将字符串列转换为时间序列，可以使用 dt.strftime 方法：

If you convert the column of strings to a time series, you could use the dt.strftime method:

import numpy as np
import pandas as pd
nan = np.nan
df = pd.DataFrame({'TBD': [nan, nan, nan], 'TBD.1': [nan, nan, nan], 'TBD.2': [nan, nan, nan], 'TimeStamp': ['2016/06/08 17:19:53', '2016/06/08 17:19:54', '2016/06/08 17:19:54'], 'Value': [0.062941999999999998, 0.062941999999999998, 0.062941999999999998]})
df['TimeStamp'] = pd.to_datetime(df['TimeStamp']).dt.strftime('%m/%d/%Y %H:%M:%S')
print(df)

收益

   TBD  TBD.1  TBD.2            TimeStamp     Value
0  NaN    NaN    NaN  06/08/2016 17:19:53  0.062942
1  NaN    NaN    NaN  06/08/2016 17:19:54  0.062942
2  NaN    NaN    NaN  06/08/2016 17:19:54  0.062942

由于要将一列字符串转换为另一（不同的）字符串列，因此也可以使用向量化的 str.replace 方法：

import numpy as np
import pandas as pd
nan = np.nan
df = pd.DataFrame({'TBD': [nan, nan, nan], 'TBD.1': [nan, nan, nan], 'TBD.2': [nan, nan, nan], 'TimeStamp': ['2016/06/08 17:19:53', '2016/06/08 17:19:54', '2016/06/08 17:19:54'], 'Value': [0.062941999999999998, 0.062941999999999998, 0.062941999999999998]})
df['TimeStamp'] = df['TimeStamp'].str.replace(r'(\d+)/(\d+)/(\d+)(.*)', r'\2/\3/\1\4')
print(df)

因为

In [32]: df['TimeStamp'].str.replace(r'(\d+)/(\d+)/(\d+)(.*)', r'\2/\3/\1\4')
Out[32]: 
0    06/08/2016 17:19:53
1    06/08/2016 17:19:54
2    06/08/2016 17:19:54
Name: TimeStamp, dtype: object

这使用正则表达式重新排列了strin g ，而无需先将
字符串解析为日期。这比第一种方法要快（主要是因为它跳过了
的解析步骤），但是它也具有不检查
日期字符串是否为有效日期的缺点。

This uses regex to rearrange pieces of the string without first parsing the string as a date. This is faster than the first method (mainly because it skips the parsing step), but it also has the disadvantage of not checking that the date strings are valid dates.