pandas - 每个点的不同颜色图例的散点图

Question

从以下示例开始：

fig, ax = plt.subplots()

df = pd.DataFrame({'n1':[1,2,1,3], 'n2':[1,3,2,1], 'l':['a','b','c','d']})

for label in df['l']:

    df.plot('n1','n2', kind='scatter', ax=ax, s=50, linewidth=0.1, label=label)

我获得的是以下散点图：

我现在正尝试为这四个点中的每一个设置不同的颜色。我知道我可以在列表中循环一组，例如4种颜色，如：

colorlist = ['b','r','c','y']

但由于我的真实数据集包含至少20个不同的点，我正在寻找一种“颜色生成器”来循环它。

Answer 1

以下方法将创建一个颜色列表，只要您的数据帧，然后使用每种颜色的标签绘制一个点：

import matplotlib.pyplot as plt
import matplotlib.cm as cm
import matplotlib.colors as colors
import numpy as np
import pandas as pd

fig, ax = plt.subplots()

df = pd.DataFrame({'n1':[1,2,1,3], 'n2':[1,3,2,1], 'l':['a','b','c','d']})

colormap = cm.viridis
colorlist = [colors.rgb2hex(colormap(i)) for i in np.linspace(0, 0.9, len(df['l']))]

for i,c in enumerate(colorlist):

    x = df['n1'][i]
    y = df['n2'][i]
    l = df['l'][i]

    ax.scatter(x, y, label=l, s=50, linewidth=0.1, c=c)

ax.legend()

plt.show()

Answer 2

这个怎么样，

这是源代码，

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
from matplotlib import cm

fig, ax = plt.subplots()

df = pd.DataFrame({'n1':[1,2,1,3], 'n2':[1,3,2,1], 'l':['a','b','c','d']})

#colors = ['b','r','c','y']
nrof_labels = len(df['l'])
colors = cm.rainbow(np.linspace(0, 1, nrof_labels))     # create a bunch of colors

for i, r in df.iterrows():
    ax.plot(r['n1'], r['n2'], 'o', markersize=10, color=colors[i], linewidth=0.1, label=r['l'])

ax.set_xlim(0.5, 3.5)
ax.set_ylim(0.5, 3.5)
plt.legend(loc='best')

plt.show()

Answer 3

IIUC你可以这样做：

import matplotlib.pyplot as plt
from matplotlib import colors
import pandas as pd

colorlist = list(colors.ColorConverter.colors.keys())
fig, ax = plt.subplots()
[df.iloc[[i]].plot.scatter('n1', 'n2', ax=ax, s=50, label=l,
                         color=colorlist[i % len(colorlist)])
 for i,l in enumerate(df.l)]

colorlist：

In [223]: colorlist
Out[223]: ['m', 'b', 'g', 'r', 'k', 'y', 'c', 'w']

PS colorlist[i % len(colorlist)] - 应始终保留在列表边界

中