2 回答
TA贡献1893条经验 获得超10个赞
由于添加图例matplotlib
是一个复杂而广泛的步骤,因此请考虑使用您引用的 @jrjc函数解决方案的链接。但是,您需要根据水平条形图的需要调整功能。具体来说:
为颜色图和
DataFrame.plot
调用添加参数将条形图从
kind='bar'
到调整kind='barh'
为水平版本将 x 换成 y 行:
rect.set_y(rect.get_y() + 1 / float(n_df + 1) * i / float(n_col))
换线
width
:height
rect.set_height(1 / float(n_df + 1))
调整
axe.set_xticks
和axe.set_xticklabels
值np.arange(0, 120, 20)
_
功能
import numpy as np
import pandas as pd
import matplotlib.cm as cm
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap as coloring
def plot_clustered_stacked(dfall, labels=None, title="multiple stacked bar plot", H="//",
colors_ABCDE=['tomato', 'gold', 'greenyellow', 'forestgreen', 'palevioletred'], **kwargs):
"""
CREDIT: @jrjc (https://stackoverflow.com/a/22845857/1422451)
Given a list of dataframes, with identical columns and index, create a clustered stacked bar plot.
labels is a list of the names of the dataframe, used for the legend
title is a string for the title of the plot
H is the hatch used for identification of the different dataframe
"""
n_df = len(dfall)
n_col = len(dfall[0].columns)
n_ind = len(dfall[0].index)
axe = plt.subplot(111)
for df in dfall : # for each data frame
axe = df.plot(kind="barh",
linewidth=0,
stacked=True,
ax=axe,
legend=False,
grid=False,
colormap=coloring.from_list("my_colormap", colors_ABCDE),
edgecolor="darkgrey",
**kwargs) # make bar plots
h,l = axe.get_legend_handles_labels() # get the handles we want to modify
for i in range(0, n_df * n_col, n_col): # len(h) = n_col * n_df
for j, pa in enumerate(h[i:i+n_col]):
for rect in pa.patches: # for each index
rect.set_y(rect.get_y() + 1 / float(n_df + 2) * i / float(n_col))
rect.set_hatch(H * int(i / n_col)) #edited part
rect.set_height(1 / float(n_df + 2))
axe.set_xticks(np.arange(0, 125, 20))
axe.set_xticklabels(np.arange(0, 125, 20).tolist(), rotation = 0)
axe.margins(x=0, tight=None)
axe.set_title(title)
# Add invisible data to add another legend
n=[]
for i in range(n_df):
n.append(axe.bar(0, 0, color="gray", hatch=H * i, edgecolor="darkgrey"))
l1 = axe.legend(h[:n_col], l[:n_col], loc=[1.01, 0.5])
if labels is not None:
l2 = plt.legend(n, labels, loc=[1.01, 0.1])
axe.add_artist(l1)
return axe
称呼
plt.figure(figsize=(10, 4))
plot_clustered_stacked([df_1, df_2],["df_1", "df_2"])
plt.show()
plt.clf()
plt.close()
输出
TA贡献1856条经验 获得超17个赞
我认为@jrjc 的这个功能解决方案对我的理解来说相当令人困惑,因此,我更愿意稍微改变我自己的东西并进行调整。
所以,我花了一些时间才明白,当为情节创建第二个图例时,python 会自动删除第一个图例,这是add_artist()必须使用的时候。
添加第二个图例的另一个先决条件是命名图并将 .add_artist() 方法应用于该特定图,以便 python 知道将新部分粘贴在哪里。
简而言之,这就是我设法创建我想到的情节的方式,我希望这些评论能让它更清晰,对任何人都有用。
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap as coloring
import matplotlib.patches as mpatches
# copy the dfs below and use pd.read_clipboard() to reproduce
df_1
A B C D E
Mg 10 15 23 25 27
Ca 30 33 0 20 17
df_2
A B C D E
Mg 20 12 8 40 10
Ca 7 26 12 22 16
hatches=(' ', '//')
colors_ABCDE=['tomato', 'gold', 'greenyellow', 'forestgreen', 'palevioletred']
dfs=[df_1,df_2]
for each_df, df in enumerate(dfs):
#I name the plot as "figure"
figure=df.plot(ax=plt.subplot(111), kind="barh", \
stacked=True, hatch=hatches[each_df], \
colormap=coloring.from_list("my_colormap", colors_ABCDE), \
figsize=(7,2.5), position=len(dfs)-each_df-1, \
align='center', width=0.2, edgecolor="darkgrey", \
legend=False) #I had to False the legend too
legend_1=plt.legend(df_1.columns, loc='center left', bbox_to_anchor=(1.0, 0.5), fontsize=12)
patch_hatched = mpatches.Patch(facecolor='beige', hatch='///', edgecolor="darkgrey", label='hatched')
patch_unhatched = mpatches.Patch(facecolor='beige', hatch=' ', edgecolor="darkgrey", label='non-hatched')
legend_2=plt.legend(handles=[patch_hatched, patch_unhatched], loc='center left', bbox_to_anchor=(1.15, 0.5), fontsize=12)
# as soon as a second legend is made, the first disappears and needs to be added back again
figure.add_artist(legend_1) #python now knows that "figure" must take the "legend_1" along with "legend_2"
有两个传说的情节
我很确定它可以更加优雅和自动化。
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