Matplotlib基于另一个变量添加带有颜色值的颜色图例 [英] Matplotlib add color legend with value based on another variable

问题描述

我想像这样添加颜色图例:

I want to add color legend like this:

Green - Tier 1
Gold - Tier 2
Silver - Tier 3
Chocolate - Tier 4. 

值层1"，层2"，层3"，层4"基于另一个称为 RFM ['Tier'] 的列.

The values 'Tier 1', 'Tier 2', 'Tier 3', 'Tier 4' are based on another column called RFM['Tier'].

plt.style.use('classic')
#Scatter plot monetary and recency
Color = ['green','gold','silver','chocolate']
RFM['Color']= RFM['Cluster'].map(lambda p: Color[p])
ax = RFM.plot(
    kind='scatter',
    x='Monetary', y='Recency',
    figsize=(10,8),
    c= RFM['Color']
)
ax.set_title('Monetary and Recency Distribution',color='darkslategray')

推荐答案

Seaborn的散点图将自动创建图例.hue= 告诉要使用哪一列着色，palette= 参数告诉要使用的颜色(列表或颜色图).对于非数字数据，hue_order= 可以固定某种顺序.

Seaborn's scatterplot would automatically create a legend. The hue= tells which column to use for coloring, the palette= parameter tells the colors to use (either a list or a colormap). For non-numeric data, hue_order= can fix a certain ordering.

以下是玩具数据的示例:

Here is an example with toy data:

import matplotlib.pyplot as plt
import seaborn as sns
import pandas as pd
import numpy as np

plt.style.use('classic')
N = 100
RFM = pd.DataFrame({'Tier': np.random.randint(1, 5, N),
                    'Monetary': np.random.uniform(1, 100, N),
                    'Recency': np.random.uniform(1, 500, N)})
color_palette = ['green', 'gold', 'silver', 'chocolate']
plt.figure(figsize=(10, 8))
ax = sns.scatterplot(data=RFM, x='Monetary', y='Recency', hue='Tier', palette=color_palette)
ax.set_title('Monetary and Recency Distribution', color='darkslategray')
plt.show()

PS:要更改图例标签，可以显式调用 ax.legend ，例如:

PS: To change the legend labels, ax.legend can be called explicitly, e.g.:

ax.legend([f'Tier {i}' for i in range(1, 5)])

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