pandas 如何在熊猫中减去天数后获取日期
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/40104946/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to get date after subtracting days in pandas
提问by rey
I have a dataframe:
我有一个数据框：
In [15]: df
Out[15]:
date day
0 2015-10-10 23
1 2015-12-19 9
2 2016-03-05 34
3 2016-09-17 23
4 2016-04-30 2
I want to subtract the number of days from the date and create a new column.
我想从日期中减去天数并创建一个新列。
In [16]: df.dtypes
Out[16]:
date datetime64[ns]
day int64
Desired output something like:
所需的输出类似于：
In [15]: df
Out[15]:
date day date1
0 2015-10-10 23 2015-09-17
1 2015-12-19 9 2015-12-10
2 2016-03-05 34 2016-01-29
3 2016-09-17 23 2016-08-25
4 2016-04-30 2 2016-04-28
I tried but this does not work:
我试过了，但这不起作用：
df['date1']=df['date']+pd.Timedelta(df['date'].dt.day-df['day'])
it throws error :
它抛出错误：
TypeError: unsupported type for timedelta days component: Series
类型错误：timedelta 天组件不支持的类型：系列
回答by jezrael
You can use to_timedelta:
您可以使用to_timedelta：
df['date1'] = df['date'] - pd.to_timedelta(df['day'], unit='d')
print (df)
date day date1
0 2015-10-10 23 2015-09-17
1 2015-12-19 9 2015-12-10
2 2016-03-05 34 2016-01-31
3 2016-09-17 23 2016-08-25
4 2016-04-30 2 2016-04-28
If need Timedeltause apply, but it is slowier:
如果需要Timedelta使用apply，但速度较慢：
df['date1'] = df['date'] - df.day.apply(lambda x: pd.Timedelta(x, unit='D'))
print (df)
date day date1
0 2015-10-10 23 2015-09-17
1 2015-12-19 9 2015-12-10
2 2016-03-05 34 2016-01-31
3 2016-09-17 23 2016-08-25
4 2016-04-30 2 2016-04-28
Timings:
时间：
#[5000 rows x 2 columns]
df = pd.concat([df]*1000).reset_index(drop=True)
In [252]: %timeit df['date'] - df.day.apply(lambda x: pd.Timedelta(x, unit='D'))
10 loops, best of 3: 45.3 ms per loop
In [253]: %timeit df['date'] - pd.to_timedelta(df['day'], unit='d')
1000 loops, best of 3: 1.71 ms per loop
回答by sachin saxena
import dateutil.relativedelta
def calculate diff(v):
return v['date'] - dateutil.relativedelta.relativedelta(day=v['day'])
df['date1']=df.apply(calculate_diff, axis=1)
given that v['date'] is datetime object
鉴于 v['date'] 是日期时间对象
