如何使用matplotlib为许多子图创建单个图例？

Question

我正在绘制相同类型的信息，但对于不同的国家/地区，使用matplotlib的多个子图。也就是说，我在3x3网格上有9个图，所有线都相同（当然，每行不同的值）。

但是，我还没想出如何在图上放一个单一的图例（因为所有9个子图都有相同的线）。

我该怎么做？

Answer 1

figlegend可能是您正在寻找的：http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.figlegend

此处示例：http://matplotlib.org/examples/pylab_examples/figlegend_demo.html

另一个例子：

plt.figlegend( lines, labels, loc = 'lower center', ncol=5, labelspacing=0. )

或：

fig.legend( lines, labels, loc = (0.5, 0), ncol=5 )

Answer 2

还有一个很好的函数get_legend_handles_labels()，你可以调用最后一个轴（如果你迭代它们），它们将从label=个参数收集你需要的所有东西：

handles, labels = ax.get_legend_handles_labels()
fig.legend(handles, labels, loc='upper center')

Answer 3

你只需要在循环之外询问一次传奇。

例如，在这种情况下，我有4个子图，具有相同的线和一个图例。

from matplotlib.pyplot import *

ficheiros = ['120318.nc', '120319.nc', '120320.nc', '120321.nc']

fig = figure()
fig.suptitle('concentration profile analysis')

for a in range(len(ficheiros)):
    # dados is here defined
    level = dados.variables['level'][:]

    ax = fig.add_subplot(2,2,a+1)
    xticks(range(8), ['0h','3h','6h','9h','12h','15h','18h','21h']) 
    ax.set_xlabel('time (hours)')
    ax.set_ylabel('CONC ($\mu g. m^{-3}$)')

    for index in range(len(level)):
        conc = dados.variables['CONC'][4:12,index] * 1e9
        ax.plot(conc,label=str(level[index])+'m')

    dados.close()

ax.legend(bbox_to_anchor=(1.05, 0), loc='lower left', borderaxespad=0.)
         # it will place the legend on the outer right-hand side of the last axes

show()

Answer 4

对于具有多个轴的figure中的单个图例的自动定位，如使用subplots()获得的那些，以下解决方案效果非常好：

plt.legend( lines, labels, loc = 'lower center', bbox_to_anchor = (0,-0.1,1,1),
            bbox_transform = plt.gcf().transFigure )

使用bbox_to_anchor和bbox_transform=plt.gcf().transFigure，您定义的figure大小的新边界框将成为loc的引用。使用(0,-0.1,1,1)稍微向下移动此布线框以防止图例被放置在其他艺术家之上。

OBS：在您使用fig.set_size_inches()后使用此解决方案之前使用fig.tight_layout()

Answer 5

我注意到没有答案显示带有单个图例的图像，该图例引用了不同子图中的许多曲线，因此我必须向您展示一个...以使您感到好奇...

现在，您想要查看代码，不是吗？

from numpy import linspace
import matplotlib.pyplot as plt

# Calling the axes.prop_cycle returns an itertoools.cycle

color_cycle = plt.rcParams['axes.prop_cycle']()

# I need some curves to plot

x = linspace(0, 1, 51)
f1 = x*(1-x)   ; lab1 = 'x - x x'
f2 = 0.25-f1   ; lab2 = '1/4 - x + x x' 
f3 = x*x*(1-x) ; lab3 = 'x x - x x x'
f4 = 0.25-f3   ; lab4 = '1/4 - x x + x x x'

# let's plot our curves (note the use of color cycle, otherwise the curves colors in
# the two subplots will be repeated and a single legend becomes difficult to read)
fig, (a13, a24) = plt.subplots(2)

a13.plot(x, f1, label=lab1, **next(color_cycle))
a13.plot(x, f3, label=lab3, **next(color_cycle))
a24.plot(x, f2, label=lab2, **next(color_cycle))
a24.plot(x, f4, label=lab4, **next(color_cycle))

# so far so good, now the trick

lines_labels = [ax.get_legend_handles_labels() for ax in fig.axes]
lines, labels = [sum(lol, []) for lol in zip(*lines_labels)]

# finally we invoke the legend (that you probably would like to customize...)

fig.legend(lines, labels)
plt.show()

这两行

lines_labels = [ax.get_legend_handles_labels() for ax in fig.axes]
lines, labels = [sum(lol, []) for lol in zip(*lines_labels)]

值得解释-为此，我将棘手的部分封装在一个函数中，只有4行代码，但是沉重注释了

def fig_legend(fig, **kwdargs):

    # generate a sequence of tuples, each contains
    #  - a list of handles (lohand) and
    #  - a list of labels (lolbl)
    tuples_lohand_lolbl = (ax.get_legend_handles_labels() for ax in fig.axes)
    # e.g. a figure with two axes, ax0 with two curves, ax1 with one curve
    # yields:   ([ax0h0, ax0h1], [ax0l0, ax0l1]) and ([ax1h0], [ax1l0])

