python - 如何计算 Pandas 中每月分成几天的两个日期之间的天数

Question

我想从“B”中的日期中减去“A”中的日期，并获得日期之间每个月的天数差:

df
      A        B
 2014-01-01  2014-02-28 
 2014-02-03  2014-03-01

df['A'] = pd.to_datetime(df['A'])
df['B'] = pd.to_datetime(df['B'])
#df['A'] - df['B']

Desired Output:
=================
01(Jan)    02(Feb)      03(Mar)
================================
31days     28days       0days
0days      26days       1day     

如何使用 pandas 实现这一点？

Answer 1

有趣的问题，感谢分享。这里介绍的基本思想是构建一个可以在开始日期和结束日期之间迭代的函数，并返回一个包含年/月键和该月天数值的字典。

代码:

import calendar
import datetime as dt

def year_month(date):
    """ return year/month tuple from date """
    return date.year, date.month

def next_year_month(date):
    """ given a year/month tuple, return the next year/month """
    if date[1] == 12:
        return date[0] + 1, 1
    else:
        return date[0], date[1] + 1

def days_per_month(start_date, end_date):
    """ return dict keyed with year/month tuples and valued with days in month """
    assert isinstance(start_date, (dt.datetime, dt.date))
    assert isinstance(end_date, (dt.datetime, dt.date))

    start = year_month(start_date)
    end = year_month(end_date)
    days_in_month = (
        calendar.monthrange(*start)[1] - start_date.day + 1)

    result = {}
    while start != end:
        result[start] = days_in_month
        start = next_year_month(start)
        days_in_month = calendar.monthrange(*start)[1]
    result[end] = (
        end_date.day - calendar.monthrange(*end)[1] + days_in_month)
    return result

测试代码:

import pandas as pd
data = [x.strip().split() for x in """
        A          B
    2014-01-01  2014-02-28
    2014-02-03  2014-03-01
    2014-02-03  2014-02-05
""".split('\n')[1:-1]]
df = pd.DataFrame(data=data[1:], columns=data[0])
df['A'] = pd.to_datetime(df['A'])
df['B'] = pd.to_datetime(df['B'])

result = pd.DataFrame.from_records(
    (days_per_month(a, b) for a, b in zip(df['A'], df['B']))
).fillna(0).astype(int)

print(result)

结果:

   (2014, 1)  (2014, 2)  (2014, 3)
0         31         28          0
1          0         26          1
2          0          3          0