往返传奇

Question

这是一个小代码sn-p：

import matplotlib.pyplot as plt

fig, ax = plt.subplots()

ax.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')
ax.legend()
ax.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')
leg = ax.get_legend()

看起来像这样：

现在，如果给了我leg，我怎么能重现传奇？

我试过了

ax.legend(leg.legendHandles, [i.get_text() for i in leg.get_texts()])

但是，这不会保留制造商信息（注意现在图例中的线只是一条直线，而不是一条带有错误栏的线）

我也试过

ax.legend(*ax.get_legend_handles_labels());

但是，这会添加一个在原始图例中不可见的新行。

---

编辑

如果原来的情节是

import matplotlib.pyplot as plt

fig, ax = plt.subplots()

ax.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')
ax.plot([1, 2, 3], [3, 3, 3], c='green', label='green')
ax.legend()
ax.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')
leg = ax.get_legend()

，如下所示：

然后我想保留橙色和绿色的线。基本上，我只想保留图例中已经可见的内容，而 ax.get_legend_handles_labels 将所有内容归还给我。

编辑2

leg.legendHandles 和 ax.get_legend_handles_labels() 之间的 1-1 映射可以实现这一点，可以做到吗？

Answer 1

leg.legendHandles 和 ax.get_legend_handles_labels() 之间的 1-1 映射可以实现这一点，可以做到吗？

是的，看下面的例子

import matplotlib.pyplot as plt

figure, (axes1, axes2) = plt.subplots(nrows=1, ncols=2, figsize=(8, 4))

orange = axes1.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')

axes1.legend()

axes1.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')

handles, labels = axes1.get_legend_handles_labels()

axes2.legend([handles[1]], [labels[1]])

plt.show()

您还可以创建返回 matplotlib.pyplot.plot() 的图例。

import matplotlib.pyplot as plt

figure, (axes1, axes2) = plt.subplots(nrows=1, ncols=2, figsize=(8, 4))

orange = axes1.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')

axes1.legend()

axes1.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')

axes2.legend([orange], ['orange'])

plt.show()

现在，如果给了我leg，我怎么能重现传奇？

我试过了

ax.legend(leg.legendHandles, [i.get_text() for i in leg.get_texts()])

我在sourcecode of legend 中找到的与句柄相关的属性也是legendHandles。该属性在_init_legend_box() 中初始化。相关代码是

    def _init_legend_box(self, handles, labels, markerfirst=True):
        ...

        text_list = []  # the list of text instances
        handle_list = []  # the list of text instances  <-- I think this might be typo
        handles_and_labels = []

        ...

        for orig_handle, lab in zip(handles, labels):
            handler = self.get_legend_handler(legend_handler_map, orig_handle)
            if handler is None:
                _api.warn_external(
                    "Legend does not support {!r} instances.\nA proxy artist "
                    "may be used instead.\nSee: "
                    "https://matplotlib.org/users/legend_guide.html"
                    "#creating-artists-specifically-for-adding-to-the-legend-"
                    "aka-proxy-artists".format(orig_handle))
                # We don't have a handle for this artist, so we just defer
                # to None.
                handle_list.append(None)
            else:
                textbox = TextArea(lab, textprops=label_prop,
                                   multilinebaseline=True)
                handlebox = DrawingArea(width=self.handlelength * fontsize,
                                        height=height,
                                        xdescent=0., ydescent=descent)

                text_list.append(textbox._text)
                # Create the artist for the legend which represents the
                # original artist/handle.
                handle_list.append(handler.legend_artist(self, orig_handle,
                                                         fontsize, handlebox))
                handles_and_labels.append((handlebox, textbox))

        ...

        self.texts = text_list
        self.legendHandles = handle_list

正如我们所见，handle_list 最终分配给了self.legendHandles。在for循环中将值附加到handle_list，如下所示

# Create the artist for the legend which represents the
# original artist/handle.
handle_list.append(handler.legend_artist(self, orig_handle,
                                         fontsize, handlebox))

