堆積餅圖-如何根據外部餅圖設定內部餅圖示簽

我有一個帶有百分比標簽的堆積餅圖。外部餅圖有我想要的標簽；但是，內部圖表的標簽應基于它在外部餅圖下的位置。例如，在下圖中，對于外部餅圖的每種顏色，內部餅圖的百分比標簽的總和應為 100%。目前，對于紅色的外部餅圖，內部餅圖有11.0%、12.0%、7.0%，這是沒有考慮外部圖表的自身百分比。我希望他們為外部餅圖的綠色和黃色顏色顯示 35%、38%、27% 和相同的值，其中內部圖表標簽取決于外部餅圖顏色。

這是一個可重現的示例

import pandas as pd
import numpy as np
import random
import matplotlib.pyplot as plt

options = ['RED','YELLOW','GREEN']
color_list1 = []
color_list2 = []

for i in np.arange(100):
    color1 = random.choice(options)
    color2 = random.choice(options)
    color_list1.append(color1)
    color_list2.append(color2)

test_df = pd.DataFrame(list(zip(color_list1, color_list2)), columns =['Outer_Color', 'Inner_Color'])

outer = pd.DataFrame(test_df.groupby('Outer_Color').size())
inner = pd.DataFrame(test_df.groupby(['Outer_Color', 'Inner_Color']).size())
inner_labels = inner.index.get_level_values(1)


my_pal = {"RED": "r", "YELLOW": "yellow", "GREEN":"g"}

fig, ax = plt.subplots(figsize=(14,7))
size = 0.3

ax.pie(outer.values.flatten(), radius=1,
#        labels=outer.index,
       autopct='%1.1f%%', pctdistance=0.85, # labeldistance=0.2,
       wedgeprops=dict(width=size, edgecolor='w'),
       colors=[my_pal[key] for key in outer.index])

ax.pie(inner.values.flatten(), radius=1-size, 
#        labels = inner_labels,
       autopct='%1.1f%%', pctdistance=0.7, # labeldistance=1.2,
       wedgeprops=dict(width=size, edgecolor='w'),
       colors=[my_pal[key] for key in inner_labels])

plt.show()

uj5u.com熱心網友回復：

您可以采取的一種方法是計算內部楔形相對于外部楔形的百分比，然后制作新標簽以在您這樣做時通過ax.pie。例如：

# Construct inner labels
    inner_str_label = []  # new labels storage
    n_inside = 3  # number of inner wedges per outer wedge
    for idx, outer_value in enumerate(outer.values.flatten()):
        # Get inner values for each outer wedge
        for inner_value in inner.values.flatten()[idx*n_inside:idx*n_inside n_inside]:
            # format for only 1 decimal place
            new_label = f"{'{0:.1f}'.format(inner_value/outer_value*100)}%"
            inner_str_label.append(new_label)

ax.pie現在我們可以在using中傳遞新標簽labels：

import pandas as pd
import numpy as np
import random
import matplotlib.pyplot as plt


if __name__ == '__main__':

    options = ['RED', 'YELLOW', 'GREEN']
    color_list1 = []
    color_list2 = []

    for i in np.arange(100):
        color1 = random.choice(options)
        color2 = random.choice(options)
        color_list1.append(color1)
        color_list2.append(color2)

    test_df = pd.DataFrame(list(zip(color_list1, color_list2)), columns =['Outer_Color', 'Inner_Color'])

    outer = pd.DataFrame(test_df.groupby('Outer_Color').size())
    inner = pd.DataFrame(test_df.groupby(['Outer_Color', 'Inner_Color']).size())

    my_pal = {"RED": "r", "YELLOW": "yellow", "GREEN": "g"}

    fig, ax = plt.subplots(figsize=(14, 7))
    size = 0.3

    ax.pie(outer.values.flatten(), radius=1,
           #        labels=outer.index,
           autopct='%1.1f%%', pctdistance=0.85,  # labeldistance=0.2,
           wedgeprops=dict(width=size, edgecolor='w'),
           colors=[my_pal[key] for key in outer.index])

    # Construct inner labels
    inner_str_label = []  # new labels storage
    n_inside = 3  # number of inner wedges per outer wedge
    for idx, outer_value in enumerate(outer.values.flatten()):
        # Get inner values for each outer wedge
        for inner_value in inner.values.flatten()[idx*n_inside:idx*n_inside n_inside]:
            # format for only 1 decimal place
            new_label = f"{'{0:.1f}'.format(inner_value/outer_value*100)}%"
            inner_str_label.append(new_label)

    wedges, labels = ax.pie(inner.values.flatten(), radius=1-size, labels=inner_str_label, labeldistance=0.8,
                            wedgeprops=dict(width=size, edgecolor='w'),
                            colors=[my_pal[key] for key in inner.index.get_level_values(1)])

    # Fix inner label position
    for string_label in labels:
        string_label.update({"rotation": 0, "horizontalalignment": "center", "verticalalignment": "center"})

    plt.show()

請注意，這是假設每個外楔具有相同數量的內楔。如果內部楔形的數量不規則，則方法類似，但您在串列中設定每個外部楔形的內部楔形數量：

    # Construct inner labels
    inner_str_label = []  # new labels storage
    n_inside_list = [2, 3, 2]  # number of inner wedges per outer wedge
    for idx, outer_value in enumerate(outer.values.flatten()):
        # Get inner values for each outer wedge
        n_inside = n_inside_list[idx]
        for inner_value in inner.values.flatten()[idx*n_inside:idx*n_inside n_inside]:
            # format for only 1 decimal place
            new_label = f"{'{0:.1f}'.format(inner_value/outer_value*100)}%"
            inner_str_label.append(new_label)

希望這會有所幫助 - 干杯！

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