Question

我有一个存储为Pandas Period对象的日期：

date1 = Period(1989, 'Y')

我想将它与日期时间对象进行比较，该对象是：

date2 = datetime.strftime('1988-09-09', '%Y-%m-%d')

我希望可以进行如下减法运算：

diff = date2 - date1

为此，我需要将Period转换为datetime对象。有办法吗？

Answer 1

使用pandas.Period.to_timestamp
- 使用freq参数指定开始或结束
date2 = datetime.strftime('1988-09-09', '%Y-%m-%d')导致TypeError
- 使用date2 = pd.to_datetime('1988-09-09')获取日期时间格式

import pandas as pd

date1 = pd.Period(1989, 'Y')

date1_ts = date1.to_timestamp()

print(date1_ts)
[out]:
Timestamp('1989-01-01 00:00:00')

# create date2 as a datetime, not a string
date2 = pd.to_datetime('1988-09-09')

# take the difference
diff = date2 - date1_ts

print(diff)
[out]: 
Timedelta('-114 days +00:00:00')

Answer 2

也许您可以做到：

from datetime import datetime
from pandas import DataFrame, Period

if __name__ == '__main__':
    data = {'Year': [Period(1989, 'Y')]}

    df = DataFrame(data,
                   columns=['Year'])

    date2 = datetime.now().strftime('%Y-%m-%d').split('-')[0]
    date1 = df['Year'].values[0].year
    diff = int(date2) - date1
    print(diff)

date2 = datetime.strftime('1988-09-09', '%Y-%m-%d')输入了错误，因此我替换为

date2 = datetime.now().strftime('%Y-%m-%d').split('-')[0]

输出：

如何将熊猫期间转换为时间戳或日期时间对象？

2 个答案: