在Pandas数据框中将字符串日期转换为其他格式 [英] Convert string date to a different format in pandas dataframe

问题描述

到目前为止，我一直在社区中寻找这个答案，

I have been looking for this answer in the community so far, could not have.

我在python 3.5.1中有一个数据框，其中包含带有日期的列

I have a dataframe in python 3.5.1 that contains a column with dates in string imported from a CSV file.

数据框看起来像这样

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

我需要的是更改TimeStamp列格式为
％m /％d /％y％H：％M：％D

what I need is to change the TimeStamp column format to be %m/%d/%y %H:%M:%D

                  TimeStamp  TBD  TBD     Value  TBD
0       06/08/2016 17:19:53  NaN  NaN  0.062942  NaN

到目前为止，我已经找到了一些适用于字符串而非序列的解决方案

So far I have found some solutions that works but for string and not for series

任何帮助将不胜感激

谢谢

推荐答案

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.

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