在新的 pandas 数据框列中计算以年，月等为单位的日期时间差异 [英] calculate datetime-difference in years, months, etc. in a new pandas dataframe column

问题描述

我有一个如下所示的熊猫数据框:

I have a pandas dataframe looking like this:

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

我想添加一个新列，以提供start和end列之间的年，月，日之间的差异.

I want to add a new column providing the difference between the start and end column in years, months, days.

所以结果应该像这样:

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

diff列也可以是datetime对象或timedelta对象，但是对我来说，关键点是，我可以轻松获得 Year 和 Month .

the diff column may also be a datetime object or a timedelta object, but the key point for me is, that I can easily get the Year and Month out of it.

到目前为止，我尝试过的是:

What I tried until now is:

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

这将导致新列包含10848 days.但是，我不知道如何将日期转换为 29y9m等.

This results in the new column containing 10848 days. However, I do not know how to convert the days to 29y9m etc.

推荐答案

使用简单的功能，您就可以实现自己的目标.

With a simple function you can reach your goal.

该功能可以通过简单的计算来计算年差和月差.

The function calculates the years difference and the months difference with a simple calculation.

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

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