pandas.plot() x 轴刻度频率——如何显示更多刻度？

Question

I am plotting time series using pandas.plot() and want to see every month shown as an x-tick.

Here is the dataset structure

Here is the result of the.plot()

I was trying to use examples from other posts and matplotlib documentation and do something like

ax.xaxis.set_major_locator(
   dates.MonthLocator(revenue_pivot.index, bymonthday=1,interval=1))

But that removed all the ticks:(

I also tried to pass xticks = df.index , but it has not changed anything.

What would be the rigth way to show more ticks on x-axis?

Answer 1

No need to pass any args to MonthLocator . Make sure to use x_compat in the df.plot() call per @Rotkiv's answer.

import pandas as pd
import numpy as np
import matplotlib.pylab as plt
import matplotlib.dates as mdates

df = pd.DataFrame(np.random.rand(100,2), index=pd.date_range('1-1-2018', periods=100))
ax = df.plot(x_compat=True)
ax.xaxis.set_major_locator(mdates.MonthLocator())
plt.show()

Answer 2

You could also format the x-axis ticks and labels of a pandas DateTimeIndex "manually" using the attributes of a pandas Timestamp object.

I found that much easier than using locators from matplotlib.dates which work on other datetime formats than pandas (if I am not mistaken) and thus sometimes show an odd behaviour if dates are not converted accordingly.

Here's a generic example that shows the first day of each month as a label based on attributes of pandas Timestamp objects:

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


# data
dim = 8760
idx = pd.date_range('1/1/2000 00:00:00', freq='h', periods=dim)
df = pd.DataFrame(np.random.randn(dim, 2), index=idx)

# select tick positions based on timestamp attribute logic. see:
# https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.html
positions = [p for p in df.index
             if p.hour == 0
             and p.is_month_start
             and p.month in range(1, 13, 1)]
# for date formatting, see:
# https://docs.python.org/2/library/datetime.html#strftime-and-strptime-behavior
labels = [l.strftime('%m-%d') for l in positions]

# plot with adjusted labels
ax = df.plot(kind='line', grid=True)
ax.set_xlabel('Time (h)')
ax.set_ylabel('Foo (Bar)')
ax.set_xticks(positions)
ax.set_xticklabels(labels)

plt.show()

yields:

Hope this helps!

Answer 3

这里描述的正确方法使用x_compat参数，可以抑制自动刻度分辨率调整

df.A.plot(x_compat=True)

Answer 4

If you want to just show more ticks, you can also dive deep into the structure of pd.plotting._converter :

dai = ax.xaxis.minor.formatter.plot_obj.date_axis_info
dai['fmt'][dai['fmt'] == b''] = b'%b'

After plotting, the formatter is a TimeSeries_DateFormatter and _set_default_format has been called, so self.plot_obj.date_axis_info is not None . You can now manipulate the structured array .date_axis_info to be to your liking, namely contain less b'' and more b'%b'

Answer 5

Remove tick labels:

ax = df.plot(x='date', y=['count'])
every_nth = 10
for n, label in enumerate(ax.xaxis.get_ticklabels()):
  if n % every_nth != 0:
    label.set_visible(False)

Lower every_nth to include more labels, raise to keep fewer.