使用数据框中的列为海洋图的背景着色 [英] Colorize the background of a seaborn plot using a column in dataframe

问题描述

问题

如何为

所需的输出

我想拥有的东西新列背景 和任何

解决方案

ax.axvspan（）可以工作对您来说，假设背景不会在时间点上重叠。

 将numpy导入为np 
进口seaborn为sns； sns.set（）
导入matplotlib.pyplot as plt 
 fmri = sns.load_dataset（ fmri）
 fmri.sort_values（'timepoint'，inplace = True）
 arr = np.ones（len（fmri））
 arr [：300] = 0 
 arr [600：] = 2 
 fmri ['background'] = arr 
 fmri [ 'background'] = fmri ['background']。astype（int）.astype（str）.map（lambda x：'C'+ x）
 
 ax = sns.lineplot（x = timepoint，y = signal，hue = event，data = fmri）
范围= fmri.groupby（'background'）['timepoint']。agg（['min'，'max'] ）
 for i，range.iterrows（）中的行：
 ax.axvspan（xmin = row ['min']，xmax = row ['max']，facecolor = i，alpha = 0.3） 
 
  

Question

How to shade or colorize the background of a seaborn plot using a column of a dataframe?

Code snippet

import numpy as np
import seaborn as sns; sns.set()
import matplotlib.pyplot as plt
fmri = sns.load_dataset("fmri")
fmri.sort_values('timepoint',inplace=True)
ax = sns.lineplot(x="timepoint", y="signal", data=fmri)
arr = np.ones(len(fmri))
arr[:300] = 0
arr[600:] = 2
fmri['background'] = arr

ax = sns.lineplot(x="timepoint", y="signal", hue="event", data=fmri)

Which produced this graph:

Desired output

What I'd like to have, according to the value in the new column 'background' and any palette or user defined colors, something like this:

解决方案

ax.axvspan() could work for you, assuming backgrounds don't overlap over timepoints.

import numpy as np
import seaborn as sns; sns.set()
import matplotlib.pyplot as plt
fmri = sns.load_dataset("fmri")
fmri.sort_values('timepoint',inplace=True)
arr = np.ones(len(fmri))
arr[:300] = 0
arr[600:] = 2
fmri['background'] = arr
fmri['background'] = fmri['background'].astype(int).astype(str).map(lambda x: 'C'+x)

ax = sns.lineplot(x="timepoint", y="signal", hue="event", data=fmri)
ranges = fmri.groupby('background')['timepoint'].agg(['min', 'max'])
for i, row in ranges.iterrows():
    ax.axvspan(xmin=row['min'], xmax=row['max'], facecolor=i, alpha=0.3)

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