我想在A' A'中删除日期。来自B' B'并获得日期之间每个月的天数差异:
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
如何使用熊猫来实现这一目标?
答案 0 :(得分:2)
有趣的问题,谢谢分享。这里介绍的基本思想是构建一个函数,它可以在开始日期和结束日期之间进行迭代,并返回一个带有年/月密钥的dict,以及该月份的天数值。
<强>代码:强>
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)
<强>结果:强>
(2014, 1) (2014, 2) (2014, 3)
0 31 28 0
1 0 26 1
2 0 3 0