控制pyplot中的x刻度数

Question

我想显示所有13个刻度，但图表只显示其中7个有两个间隔。

plt.locator_params(axis='x',nbins=13)

为什么代码上面没有工作？

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import matplotlib.dates as dates

y = [0, 0.86, 0.826, 0.816, 0.807, 0.803, 0.804, 0.803, 0.802,0.81, 0.813, 0.813, 0.813]

times = pd.date_range('2015-02-25', periods=13)
fig, ax = plt.subplots(1)
fig.autofmt_xdate()

xfmt = dates.DateFormatter('%d-%m-%y')
ax.xaxis.set_major_formatter(xfmt)

plt.locator_params(axis='x',nbins=13)

ax.plot_date(times.to_pydatetime(), y, 'v-')
ax.xaxis.set_minor_locator(dates.WeekdayLocator(byweekday=(1),
                                            interval=1))
ax.xaxis.set_minor_formatter(dates.DateFormatter('%d\n%a'))
ax.xaxis.grid(True, which="minor")
ax.yaxis.grid()
plt.tight_layout()
plt.show()

Answer 1

警告应该会给你一些线索，说明为什么会发生这种情况：

UserWarning: 'set_params()' not defined for locator of type <class 'pandas.tseries.converter.PandasAutoDateLocator'>
  str(type(self)))

改为使用plt.xticks(times.to_pydatetime())：

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import matplotlib.dates as dates

y = [0, 0.86, 0.826, 0.816, 0.807, 0.803, 0.804, 0.803, 0.802,0.81, 0.813, 0.813, 0.813]

times = pd.date_range('2015-02-25', periods=13)
fig, ax = plt.subplots(1)
fig.autofmt_xdate()

xfmt = dates.DateFormatter('%d-%m-%y')
ax.xaxis.set_major_formatter(xfmt)


ax.plot_date(times.to_pydatetime(), y, 'v-')
ax.xaxis.set_minor_locator(dates.WeekdayLocator(byweekday=(1),
                                            interval=1))
plt.xticks(times.to_pydatetime())
ax.xaxis.set_minor_formatter(dates.DateFormatter('%d\n%a'))
ax.xaxis.grid(True, which="minor")
ax.yaxis.grid()
plt.tight_layout()
plt.show()