# print(f'{orig_handle} - {type(orig_handle}') gives
"""
<ErrorbarContainer object of 3 artists> - <class 'matplotlib.container.ErrorbarContainer'>
"""

评论说Create the artist for the legend which represents the original artist/handle。但是从下面的例子中，我们可以看到返回的艺术家并不代表原始的句柄。我认为这可能是 matplotlib 的一个错误。

import matplotlib.pyplot as plt

figure, (axes1, axes2) = plt.subplots(nrows=1, ncols=2, figsize=(8, 4))

orange = axes1.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')

axes1.legend()

axes1.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')

leg = axes1.get_legend()

handles, labels = axes1.get_legend_handles_labels()

print(f'{orange} - {type(orange)}')
print(f'{handles[1]} - {type(handles[1])}')
print(f'{leg.legendHandles} - {type(leg.legendHandles)}')

"""
<ErrorbarContainer object of 3 artists> - <class 'matplotlib.container.ErrorbarContainer'>
<ErrorbarContainer object of 3 artists> - <class 'matplotlib.container.ErrorbarContainer'>
[<matplotlib.collections.LineCollection object at 0x7f4c79919040>] - <class 'list'>
"""

Answer 2

坐标轴图例ax 存储在ax.legend_ 中。要覆盖它，只需分配ax.legend_ = leg。然后，您需要使用 ax.update_from(ax) 更新轴。

一些示例代码来演示，使用你的情节：

from matplotlib import pyplot as plt


#### SECTION 1 ####

fig, ax = plt.subplots()
ax.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')
ax.plot([1, 2, 3], [3, 3, 3], c='green', label='green')
# Modifying your code a little to highlight what ax.legend() returns 
# vs what ax.get_legend() returns
leg_orig = ax.legend()
ax.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')
leg = ax.get_legend()

# ax.get_legend() is just returning ax.legend_
assert(leg==leg_orig)


#### SECTION 2 ####

# calling ax.legend() overwrites ax.legend_ with a new legend object,
# updates the axes, and returns the new legend object
leg = ax.legend()
assert(leg==ax.legend_)
assert(leg==ax.get_legend())
assert(leg!=leg_orig)


#### SECTION 3 ####

# we can manually overwite and update with the original legend
ax.legend_ = leg_orig
ax.update_from(ax) 

尝试分段运行代码，看看每次更新图例的效果。

重要提示

像 Python 中的大多数可变对象一样，当您覆盖 ax.legend_ 时，您只是覆盖了一个指针，而不是创建一个新对象。该对象独立于ax.legend_ 存在，这意味着您对图例对象所做的任何更改都会影响共享该图例对象的所有轴。

Answer 3

以下函数可以将图例复制到选定的轴：

import matplotlib.pyplot as plt

def duplicate_legend(leg, axis_to):
    """Duplicate a legend to another axis"""
    labels_in = [l.get_text() for l in leg.texts]
    handles, labels = [], []
    for h, l in zip(*leg.axes.get_legend_handles_labels()):
        if l in labels_in:
            handles.append(h)
            labels.append(l)
            labels_in.remove(l)
    axis_to.legend(handles, labels)

这里是一个例子：

fig, (ax1, ax2) = plt.subplots(2, 1)

ax1.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')
ax1.plot([1, 2, 3], [3, 3, 3], c='green', label='green')
ax1.legend()
ax1.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')
leg = ax1.get_legend()

duplicate_legend(leg, ax2)
plt.show()

在多个标签的情况下，有的隐藏有的可见，只会添加可见的：

fig, (ax1, ax2) = plt.subplots(2, 1)

ax1.errorbar([1, 2, 3], [1, 1, 1], yerr=[.1, .1, .1], c='orange', label='orange')
ax1.plot([1, 2, 3], [3, 3, 3], c='green', label='green')
ax1.plot([1, 2, 3], [4, 4, 4], c='green', label='green')
ax1.legend()
ax1.plot([1, 2, 3], [5, 5, 5], c='green', label='green')
ax1.plot([1, 2, 3], [2, 2, 2], c='blue', label='blue')
leg = ax1.get_legend()

duplicate_legend(leg, ax2)
plt.show()