使用matplotlib和python绘制datetime.timedelta

Question

我正在执行一项任务，我需要计算每天花费的时间，然后使用条形图表示该时间。因此，对于此任务，我使用python并能够获取每天花费的时间，并将其存储在列表中“
time_list”，现在我不明白如何使用matplotlib函数来绘制它。问题在于，此列表包含datetime.timedelta类值。例：

time_list
[datetime.timedelta(0, 23820), datetime.timedelta(0, 27480), datetime.timedelta(0, 28500), datetime.timedelta(0, 24180), datetime.timedelta(0, 27540), datetime.timedelta(0, 28920), datetime.timedelta(0, 28800), datetime.timedelta(0, 29100), datetime.timedelta(0, 29100), datetime.timedelta(0, 24480), datetime.timedelta(0, 27000)]

这些值的含义如下：

Total Time Spent on  2  is  6:37:00
Total Time Spent on  3  is  7:38:00
Total Time Spent on  4  is  7:55:00
Total Time Spent on  5  is  6:43:00
Total Time Spent on  8  is  7:39:00
Total Time Spent on  9  is  8:02:00
Total Time Spent on  10  is  8:00:00
Total Time Spent on  11  is  8:05:00
Total Time Spent on  12  is  8:05:00
Total Time Spent on  15  is  6:48:00
Total Time Spent on  16  is  7:30:00

有人可以帮我画图吗？提前致谢

Answer 1

尽管matplotlib原则上可以处理日期时间对象，但条形图无法直接解释它们。因此，您可以在timedelta上添加任意日期，然后使用转换为数字matplotlib.dates.date2num()。然后使用DateFormatter启用漂亮的ticklabels。

import numpy as np
import datetime
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

days = [2, 3, 4, 5, 8, 9, 10, 11, 12, 15, 16]

time_list = [datetime.timedelta(0, 23820), datetime.timedelta(0, 27480), 
             datetime.timedelta(0, 28500), datetime.timedelta(0, 24180), 
             datetime.timedelta(0, 27540), datetime.timedelta(0, 28920), 
             datetime.timedelta(0, 28800), datetime.timedelta(0, 29100), 
             datetime.timedelta(0, 29100), datetime.timedelta(0, 24480), 
             datetime.timedelta(0, 27000)]

# specify a date to use for the times
zero = datetime.datetime(2018,1,1)
time = [zero + t for t in time_list]
# convert datetimes to numbers
zero = mdates.date2num(zero)
time = [t-zero for t in mdates.date2num(time)]

f = plt.figure()
ax = f.add_subplot(1,1,1)

ax.bar(days, time, bottom=zero)
ax.yaxis_date()
ax.yaxis.set_major_formatter(mdates.DateFormatter("%H:%M"))

# add 10% margin on top (since ax.margins seems to not work here)
ylim = ax.get_ylim()
ax.set_ylim(None, ylim[1]+0.1*np.diff(ylim))

plt.show()