    # legend needs a list of handles and a list of labels, 
    # so our first step is to transpose our data,
    # generating two tuples of lists of homogeneous stuff(tolohs), i.e
    # we yield ([ax0h0, ax0h1], [ax1h0]) and ([ax0l0, ax0l1], [ax1l0])
    tolohs = zip(*tuples_lohand_lolbl)

    # finally we need to concatenate the individual lists in the two
    # lists of lists: [ax0h0, ax0h1, ax1h0] and [ax0l0, ax0l1, ax1l0]
    # a possible solution is to sum the sublists - we use unpacking
    handles, labels = (sum(list_of_lists, []) for list_of_lists in tolohs)

    # call fig.legend with the keyword arguments, return the legend object

    return fig.legend(handles, labels, **kwdargs)

---

PS我认识到sum(list_of_lists, [])是拉平列表列表的一种非常低效的方法，但是①我喜欢它的紧凑性，②通常在几个子图中有一些曲线，并且③Matplotlib和效率？ ;-）

Answer 6

虽然游戏时间已经很晚了，但我还是会给出另一个解决方案，因为这仍然是谷歌上显示的第一个链接之一。使用matplotlib 2.2.2，可以使用gridspec功能实现。在下面的示例中，目标是以2x2方式排列四个子图，底部显示图例。 A＆＃39; faux＆＃39;在底部创建轴以将图例放置在固定点中。 ＆＃39; faux＆＃39;然后关闭轴，只显示图例。结果：https://i.stack.imgur.com/5LUWM.png。

import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec

#Gridspec demo
fig = plt.figure()
fig.set_size_inches(8,9)
fig.set_dpi(100)

rows   = 17 #the larger the number here, the smaller the spacing around the legend
start1 = 0
end1   = int((rows-1)/2)
start2 = end1
end2   = int(rows-1)

gspec = gridspec.GridSpec(ncols=4, nrows=rows)

axes = []
axes.append(fig.add_subplot(gspec[start1:end1,0:2]))
axes.append(fig.add_subplot(gspec[start2:end2,0:2]))
axes.append(fig.add_subplot(gspec[start1:end1,2:4]))
axes.append(fig.add_subplot(gspec[start2:end2,2:4]))
axes.append(fig.add_subplot(gspec[end2,0:4]))

line, = axes[0].plot([0,1],[0,1],'b')           #add some data
axes[-1].legend((line,),('Test',),loc='center') #create legend on bottommost axis
axes[-1].set_axis_off()                         #don't show bottommost axis

fig.tight_layout()
plt.show()

Answer 7

这个答案是@Evert在传奇位置上的补充。

我对@Evert的解决方案的第一次尝试失败了，因为传奇和子剧情的标题重叠。

实际上，重叠是由fig.tight_layout()引起的，它会更改子图的布局而不考虑图例。但是，fig.tight_layout()是必要的。

为了避免重叠，我们可以告诉fig.tight_layout() fig.tight_layout(rect=(0,0,1,0.9))为图形的图例留出空格。

Description of tight_layout() parameters

Answer 8

如果要对条形图使用子图，则每个条形具有不同的颜色。使用mpatches

自己创建文物可能会更快

假设您有四个颜色不同的条形，例如r m c k，您可以按照以下方式设置图例

import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
labels = ['Red Bar', 'Magenta Bar', 'Cyan Bar', 'Black Bar']


#####################################
# insert code for the subplots here #
#####################################


# now, create an artist for each color
red_patch = mpatches.Patch(facecolor='r', edgecolor='#000000') #this will create a red bar with black borders, you can leave out edgecolor if you do not want the borders
black_patch = mpatches.Patch(facecolor='k', edgecolor='#000000')
magenta_patch = mpatches.Patch(facecolor='m', edgecolor='#000000')
cyan_patch = mpatches.Patch(facecolor='c', edgecolor='#000000')
fig.legend(handles = [red_patch, magenta_patch, cyan_patch, black_patch],labels=labels,
       loc="center right", 
       borderaxespad=0.1)
plt.subplots_adjust(right=0.85) #adjust the subplot to the right for the legend

Answer 9

要建立在@gboffi和Ben Usman的答案之上：

在不同的子图中具有相同颜色和标签的不同线条的情况下，人们可以沿...的线条做一些事情

labels_handles = {
  label: handle for ax in fig.axes for handle, label in zip(*ax.get_legend_handles_labels())
}

fig.legend(
  labels_handles.values(),
  labels_handles.keys(),
  loc="upper center",
  bbox_to_anchor=(0.5, 0),
  bbox_transform=plt.gcf().transFigure,
)