如何根据matplotlib中的colormap对散点中的点进行着色?

时间:2012-08-09 13:55:06

标签: python numpy matplotlib scipy

我正在尝试根据从已定义的颜色贴图(如蓝调或红色)中选取的一组值(从0到1)对散点图中的点进行着色。我试过这个:

import matplotlib
import matplotlib.pyplot as plt
from numpy import *
from scipy import *
fig = plt.figure()
mymap = plt.get_cmap("Reds")
x = [8.4808517662594909, 11.749082788323497, 5.9075039082855652, 3.6156231827873615, 12.536817102137768, 11.749082788323497, 5.9075039082855652, 3.6156231827873615, 12.536817102137768]
spaced_colors = linspace(0, 1, 10)
print spaced_colors
plt.scatter(x, x,
            color=spaced_colors,
            cmap=mymap)
# this does not work either
plt.scatter(x, x,
            color=spaced_colors,
            cmap=plt.get_cmap("gray"))

但它不起作用,使用红色或灰色地图。怎么办呢?

编辑:如果我想分别绘制每个点,以便它可以有一个单独的图例,我该怎么办?我试过了:

fig = plt.figure()
mymap = plt.get_cmap("Reds")
data = np.random.random([10, 2])
colors = list(linspace(0.1, 1, 5)) + list(linspace(0.1, 1, 5))
print "colors: ", colors
plt.subplot(1, 2, 1)
plt.scatter(data[:, 0], data[:, 1],
           c=colors,
           cmap=mymap)
plt.subplot(1, 2, 2)
# attempt to plot first five points in five shades of red,
# with a separate legend for each point
for n in range(5):
    plt.scatter([data[n, 0]], [data[n, 1]],
               c=[colors[n]],
               cmap=mymap,
               label="point %d" %(n))
plt.legend()

但它失败了。我需要调用每个点的散射,以便它可以有一个单独的label =,但仍然希望每个点的颜色映射的颜色与颜色不同。 感谢。

3 个答案:

答案 0 :(得分:20)

如果你真的想这样做(你在编辑中描述的内容),你必须“拉”你的色彩图中的颜色(我已经评论了我对你的代码所做的所有更改):

import numpy as np
import matplotlib.pyplot as plt 

# plt.subplots instead of plt.subplot
# create a figure and two subplots side by side, they share the
# x and the y-axis
fig, axes = plt.subplots(ncols=2, sharey=True, sharex=True)
data = np.random.random([10, 2]) 
# np.r_ instead of lists
colors = np.r_[np.linspace(0.1, 1, 5), np.linspace(0.1, 1, 5)] 
mymap = plt.get_cmap("Reds")
# get the colors from the color map
my_colors = mymap(colors)
# here you give floats as color to scatter and a color map
# scatter "translates" this
axes[0].scatter(data[:, 0], data[:, 1], s=40,
                c=colors, edgecolors='None',
                cmap=mymap)
for n in range(5):
    # here you give a color to scatter
    axes[1].scatter(data[n, 0], data[n, 1], s=40,
                    color=my_colors[n], edgecolors='None',
                    label="point %d" %(n))
# by default legend would show multiple scatterpoints (as you would normally
# plot multiple points with scatter)
# I reduce the number to one here
plt.legend(scatterpoints=1)
plt.tight_layout()
plt.show()

colored-scatter

但是,如果您只想绘制10个值并想要为每个值命名, 你应该考虑使用不同的东西,比如这里的条形图 example。另一个机会是将plt.plot与自定义颜色循环一起使用,就像在此example中一样。

答案 1 :(得分:13)

As per the documentation,您需要c关键字参数而不是color。 (我同意这有点令人困惑,但在这种情况下,“c”和“s”术语继承自matlab。)

E.g。

import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl

x, y, colors = np.random.random((3,10))

fig, ax = plt.subplots()
ax.scatter(x, y, c=colors, s=50, cmap=mpl.cm.Reds)

plt.show()

enter image description here

答案 2 :(得分:7)

怎么样:

import matplotlib.pyplot as plt
import numpy as np


reds = plt.get_cmap("Reds")
x = np.linspace(0, 10, 10)
y = np.log(x)

# color by value given a cmap
plt.subplot(121)
plt.scatter(x, y, c=x, s=100, cmap=reds)


# color by value, and add a legend for each
plt.subplot(122)
norm = plt.normalize()
norm.autoscale(x)
for i, (x_val, y_val) in enumerate(zip(x, y)):
    plt.plot(x_val, y_val, 'o', markersize=10, 
             color=reds(norm(x_val)),
             label='Point %s' % i
             )
plt.legend(numpoints=1, loc='lower right')

plt.show()

code output

代码应该都是相当自我解释的,但是如果你想让我重复一遍,那就大喊吧。