Reputation: 1339
I have a dataframe with one index as datetime like below and I am looking to add a first columns index (see "target" below) where any dates are crossed to it (First_column).
First_column = ['s0000', 's0001', 's0002', 's0003', 's0004', ...]
Has someone any idea on how to proceed ?
Thank you very much. Alexis
My dataframe :
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 17544 entries, 2015-01-01 00:00:00 to 2016-12-31 23:00:00
Data columns (total 12 columns):
Target:
<class 'pandas.core.frame.DataFrame'>
MultiIndex: 996000 entries, (s0000, 2015-01-01 00:00:00) to (s0999, 2012-12-31 00:00:00)
Data columns (total 8 columns):
SCENARIO DATE
s0000 2015-02-28
2015-03-03
2015-03-04
2015-03-05
2015-03-06
2015-03-07
2015-03-10
2015-03-11
2015-03-12
2015-03-13
s0001 2015-02-28
2015-03-03
2015-03-04
2015-03-05
2015-03-06
2015-03-07
2015-03-10
2015-03-11
2015-03-12
2015-03-13
s0002 2015-02-28
2015-03-03
2015-03-04
2015-03-05
2015-03-06
2015-03-07
2015-03-10
2015-03-11
2015-03-12
2015-03-13
s0003 ...
Upvotes: 1
Views: 337
Reputation: 7131
You could do something like this...
import pandas as pd
first_col = ['s0001', 's0002', 's0003', 's0004']
# Make your datetime index
dt_index = pd.date_range('2015-2-27', freq='B', periods=10)
# Make your first_col index - must be same length as dt_index
first_col_index = len(dt_index)*first_col
first_col_index.sort()
# Make a dateframe with a hierarchical index
df = pd.DataFrame(range(len(first_col)*len(dt_index)), index=[first_col_index,
dt_index.repeat(len(first_col))])
Upvotes: 0
Reputation: 879501
You could use pd.concat
with the keys
parameter:
import pandas as pd
df = pd.DataFrame(range(10), index=pd.date_range('2015-2-27', freq='B', periods=10))
# 0
# 2015-02-27 0
# 2015-03-02 1
# 2015-03-03 2
# 2015-03-04 3
# 2015-03-05 4
# 2015-03-06 5
# 2015-03-09 6
# 2015-03-10 7
# 2015-03-11 8
# 2015-03-12 9
first_col = ['s{:04d}'.format(i) for i in range(1,5)]
# ['s0001d', 's0002d', 's0003d', 's0004d']
newdf = pd.concat([df]*len(first_col), keys=first_col)
print(newdf)
yields
0
s0001 2015-02-27 0
2015-03-02 1
2015-03-03 2
2015-03-04 3
2015-03-05 4
2015-03-06 5
2015-03-09 6
2015-03-10 7
2015-03-11 8
2015-03-12 9
s0002 2015-02-27 0
2015-03-02 1
2015-03-03 2
2015-03-04 3
2015-03-05 4
2015-03-06 5
2015-03-09 6
2015-03-10 7
2015-03-11 8
2015-03-12 9
s0003 2015-02-27 0
2015-03-02 1
2015-03-03 2
2015-03-04 3
2015-03-05 4
2015-03-06 5
2015-03-09 6
2015-03-10 7
2015-03-11 8
2015-03-12 9
s0004 2015-02-27 0
2015-03-02 1
2015-03-03 2
2015-03-04 3
2015-03-05 4
2015-03-06 5
2015-03-09 6
2015-03-10 7
2015-03-11 8
2015-03-12 9
Happily, I just learned this yesterday from Joris.
Upvotes: 1