我在散點圖中顯示圖例時遇到問題。向您展示我的問題的一個簡單示例是：從 1 到 10 的每個整數都有顏色。我想要一個顯示十個數字及其顏色的標簽（基本上：顏色及其對應的數字）
我有一個資料幀中的所有值（我向您展示的資料幀只是一個例子，真實的例子由數百行組成）
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
x = np.array([1,2,3,4,5,6,7,8,9,10])
y = 2*x
df = pd.DataFrame()
palette = {1: "blue", 2:"orange", 3:"green", 4:"red", 5:"purple", 6:"brown", 7:"pink", 8:"gray", 9:"olive", 10:"cyan"}
df["first"] = x
df["second"] = y
df["third"] = df["first"].apply(lambda x: palette[x])
plt.scatter(df["first"], df["second"], c=df["third"])
plt.legend()
plt.show()
將引數添加到散點線無濟于事（legend = c=df2["third"]）
我找不到解決方案。
如有指點，謝謝
uj5u.com熱心網友回復：
標準圖例中的每條線對應一個帶有標簽的圖。您可以一次繪制一種顏色的散點圖，并分配相應的標簽。
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
y = 2 * x
df = pd.DataFrame()
palette = {1: "blue", 2: "orange", 3: "green", 4: "red", 5: "purple", 6: "brown", 7: "pink", 8: "gray", 9: "olive", 10: "cyan"}
df["first"] = x
df["second"] = y
df["third"] = df["first"].apply(lambda x: palette[x])
for number, color in palette.items():
plt.scatter(x="first", y="second", c=color, data=df[df["first"] == number], label=number)
plt.legend()
plt.show()
這個程序可以簡化很多，在 Seaborn 中使用色調：
import matplotlib.pyplot as plt
import seaborn as sns
import pandas as pd
import numpy as np
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
y = 2 * x
df = pd.DataFrame()
palette = {1: "blue", 2: "orange", 3: "green", 4: "red", 5: "purple", 6: "brown", 7: "pink", 8: "gray", 9: "olive", 10: "cyan"}
df["first"] = x
df["second"] = y
df["third"] = df["first"].apply(lambda x: palette[x])
sns.set()
sns.scatterplot(data=df, x="first", y="second", hue="first", palette=palette)
plt.show()
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