Seaborn regplot与colorbar？

Question

我正在用seaborn's密谋regplot.据我所知,它pyplot.scatter在幕后使用.所以我假设如果我将散点图的颜色指定为序列,那么我就可以调用plt.colorbar,但它似乎不起作用:

sns.regplot('mapped both', 'unique; repeated at least once', wt, ci=95, logx=True, truncate=True, line_kws={"linewidth": 1, "color": "seagreen"}, scatter_kws={'c':wt['Cis/Trans'], 'cmap':'summer', 's':75})
plt.colorbar()

Traceback (most recent call last):

  File "<ipython-input-174-f2d61aff7c73>", line 2, in <module>
    plt.colorbar()

  File "/usr/local/lib/python2.7/dist-packages/matplotlib/pyplot.py", line 2152, in colorbar
    raise RuntimeError('No mappable was found to use for colorbar '

RuntimeError: No mappable was found to use for colorbar creation. First define a mappable such as an image (with imshow) or a contour set (with contourf).

为什么它不起作用,有没有办法绕过它？

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如果有一种简单的方法来生成尺寸的图例,我会使用点的大小而不是颜色

Answer 1

另一种方法是

import seaborn as sns
import matplotlib.pyplot as plt

tips = sns.load_dataset("tips")

points = plt.scatter(tips["total_bill"], tips["tip"],
                     c=tips["size"], s=75, cmap="BuGn")
plt.colorbar(points)

sns.regplot("total_bill", "tip", data=tips, scatter=False, color=".1")

Answer 2

regplot的颜色参数将单一颜色应用于regplot元素(这在seaborn文档中).要控制散点图,您需要传递kwargs:

import pandas as pd
import seaborn as sns
import numpy.random as nr
import matplotlib.pyplot as plt

fig, ax = plt.subplots()

data = nr.random((9,3))
df = pd.DataFrame(data, columns=list('abc'))
out = sns.regplot('a','b',df, scatter=True,
                  ax=ax,
                  scatter_kws={'c':df['c'], 'cmap':'jet'})

然后你从AxesSubplot seaborn返回中获取可映射的东西(由scatter创建的集合),并指定你想要一个用于mappable的颜色条.请注意我的TODO评论,如果您计划通过对图表的其他更改来运行此评论.

outpathc = out.get_children()[3] 
#TODO -- don't assume PathCollection is 4th; at least check type

plt.colorbar(mappable=outpathc)

plt.show()