如何在 seaborn 的 facetgrid 中设置可读的 xticks？

Question

我有这个带有 seaborn 的 facetgrid 的数据框图：

import seaborn as sns
import matplotlib.pylab as plt
import pandas
import numpy as np

plt.figure()
df = pandas.DataFrame({"a": map(str, np.arange(1001, 1001 + 30)),
                       "l": ["A"] * 15 + ["B"] * 15,
                       "v": np.random.rand(30)})
g = sns.FacetGrid(row="l", data=df)
g.map(sns.pointplot, "a", "v")
plt.show()

seaborn 绘制了所有 xtick 标签，而不是只挑选了几个标签，它看起来很糟糕：

有没有办法对其进行自定义，使其在 x 轴上绘制每个第 n 个刻度而不是所有刻度？

Answer 1

您必须像本例中那样手动跳过 x 个标签：

import seaborn as sns
import matplotlib.pylab as plt
import pandas
import numpy as np

df = pandas.DataFrame({"a": range(1001, 1031),
                       "l": ["A",] * 15 + ["B",] * 15,
                       "v": np.random.rand(30)})
g = sns.FacetGrid(row="l", data=df)
g.map(sns.pointplot, "a", "v")

# iterate over axes of FacetGrid
for ax in g.axes.flat:
    labels = ax.get_xticklabels() # get x labels
    for i,l in enumerate(labels):
        if(i%2 == 0): labels[i] = '' # skip even labels
    ax.set_xticklabels(labels, rotation=30) # set new labels
plt.show()

Answer 2

seaborn.pointplot 不是该图的正确工具。但答案很简单：使用基本的 matplotlib.pyplot.plot 函数：

import seaborn as sns
import matplotlib.pylab as plt
import pandas
import numpy as np

df = pandas.DataFrame({"a": np.arange(1001, 1001 + 30),
                       "l": ["A"] * 15 + ["B"] * 15,
                       "v": np.random.rand(30)})
g = sns.FacetGrid(row="l", data=df)
g.map(plt.plot, "a", "v", marker="o")
g.set(xticks=df.a[2::8])

如何在 seaborn 的 facetgrid 中设置可读的 xticks？

how to set readable xticks in seaborn's facetgrid?

python

matplotlib

pandas

seaborn