假设我有一个Dataframe:
df = {'date':['20170101', '20170102', '20170103', '20170104']}
我想将'20170101'
转换为'2017-01-01 00:00:00'
我怎样才能实现它?
答案 0 :(得分:4)
您可以将日期时间字符串转换为jump
对象。然后得到你想要的任何格式:
datetime
输出:
from datetime import datetime
datetime.strptime("20170103", "%Y%m%d").strftime("%Y-%m-%d %H:%M:%S")
答案 1 :(得分:3)
如果你有@SuppressWarnings("unchecked")
@Override
public void onApplicationEvent(ApplicationReadyEvent arg0) {
ECServiceFactory svcFactory = new ECServiceFactory(application, documentService,new SuperUser(),AccessDeniedException.class,config);
}
,你可以用:
pandas.DataFrame
pd.to_datetime(df['date'], format='%Y%m%d')
df = pd.DataFrame({'date':['20170101', '20170102', '20170103', '20170104']})
df['date2'] = df['date'].apply(lambda d: dt.datetime.strptime(d, '%Y%m%d'))
df['date3'] = pd.to_datetime(df['date'], format='%Y%m%d')
print(df)
答案 2 :(得分:0)
如果你知道如何构建字符串(即YYYYMMDD
),你可以简单地使用它:
def f(s):
return '{}-{}-{} 00:00:00'.format(s[:4],s[4:6],s[6:])
s = '20170101'
f(s)
Out[7]: '2017-01-01 00:00:00'