Seaborn lmplot（python）中具有多个绘图的标签点

我正在尝试向lmplot中的每个数据点添加标签。我想用索引标记每个数据点。现在我的代码如下：

p1=sns.lmplot(x="target", y="source", col="color", hue="color", 
              data=ddf, col_wrap=2, ci=None, palette="muted",
              scatter_kws={"s": 50, "alpha": 1})

def label_point(x, y, val, ax):
    a = pd.concat({'x': x, 'y': y, 'val': val}, axis=1)
    for i, point in a.iterrows():
    ax.text(point['x']+.02, point['y'], str(point['val']))

label_point(ddf.target, ddf.source, ddf.chip, plt.gca())

这会将所有标签打印到最后一个打印上

我尝试了

label_point（ddf.target，ddf.source，ddf.chip，plt.gcf（））

而不是使用整个图形而不是当前轴，但它抛出了一个错误

ValueError: Image size of 163205x147206 pixels is too large. 
It must be less than 2^16 in each direction.

---

问题是，如果将整个数据集传递给标签函数，标签函数如何知道要标记哪个绘图

例如，您可以使用pandas的

.groupby

循环遍历唯一的颜色，并为每个颜色创建

seaborn.regplot

。然后很容易单独标记每个轴

import matplotlib.pyplot as plt
import numpy as np; np.random.seed(42)
import pandas as pd
import seaborn as sns

def label_point(df, ax):
    for i, point in df.iterrows():
        ax.annotate("{:.1f}".format(point['val']), xy = (point['x'], point['y']),
                    xytext=(2,-2), textcoords="offset points")

df = pd.DataFrame({"x": np.sort(np.random.rand(50)),
                   "y": np.cumsum(np.random.randn(50)),
                   "val" : np.random.randint(10,31, size=50),
                   "color" : np.random.randint(0,3,size=50 )})
colors = ["crimson", "indigo", "limegreen"]

fig, axes = plt.subplots(2,2, sharex=True, sharey=True)

for (c, grp), ax in zip(df.groupby("color"), axes.flat):
    sns.regplot(x="x", y="y", data=grp, color=colors[c], ax=ax,
                scatter_kws={"s": 25, "alpha": 1})
    label_point(grp, ax)

axes.flatten()[-1].remove()
plt.show()