我想在 Seaborn FacetGrid 上的 pandas 数据框中的列中绘制误差线
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar']*2,
'B' : ['one', 'one', 'two', 'three',
'two', 'two', 'one', 'three'],
'C' : np.random.randn(8),
'D' : np.random.randn(8)})
df
示例数据框
A B C D
0 foo one 0.445827 -0.311863
1 bar one 0.862154 -0.229065
2 foo two 0.290981 -0.835301
3 bar three 0.995732 0.356807
4 foo two 0.029311 0.631812
5 bar two 0.023164 -0.468248
6 foo one -1.568248 2.508461
7 bar three -0.407807 0.319404
此代码适用于固定大小的误差线:
g = sns.FacetGrid(df, col="A", hue="B", size =5)
g.map(plt.errorbar, "C", "D",yerr=0.5, fmt='o');
但我无法使用数据框中的值使其工作
df['E'] = abs(df['D']*0.5)
g = sns.FacetGrid(df, col="A", hue="B", size =5)
g.map(plt.errorbar, "C", "D", yerr=df['E']);
或者
g = sns.FacetGrid(df, col="A", hue="B", size =5)
g.map(plt.errorbar, "C", "D", yerr='E');
两者都会产生大量错误
编辑:
经过大量 matplotlib 文档阅读和各种 stackoverflow 答案后,这是一个纯 matplotlib 解决方案
#define a color palette index based on column 'B'
df['cind'] = pd.Categorical(df['B']).labels
#how many categories in column 'A'
cats = df['A'].unique()
cats.sort()
#get the seaborn colour palette and convert to array
cp = sns.color_palette()
cpa = np.array(cp)
#draw a subplot for each category in column "A"
fig, axs = plt.subplots(nrows=1, ncols=len(cats), sharey=True)
for i,ax in enumerate(axs):
df_sub = df[df['A'] == cats[i]]
col = cpa[df_sub['cind']]
ax.scatter(df_sub['C'], df_sub['D'], c=col)
eb = ax.errorbar(df_sub['C'], df_sub['D'], yerr=df_sub['E'], fmt=None)
a, (b, c), (d,) = eb.lines
d.set_color(col)
除了标签,还有轴限制它的OK。它为“A”列中的每个类别绘制了一个单独的子图,由“B”列中的类别着色。(注意随机数据与上述不同)
如果有人有任何想法,我仍然想要一个 pandas/seaborn 解决方案?