3

数据框看起来像:

       OPENED
0  2004-07-28
1  2010-03-02
2  2005-10-26
3  2006-06-30
4  2012-09-21

我成功地将它们转换为我想要的格式,但似乎效率很低。

   OPENED
0   40728
1  100302
2   51026
3   60630
4  120921

我用于日期转换的代码是:

df['OPENED'] = pd.to_datetime(df.OPENED, format='%Y-%m-%d')
df['OPENED'] = df['OPENED'].apply(lambda x: x.strftime('%y%m%d'))
df['OPENED'] = df['OPENED'].apply(lambda i: str(i))
df['OPENED'] = df['OPENED'].apply(lambda s: s.lstrip("0"))
4

1 回答 1

4

您可以使用str.replace,然后删除前 2 个字符,str[2:]最后删除前导0str.lstrip

print (type(df.ix[0,'OPENED']))
<class 'str'>
print (df.OPENED.dtype)
object

print (df.OPENED.str.replace('-','').str[2:].str.lstrip('0'))
0     40728
1    100302
2     51026
3     60630
4    120921
Name: OPENED, dtype: object

如果 dtype 已经datetime使用strftime并且str.lstrip

print (type(df.ix[0,'OPENED']))
<class 'pandas.tslib.Timestamp'>
print (df.OPENED.dtype)
datetime64[ns]

print (df.OPENED.dt.strftime('%y%m%d').str.lstrip('0'))
0     40728
1    100302
2     51026
3     60630
4    120921
Name: OPENED, dtype: object

感谢您Jon Clements的评论:

print (df['OPENED'].apply(lambda L: '{0}{1:%m%d}'.format(L.year % 100, L)))
0     40728
1    100302
2     51026
3     60630
4    120921
Name: OPENED, dtype: object
于 2016-08-18T12:05:42.673 回答