matplotlib:按字典着色，无需标准化 [英] matplotlib: color by dict without normalization

问题描述

我的目标是使用通过字典将给定数字映射到给定颜色的颜色图.

My goal is is to use a colormap that maps via a dict, a given number to a given color.

然而，ma​​tplotlib 似乎已经规范化了这个数字.

However, matplotlib seems to have normalized the number.

例如，我首先使用seaborn创建了一个自定义颜色图，并将其输入到plt.scatter

For example, I first created a custom colormap use seaborn, and feed it into plt.scatter

import seaborn as sns

colors = ['pumpkin', "bright sky blue", 'light green', 'salmon', 'grey', 'pale grey']
pal = sns.xkcd_palette(colors)
sns.palplot(pal)

from matplotlib import pyplot as plt
from matplotlib.colors import ListedColormap

cmap = ListedColormap(pal.as_hex())
x = [0, 1, 2]
y = [0, 1, 2]
plt.scatter(x, y, c=[0, 1, 2], s=500, cmap=cmap)  # I'd like to get color ['pumpkin', "bright sky blue", 'light green']

但是，它给了我颜色 ['pumpkin', 'salmon', 'pale grey']

简而言之:颜色图:

获取颜色 0、1 和 2(期望):

getting color 0, 1, and 2 (desired):

但 matplotlib 给出:

推荐答案

颜色图始终在0到1之间进行归一化.散布图默认情况下会归一化为 c 参数提供的值，例如颜色图的范围从最小值到最大值.但是，您当然可以定义自己的规范化.在这种情况下，它应该是 vmin = 0，vmax = len(colors).

A colormap is always normalized between 0 and 1. A scatter plot will by default normalize the values given to the c argument, such that the colormap ranges from the minimum to the maximum value. However, you may of course define your own normalization. In this case it would be vmin=0, vmax=len(colors).

from matplotlib import pyplot as plt
from matplotlib.colors import ListedColormap

colors = ['xkcd:pumpkin', "xkcd:bright sky blue", 'xkcd:light green', 
          'salmon', 'grey', 'xkcd:pale grey']
cmap = ListedColormap(colors)

x = range(3)
y = range(3)
plt.scatter(x, y, c=range(3), s=500, cmap=cmap, vmin=0, vmax=len(colors))

plt.show()

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