本文介绍了如何在Seaborn的facetgrid中设置可读的xticks?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个带有seaborn的facetgrid的数据框图:
i have this plot of a dataframe with seaborn's facetgrid:
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,而不是仅仅选择了一些标签,这看起来很可怕:
seaborn plots all the xtick labels instead of just picking a few and it looks horrible:
是否可以自定义它,以便在x轴上绘制每个第n个刻度,而不是全部绘制?
Is there a way to customize it so that it plots every n-th tick on x-axis instead of all of them?
推荐答案
seaborn.pointplot
不是此绘图的正确工具.但是答案很简单:使用基本的matplotlib.pyplot.plot
函数:
The seaborn.pointplot
is not the right tool for this plot. But the answer is very simple: use the basic matplotlib.pyplot.plot
function:
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])
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