从另一列的df栏中更改列

我有数据：

date id 0 2016-06-17 06:25:05 [e-mail] 1 2016-06-17 06:25:07 [e-mail] 2 2016-06-17 06:25:10 [e-mail] 3 2016-06-17 06:25:11 [e-mail] 

其他数据

 Email,UTC shift [e-mail],5 [e-mail],3 [e-mail],3 [e-mail],2 

我需要将UTC shift添加到第一个文件date到hours 。 欲望输出：

  date id 0 2016-06-17 11:25:05 [e-mail] 1 2016-06-17 11:25:07 [e-mail] 2 2016-06-17 09:25:10 [e-mail] 3 2016-06-17 09:25:11 [e-mail] 

我将date转换为date时间，但我不知道如何将UTC shift为小时。

Answer 1

如果to_datetime不是datetime date则需要首先将to_datetime列date转换为列id和Email merge 。 然后转换to_timedelta UTC shift to_timedelta列，添加date和最后drop不必要的列：

 import pandas as pd df1 = pd.DataFrame({'date': {0: '2016-06-17 06:25:05', 1: '2016-06-17 06:25:07', 2: '2016-06-17 06:25:10', 3: '2016-06-17 06:25:11'}, 'id': {0: '[e-mail]', 1: '[e-mail]', 2: '[e-mail]', 3: '[e-mail]'}}) df2 = pd.DataFrame({'Email': {0: '[e-mail]', 1: '[e-mail]', 2: '[e-mail]', 3: '[e-mail]'}, 'UTC shift': {0: 5, 1: 3, 2: 3, 3: 2}}) print (df1) date id 0 2016-06-17 06:25:05 [e-mail] 1 2016-06-17 06:25:07 [e-mail] 2 2016-06-17 06:25:10 [e-mail] 3 2016-06-17 06:25:11 [e-mail] print (df2) Email UTC shift 0 [e-mail] 5 1 [e-mail] 3 2 [e-mail] 3 3 [e-mail] 2 

 df1['date'] = pd.to_datetime(df1.date) df = pd.merge(df1, df2, left_on='id', right_on='Email') df['date'] += pd.to_timedelta(df['UTC shift'], unit='H') df.drop(['Email','UTC shift'], axis=1, inplace=True) print (df) date id 0 2016-06-17 11:25:05 [e-mail] 1 2016-06-17 11:25:07 [e-mail] 2 2016-06-17 09:25:10 [e-mail] 3 2016-06-17 09:25:11 [e-mail] 

Answer 2

尝试这个：

 df['date'] += pd.Timedelta(df['UTC shift'], unit='H')