英文:
Pandas - Converting column of type float to date time
问题
我有一个包含两列float64对象的数据框。我尝试将它们转换为日期。
col1, col243835.0, 0.14583343835.0, 0.166667
期望输出：
col1, col205/01/2020, 3:30:00 AM05/01/2020, 4:00:00 AM
当我尝试使用pd.to_datetime(df['col1'])时，它将值转换为分别为1970-01-01 00:00:00.000043835和1970-01-01的日期。
英文:
I have a Dataframe with 2 columns that are float64 objects. I am trying to convert these to dates.
col1, col243835.0, 0.14583343835.0, 0.166667
Expected output:
col1, col205/01/2020,3:30:00 AM05/01/2020, 4:00:00 AM
When I try pd.to_datetime(df['col1']) it convert the values to 1970-01-01 00:00:00.000043835 and 1970-01-01 respectively
答案1
得分: 2
首先，可以将第一列转换为日期时间，将第二列转换为时间差：
df['col1'] = pd.to_timedelta(df['col1'], unit='d') + pd.to_datetime('1899-12-30')df['col2'] = pd.to_timedelta(df['col2'], unit='d').dt.floor('S')print(df)
输出结果：
col1 col20 2020-01-05 03:29:591 2020-01-05 04:00:00
对于自定义日期和时间的想法，但由于精度输出略有不同：
s = df['col1'] + df['col2']dates = pd.to_timedelta(s, unit='d').add(pd.to_datetime('1899-12-30')).dt.floor('S')df['col1'] = dates.dt.strftime('%d/%m/%Y')df['col2'] = dates.dt.strftime('%I:%M:%S %p')print(df)
输出结果：
col1 col20 05/01/2020 03:29:59 AM1 05/01/2020 04:00:00 AM
英文:
First is possible convert to datetimes fist column and second to timedeltas:
df['col1'] = pd.to_timedelta(df['col1'], unit='d') + pd.datetime(1899, 12, 30)df['col2'] = pd.to_timedelta(df['col2'], unit='d').dt.floor('S')print (df)col1 col20 2020-01-05 03:29:591 2020-01-05 04:00:00
One idea for custom dates and times, but becuase precision output is a bit different:
s = df['col1'] + df['col2']dates = pd.to_timedelta(s, unit='d').add(pd.datetime(1899, 12, 30)).dt.floor('S')df['col1'] = dates.dt.strftime('%d/%m/%Y')df['col2'] = dates.dt.strftime('%H:%M:%S %p')print (df)col1 col20 05/01/2020 03:29:59 AM1 05/01/2020 04:00:00 AM
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论