您可以尝试按以下方式创建一个包含年份的新列：

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

使用relativedelta非常简单：

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)

更简单的方法是使用date_range函数并计算其长度

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

只需一个简单的功能，您就可以实现您的目标。
该函数通过简单的计算来计算年差和月差。

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

我认为这是最'Pandas'的方式来做到这一点，没有使用任何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不允许对datetime对象进行简单的添加，因此必须使用map而不是apply。

您可以尝试以下函数来计算差值-

def yearmonthdiff(row):
    s = row['start']
    e = row['end']
    y = s.year - e.year
    m = s.month - e.month
    d = s.day - e.day
    if m < 0:
        y = y - 1
        m = m + 12
    if m == 0:
        if d < 0:
            m = m -1
        elif d == 0:
            s1 = s.hour*3600 + s.minute*60 + s.second
            s2 = e.hour*3600 + e.minut*60 + e.second
            if s1 < s2:
                m = m - 1
    return '{}y{}m'.format(y,m)

其中row是 Dataframe row。我假设您的start和end列是datetime对象。然后您可以使用DataFrame.apply()函数将其应用到每一行。

df

Out[92]:
                       start                        end
0 2000-01-10 00:00:00.000000 1970-04-29 00:00:00.000000
1 2015-07-18 17:54:59.070381 2014-01-11 17:55:10.053381

df['diff'] = df.apply(yearmonthdiff, axis=1)

In [97]: df
Out[97]:
                       start                        end   diff
0 2000-01-10 00:00:00.000000 1970-04-29 00:00:00.000000  29y9m
1 2015-07-18 17:54:59.070381 2014-01-11 17:55:10.053381   1y6m

与@DeepSpace的回答类似，下面是一个类似SAS的实现：

import pandas as pd
from dateutil import relativedelta

def intck_month( start, end ):
    rd = relativedelta.relativedelta( pd.to_datetime( end ), pd.to_datetime( start ) )
    return rd.years, rd.months

用法：

>> years, months = intck_month('1960-01-01', '1970-03-01')
>> print(years)
10
>> print(months)
2

你所做的基本上是减去日期，然后得到天数，将天数转换成字符串，然后用““分割，从结果列表中，天数是列表中的第一项。将其转换为整数，然后除以365。

ad['yrs']=(ad.last_dt-ad.dt).apply(lambda x: str(x).split(' ')[0]).apply(lambda x: int(x)/365)

您可以找到总秒数并计算剩余时间：

diff = pd.to_datetime('2023-01-01') - pd.to_datetime('2021-01-01')

diff.total_seconds() / (365 * 24 * 60 * 60) # years
# 2.0

diff.total_seconds() / (30 * 24 * 60 * 60) # months
# 24.333333333333332

diff.total_seconds() / (24 * 60 * 60) # days
# 730.0

对于Pandas Series，请使用dt访问器：df['diff'].dt.total_seconds() .