legendLocation = 0
            plotLocation = 1
        elif self.location == Location.SouthOutside:
            gridspec = GridSpec(2, 1, height_ratios = [1 - legendHeight, legendHeight])
            legendLocation = 1
            plotLocation = 0
        else:
            raise Exception("Unknown legend location.")

        self.legendAxes.set_position(gridspec[legendLocation].get_position(self.figure))
        self.legendAxes.set_subplotspec(gridspec[legendLocation]) # to make figure.tight_layout() work if that's desired

        self.plotAxes.set_position(gridspec[plotLocation].get_position(self.figure))
        self.plotAxes.set_subplotspec(gridspec[plotLocation]) # to make figure.tight_layout() work if that's desired

在我到目前为止测试的情况下，这或多或少都会使传说更加正常。用法就是例如。

import matplotlib.pyplot as plt

plt.ion()

figure = plt.figure()

plotAxes = figure.gca()

plotAxes.plot([1, 2, 3], [4, 5, 6], "b-", label = "Testaaaaaaaaaaaaaa 1")
plotAxes.plot([1, 2, 3], [6, 5, 4], "r-", label = "Test 2")

legend = Legend(figure, plotAxes, Location.EastOutside)

让我问一下我在评论中发布的问题......我将如何建议将其作为matplotlib开发人员的附加功能？（不是我的黑客，而是在图中外面传说的本地方式）

Answer 2

偶然发现这个问题，我想你可以使用这个图的传奇方法：

import matplotlib.pyplot as plt

# sample data
x1 = [1,2,3,4,5,6]
y1 = [0,1,0,1,0,1]
x2 = [1,2,3,4,5,6]
y2 = [9,8,8,7,8,6]

fig = plt.figure(figsize=(4, 2), dpi=100)
ax = fig.add_subplot(111)

ax.plot(x1,y1,x2,y2)

fig.legend(loc=1, mode='expand', numpoints=1, ncol=4, fancybox = True,
           fontsize='small', labels=['d1', 'd2'])
# loc=1 means at the top-right of the figure

plt.show()

虽然图例位于轴外，但如果图例太大，它仍可能与轴重叠。

Answer 3

图例的位置可以通过使用参数来控制 loc 和 bbox_to_anchor 。这是一个代码示例：

import matplotlib.pyplot as plt

#sample data
import numpy as np
x = np.linspace(0, 2, 101)

#create figure and its axes
fig = plt.figure()
ax = fig.add_axes([0,0,1,1])

#plot 3 lines and define their labels
ax.plot(x, x**2, label="square")
ax.plot(x, x**3, label="cubic")
ax.plot(x, np.sin(x), label="sinus")

#place the legend
ax.legend(loc='lower center', bbox_to_anchor=(0.5, 1.0), ncol=3)
  # The center of the lower edge of the rectangle containing the legend
  # is placed at coordinates (x,y)=(0.5,1.0) of ax. 
  # Thus, figure and legend should not overlap.

plt.show()

您现在应该看到下图： figure with legend outside

Python中情节之外的传说 - matplotlib

3 个答案: