matplotlib：具有饱和度的单色彩色图

Question

如何创建线性色图，其中单个自定义RGB颜色的饱和度从0变为1？

我需要一张类似'蓝调'或'绿色'的地图（请参阅此处：http://www.scipy.org/Cookbook/Matplotlib/Show_colormaps），但需要自定义颜色。

我认为使用LinearSegmentedColormap可以实现，但我不明白我需要如何设置参数。

Answer 1

请参阅以下示例：

from matplotlib.colors import LinearSegmentedColormap
import matplotlib.pyplot as plt
import numpy as np

def CustomCmap(from_rgb,to_rgb):

    # from color r,g,b
    r1,g1,b1 = from_rgb

    # to color r,g,b
    r2,g2,b2 = to_rgb

    cdict = {'red': ((0, r1, r1),
                   (1, r2, r2)),
           'green': ((0, g1, g1),
                    (1, g2, g2)),
           'blue': ((0, b1, b1),
                   (1, b2, b2))}

    cmap = LinearSegmentedColormap('custom_cmap', cdict)
    return cmap


fig, ax = plt.subplots(2,2, figsize=(6,6), subplot_kw={'xticks': [],'yticks': []})
fig.subplots_adjust(hspace=.1,wspace=.1)

ax = ax.ravel()

cmap1 = CustomCmap([0.00, 0.00, 0.00], [0.02, 0.75, 1.00]) # from black to +/- 5,192,255
cmap2 = CustomCmap([1.00, 1.00, 1.00], [0.02, 0.75, 1.00]) # from white to +/- 5,192,255
cmap3 = CustomCmap([1.00, 0.42, 0.04], [0.02, 0.75, 1.00]) # from +/- 255,108,10 to +/- 5,192,255
cmap4 = CustomCmap([1.00, 0.42, 0.04], [0.50, 0.50, 0.50]) # from +/- 255,108,10 to grey (128)


ax[0].imshow(np.random.rand(30,30), interpolation='none', cmap=cmap1)
ax[1].imshow(np.random.rand(30,30), interpolation='none', cmap=cmap2)
ax[2].imshow(np.random.rand(30,30), interpolation='none', cmap=cmap3)
ax[3].imshow(np.random.rand(30,30), interpolation='none', cmap=cmap4)