Matplotlib散点图：按Priority指定对应颜色的实现方法

问题背景 #

你需要基于包含多变量的DataFrame绘制散点图，具体要求：

• Y轴为Price，X轴为Weight
• 根据Priority（取值1-10）为散点指定对应颜色，预设颜色列表为：

colors=["navy","lawngreen","red","green","purple","steelblue","orange","darkred","yellow","chocolate"]

其中Priority=1对应navy，Priority=2对应lawngreen，依此类推，但尝试用字典关联颜色时未成功。

解决方案 #

方法一：构建Priority到颜色的映射字典 #

通过字典明确绑定每个Priority值和对应颜色，逻辑清晰易维护：

import pandas as pd
import matplotlib.pyplot as plt

# 预设颜色列表
colors = ["navy","lawngreen","red","green","purple","steelblue","orange","darkred","yellow","chocolate"]

# 生成映射字典：Priority值（1-10）对应颜色列表的对应项
color_mapping = {priority: colors[priority-1] for priority in range(1, 11)}

# 为DataFrame添加颜色列
df['scatter_color'] = df['Priority'].map(color_mapping)

# 绘制散点图
plt.scatter(x=df['Weight'], y=df['Price'], color=df['scatter_color'])
plt.xlabel('Weight')
plt.ylabel('Price')
plt.show()

方法二：直接通过索引匹配颜色（更简洁） #

利用Priority值减1得到颜色列表的索引，直接提取对应颜色，省去字典构建步骤：

import pandas as pd
import matplotlib.pyplot as plt

colors = ["navy","lawngreen","red","green","purple","steelblue","orange","darkred","yellow","chocolate"]

# 直接通过列表索引获取对应颜色
plt.scatter(
    x=df['Weight'], 
    y=df['Price'], 
    color=[colors[priority-1] for priority in df['Priority']]
)
plt.xlabel('Weight')
plt.ylabel('Price')
plt.show()

常见问题排查 #

如果之前用字典绑定失败，大概率是键值对不匹配：比如误将字典的键设为0-9（对应颜色列表索引），但你的Priority取值是1-10，导致无法匹配。只要确保字典的键是1-10，对应颜色列表的0-9索引位置即可解决。

内容的提问来源于stack exchange，提问作者nach