在新的pandas数据帧列中计算年,月等的日期时间差异

Question

我有一个像这样的pandas数据框:

Name    start        end
A       2000-01-10   1970-04-29

我想添加一个新列,提供年份,月份,天数start与end列之间的差异.

所以结果应该是这样的:

Name    start        end          diff
A       2000-01-10   1970-04-29   29y9m etc.

diff列也可以是一个datetime对象或一个timedelta对象,但对我而言,关键在于,我可以轻松地从中获取年份和月份.

我到现在为止尝试的是:

df['diff'] = df['end'] - df['start']

这导致新列包含10848 days.但是,我不知道如何将天数转换为29y9m等.

Answer 1

您可以尝试以这种方式创建一个带有年份的新列：

df['diff_year'] = df['diff'] / np.timedelta64(1, 'Y')

Answer 2

非常直截了当relativedelta:

from dateutil import relativedelta

>>          end      start
>> 0 1970-04-29 2000-01-10

for i in df.index:
    df.at[i, 'diff'] = relativedelta.relativedelta(df.ix[i, 'start'], df.ix[i, 'end'])

>>          end      start                                           diff
>> 0 1970-04-29 2000-01-10  relativedelta(years=+29, months=+8, days=+12)

Answer 3

我认为这是最"大熊猫"的方式,不使用任何for循环或定义外部函数:

>>> df = pd.DataFrame({'Name': ['A'], 'start': [datetime(2000, 1, 10)], 'end': [datetime(1970, 4, 29)]})
>>> df['diff'] = map(lambda td: datetime(1, 1, 1) + td, list(df['start'] - df['end']))
>>> df['diff'] = df['diff'].apply(lambda d: '{0}y{1}m'.format(d.year - 1, d.month - 1))
>>> df
  Name        end      start   diff
0    A 1970-04-29 2000-01-10  29y8m

由于pandas的timedelda64,它不允许使用map而不是apply,因为它不允许对datetime对象进行简单的添加.

Answer 4

通过简单的功能,您可以实现目标.

该函数通过简单的计算计算年份差异和月份差异.

import pandas as pd
import datetime

def parse_date(td):
    resYear = float(td.days)/364.0                   # get the number of years including the the numbers after the dot
    resMonth = int((resYear - int(resYear))*364/30)  # get the number of months, by multiply the number after the dot by 364 and divide by 30.
    resYear = int(resYear)
    return str(resYear) + "Y" + str(resMonth) + "m"

df = pd.DataFrame([("2000-01-10", "1970-04-29")], columns=["start", "end"])
df["delta"] = [parse_date(datetime.datetime.strptime(start, '%Y-%m-%d') - datetime.datetime.strptime(end, '%Y-%m-%d')) for start, end in zip(df["start"], df["end"])]
print df

        start         end  delta
0  2000-01-10  1970-04-29  29Y9m

Answer 5

一种简单得多的方法是使用date_range函数并计算相同的长度

startdt=pd.to_datetime('2017-01-01')
enddt = pd.to_datetime('2018-01-01')
len(pd.date_range(start=startdt,end=enddt,freq='M'))