基于gboffi和Ben Usman的回答:

如果在不同的子图中有不同的线，但颜色和标签相同，你可以这样做:

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,
)

Figlegend可能就是您要找的:matplotlib.pyplot.figlegend

一个例子是在图图例演示。

另一个例子:

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

Or:

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

博士TL;

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

我注意到，其他答案都没有显示一个图像，其中一个图例引用了不同子图中的许多曲线，所以我必须给你看一个……为了让你好奇……

现在，如果我已经戏弄你们够多了，这就是代码

from numpy import linspace
import matplotlib.pyplot as plt

# each Axes has a brand new prop_cycle, so to have differently
# colored curves in different Axes, we need our own prop_cycle
# Note: we CALL the axes.prop_cycle to get an itertoools.cycle
color_cycle = plt.rcParams['axes.prop_cycle']()

# I need some curves to plot
x = linspace(0, 1, 51)
functs = [x*(1-x), x**2*(1-x),
          0.25-x*(1-x), 0.25-x**2*(1-x)] 
labels = ['$x-x²$', '$x²-x³$',
          '$\\frac{1}{4} - (x-x²)$', '$\\frac{1}{4} - (x²-x³)$']

# the plot, 
fig, (a1,a2) = plt.subplots(2)
for ax, f, l, cc in zip((a1,a1,a2,a2), functs, labels, color_cycle): 
    ax.plot(x, f, label=l, **cc)
    ax.set_aspect(2) # superfluos, but nice

# So far, nothing special except the managed prop_cycle. 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, the legend (that maybe you'll customize differently)
fig.legend(lines, labels, loc='upper center', ncol=4)
plt.show()

If you want to stick with the official Matplotlib API, this is perfect, otherwise see note no.1 below (there is a private method...) The two lines lines_labels = [ax.get_legend_handles_labels() for ax in fig.axes] lines, labels = [sum(lol, []) for lol in zip(*lines_labels)] deserve an explanation, see note 2 below. I tried the method proposed by the most up-voted and accepted answer, # fig.legend(lines, labels, loc='upper center', ncol=4) fig.legend(*a2.get_legend_handles_labels(), loc='upper center', ncol=4) and this is what I've got

注1 如果您不介意使用matplotlib的私有方法。传说模块…这真的非常非常简单

from matplotlib.legend import _get_legend_handles_labels
...

fig.legend(*_get_legend_handles_and_labels(fig.axes), ...)

注2

我把这两行棘手的代码封装在一个函数中，只有四行代码，但是注释了很多

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])

    # The 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)

我认识到sum(list_of_lists，[])是一个非常低效的方法来扁平化列表的列表，但①我喜欢它的紧凑性，②通常是几个子图中的几条曲线，③Matplotlib和效率?: -)

所有之前的答案都超出了我的理解，在我的编码旅程的这个状态下，所以我只是添加了另一个Matplotlib方面，称为补丁:

import matplotlib.patches as mpatches

first_leg = mpatches.Patch(color='red', label='1st plot')
second_leg = mpatches.Patch(color='blue', label='2nd plot')
thrid_leg = mpatches.Patch(color='green', label='3rd plot')
plt.legend(handles=[first_leg ,second_leg ,thrid_leg ])

补丁方面把我需要的所有数据放在我的最终图(这是一个线状图，在Jupyter Notebook的同一个单元格中结合了三个不同的线状图)。

结果

(我更改了我自己命名的图例的名称。)

你只需要在循环之外请求一次图例。

例如，在这种情况下，我有4个子情节，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()

这个答案是对user707650在图例位置上的回答的补充。

我第一次尝试user707650的解决方案失败了，因为图例和副图的标题重叠。

事实上，重叠是由fig.tight_layout()引起的，它改变了子图的布局，而不考虑图形图例。但是，fig.tight_layout()是必要的。

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

tight_layout()参数的描述。