pandas 熊猫：增加日期时间
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Pandas: increment datetime
提问by ldevyataykina
I need to do some actions with datein df column
我需要date在 df 列中执行一些操作
buys['date_min'] = (buys['date'] - MonthDelta(1))
buys['date_min'] = (buys['date'] + timedelta(days=5))
But it return
但它返回
TypeError: incompatible type [object] for a datetime/timedelta operation
类型错误：日期时间/时间增量操作的类型 [对象] 不兼容
How can I do it to column?
我怎样才能做到这一点？
回答by jezrael
I think you need first convert column dateto_datetime, because typeod values in column dateis string:
我认为您需要先转换 column dateto_datetime，因为column 中的typeod 值date是string：
buys['date_min'] = (pd.to_datetime(buys['date']) - MonthDelta(1))
buys['date_min'] = (pd.to_datetime(buys['date']) + timedelta(days=5))
EDIT:
编辑：
You need parameter formatto to_datetimeand then another solution is with to_timedelta
您需要参数formattoto_datetime然后另一个解决方案是to_timedelta
buys = pd.DataFrame({'date':['01.01.2016','20.02.2016']})
print (buys)
date
0 01.01.2016
1 20.02.2016
buys['date']= pd.to_datetime(buys['date'],format='%d.%m.%Y')
buys['date_min'] = buys['date'] + pd.to_timedelta(5,unit='d')
print (buys)
date date_min
0 2016-01-01 2016-01-06
1 2016-02-20 2016-02-25
