MrF*_*pes 36
您可以直接使用pd.to_datetime, 关键字unit='D'和进行解析origin='1899-12-30':
import pandas as pd
df = pd.DataFrame({'xldate': [42580.3333333333]})
df['date'] = pd.to_datetime(df['xldate'], unit='D', origin='1899-12-30')
df['date']
Out[2]:
0 2016-07-29 07:59:59.999971200
Name: date, dtype: datetime64[ns]
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进一步阅读:
EdC*_*ica 13
好吧我认为最简单的方法是从浮点数构造一个TimedeltaIndex并将其添加到1900,1,1的标量日期时间:
In [85]:
import datetime as dt
import pandas as pd
df = pd.DataFrame({'date':[42580.3333333333, 10023]})
df
Out[85]:
date
0 42580.333333
1 10023.000000
In [86]:
df['real_date'] = pd.TimedeltaIndex(df['date'], unit='d') + dt.datetime(1900,1,1)
df
Out[86]:
date real_date
0 42580.333333 2016-07-31 07:59:59.971200
1 10023.000000 1927-06-12 00:00:00.000000
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好吧,看起来excel有点奇怪了它的日期谢谢@ayhan:
In [89]:
df['real_date'] = pd.TimedeltaIndex(df['date'], unit='d') + dt.datetime(1899, 12, 30)
df
Out[89]:
date real_date
0 42580.333333 2016-07-29 07:59:59.971200
1 10023.000000 1927-06-10 00:00:00.000000
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请参阅相关:如何将python datetime.datetime转换为excel序列日期编号
您可以xlrd在传递给之前使用第三方库pd.to_datetime:
import xlrd
def read_date(date):
return xlrd.xldate.xldate_as_datetime(date, 0)
df = pd.DataFrame({'date':[42580.3333333333, 10023]})
df['new'] = pd.to_datetime(df['date'].apply(read_date), errors='coerce')
print(df)
date new
0 42580.333333 2016-07-29 08:00:00
1 10023.000000 1927-06-10 00:00:00
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