how to set readable xticks in seaborn's facetg

问题:

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 plots all the xtick labels instead of just picking a few and it looks horrible:

Is there a way to customize it so that it plots every n-th tick on x-axis instead of all of them?

回答1:

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])

回答2:

You have to skip x labels manually like in this example:

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()