对于我的数据框,我想为另一列中的每个唯一值添加一个新列。新列包含几个日期时间条目,其他列的每个唯一值都应该得到。
示例:
Original Df:
ID
1
2
3
新专栏DF:
Date
2015/01/01
2015/02/01
2015/03/01
结果Df:
ID Date
1 2015/01/01
2015/02/01
2015/03/01
2 2015/01/01
2015/02/01
2015/03/01
3 2015/01/01
2015/02/01
2015/03/01
我试图坚持这个解决方案:https://stackoverflow.com/a/12394122/3856569
但它给了我以下错误:Length of values does not match length of index
任何人都有一个简单的解决方案吗?非常感谢!
答案 0 :(得分:1)
更新:复制id
次6次:
In [172]: %paste
data = """\
id
1
2
3
"""
df = pd.read_csv(io.StringIO(data))
# repeat each ID 6 times
df = pd.DataFrame(df['id'].tolist()*6, columns=['id'])
start_date = pd.to_datetime('2015-01-01')
df['date'] = start_date
df['date'] = df.groupby('id', as_index=False)\
.transform(lambda x: pd.date_range(start_date,
freq='1D',
periods=len(x)))
df.sort_values(by=['id','date'])
## -- End pasted text --
Out[172]:
id date
0 1 2015-01-01
3 1 2015-01-02
6 1 2015-01-03
9 1 2015-01-04
12 1 2015-01-05
15 1 2015-01-06
1 2 2015-01-01
4 2 2015-01-02
7 2 2015-01-03
10 2 2015-01-04
13 2 2015-01-05
16 2 2015-01-06
2 3 2015-01-01
5 3 2015-01-02
8 3 2015-01-03
11 3 2015-01-04
14 3 2015-01-05
17 3 2015-01-06
OLD更通用的答案:
准备样本DF:
start_date = pd.to_datetime('2015-01-01')
data = """\
id
1
2
2
3
1
2
3
2
1
"""
df = pd.read_csv(io.StringIO(data))
In [200]: df
Out[200]:
id
0 1
1 2
2 2
3 3
4 1
5 2
6 3
7 2
8 1
解决方案:
In [201]: %paste
df['date'] = start_date
df['date'] = df.groupby('id', as_index=False)\
.transform(lambda x: pd.date_range(start_date,
freq='1D',
periods=len(x)))
## -- End pasted text --
In [202]: df
Out[202]:
id date
0 1 2015-01-01
1 2 2015-01-01
2 2 2015-01-02
3 3 2015-01-01
4 1 2015-01-02
5 2 2015-01-03
6 3 2015-01-02
7 2 2015-01-04
8 1 2015-01-03
排序:
In [203]: df.sort_values(by='id')
Out[203]:
id date
0 1 2015-01-01
4 1 2015-01-02
8 1 2015-01-03
1 2 2015-01-01
2 2 2015-01-02
5 2 2015-01-03
7 2 2015-01-04
3 3 2015-01-01
6 3 2015-01-02
答案 1 :(得分:1)
一种相当简单的numpy
方法,利用repeat
和tile
:
import numpy as np
import pandas as pd
N = 3 # arbitrary number of IDs/dates
ID = np.arange(N) + 1
dates = pd.date_range('20160101', periods=N)
df = pd.DataFrame({'ID' : np.repeat(ID, N),
'dates' : np.tile(dates, N)})
结果DataFrame:
In [1]: df
Out[1]:
ID dates
0 1 2016-01-01
1 1 2016-01-02
2 1 2016-01-03
3 2 2016-01-01
4 2 2016-01-02
5 2 2016-01-03
6 3 2016-01-01
7 3 2016-01-02
8 3 2016-01-03
<强>更新强>
假设您已经有DataFrame
个ID
,正如MaxU指出的那样,您可以平铺ID
s
df = pd.DataFrame({'ID' : np.tile(df['ID'], N),
'dates' : np.tile(dates, N)})
# now df needs sorting
df = df.sort_values(by=['ID', 'dates'])
结果DataFrame:
In [5]: df
Out[5]:
ID dates
0 1 2016-01-01
3 1 2016-01-01
6 1 2016-01-01
1 2 2016-01-02
4 2 2016-01-02
7 2 2016-01-02
2 3 2016-01-03
5 3 2016-01-03
8 3 2016-01-03