熊猫为每一列选择不同的行

Question

I have a dataframe which consists of Date and three columns as below.

df = pd.DataFrame({'Date': ['01/01/2019', '02/01/2019', '03/01/2019', '04/01/2019', '05/01/2019', '06/01/2019', '07/01/2019', '08/01/2019'],
               'A': [1,2,3,4,5,6,7,8],
               'B': [11,12,13,14,15,16,17,18],
               'C': [21,22,23,24,25,26,27,28]})

I am trying to select parts of each column based on the date. If not selected looking to return zero. As below.

df = pd.DataFrame({'Date': ['01/01/2019', '02/01/2019', '03/01/2019', '04/01/2019', '05/01/2019', '06/01/2019', '07/01/2019', '08/01/2019'],
               'A': [1,2,0,0,5,6,7,8],
               'B': [0,0,0,0,0,16,17,18],
               'C': [21,22,0,0,0,0,0,0]})

In the example So: 'A' would be sliced 01/01/2019 to 02/01/2019 and 05/01/2019 to 08/01/2019 (or no end slice). B would be sliced 06/01/2019 to 08/01/2019 (or no end slice as that is the last data pointy. C would be sliced 01/01/2019 to 02/01/2019.

Answer 1

From what I understood, you can try and use df.where() :

df['B']=df['B'].where(df.Date.between('06/01/2019','08/01/2019'),0)
print(df)

         Date  A   B   C
0  01/01/2019  1   0  21
1  02/01/2019  2   0  22
2  03/01/2019  3   0  23
3  04/01/2019  4   0  24
4  05/01/2019  5   0  25
5  06/01/2019  6  16  26
6  07/01/2019  7  17  27
7  08/01/2019  8  18  28

You can do same operations for all column and conditions you want.

Answer 2

I would generate a Boolean mask:

B_dates = df['Dates'][-3:]
df.loc[~df['Date'].isin(B_dates), 'B'] = 0

Of course you can iterate over this for any dates and columns you choose.

Here is the output of running this code on your df and printing it:

     Date      A   B   C
0  01/01/2019  1   0  21
1  02/01/2019  2   0  22
2  03/01/2019  3   0  23
3  04/01/2019  4   0  24
4  05/01/2019  5   0  25
5  06/01/2019  6  16  26
6  07/01/2019  7  17  27
7  08/01/2019  8  18  28