[FIXED] foward fill data frame based on location of "1"s from 2 columns

Issue

    prac = pd.DataFrame(
    {"A": [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0], 
     "B": [0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0],
     "DesiredResult": [0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0]}
)

newDf = prac[['A','B']].mask(prac==0)
newDf['buySell'] =  newDf['A'].combine_first(newDf['B'])
newDf['buySell'].ffill(inplace=True)
newDf

this is what I have done/written so far:

1. I utilized mask() and create "NaN" values of columns "A" and "B"
2. next, I combined column "A" and "B" into a single column.
3. This is the part where I'm having trouble as I am trying to use ffill() to based on the range of the 1's in both columns.

Not sure how to deal w/ the ffill() based on the 2 columns from here.

Solution

IIUC, you could use a reverse cummax (or bfill if you have NaNs in place of 0) per group based on A:

prac['buySell'] = (prac.loc[::-1, 'B']
                   .groupby(prac['A'].eq(1).cumsum())
                   .cummax()[::-1]
                  )

Output:

    A  B  DesiredResult  buySell
0   0  0              0        0
1   1  0              1        1
2   0  0              1        1
3   0  1              1        1
4   0  1              1        1
5   0  1              1        1
6   0  0              0        0
7   1  0              1        1
8   0  0              1        1
9   0  0              1        1
10  0  0              1        1
11  0  0              1        1
12  0  1              1        1
13  0  1              1        1
14  0  1              1        1
15  0  0              0        0

Alternative:

newDf = prac[['A','B']].mask(prac==0)
         
newDf['buySell'] = newDf['B'].groupby(newDf['A'].eq(1).cumsum()).bfill()

Output:

      A    B  buySell
0   NaN  NaN      NaN
1   1.0  NaN      1.0
2   NaN  NaN      1.0
3   NaN  1.0      1.0
4   NaN  1.0      1.0
5   NaN  1.0      1.0
6   NaN  NaN      NaN
7   1.0  NaN      1.0
8   NaN  NaN      1.0
9   NaN  NaN      1.0
10  NaN  NaN      1.0
11  NaN  NaN      1.0
12  NaN  1.0      1.0
13  NaN  1.0      1.0
14  NaN  1.0      1.0
15  NaN  NaN      NaN

Answered By - mozway