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,每个条目都可以很长(但我不确切知道多长时间).
显然,这很容易使用
legendHandle = plt.legend(loc = "center left",bBox_to_anchor = (1,0.5))
但问题是图例被窗户边缘切断了.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",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图例是如此困难.我习惯了Matlab,就像它一样简单
legend('Location','eastoutside')
Python中是否有类似的东西我缺少?
解决方法
尝试了很多,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,width_ratios = [1 - legendWidth,legendWidth]) legendLocation = 1 plotLocation = 0 elif self.location == Location.WestOutside: gridspec = GridSpec(1,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,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,这使得图例或多或少都没有.用法是例如
import matplotlib.pyplot as plt plt.ion() figure = plt.figure() plotAxes = figure.gca() plotAxes.plot([1,3],[4,5,6],"b-",label = "Testaaaaaaaaaaaaaa 1") plotAxes.plot([1,[6,4],"r-",label = "Test 2") legend = Legend(figure,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,而是在图中有传说的本地方式)