I have a pandas dataframe that looks as follows:
ID  round   player1     player2 
1   1       A           B           
1   2       A           C
1   3       B           D
2   1       B           C           
2   2       C           D
2   3       C           E
3   1       B           C           
3   2       C           D
3   3       C           A
The dataframe contains sport match results, where the ID column denotes one tournament, the round column denotes the round for each tournament, and player1 and player2 columns contain the names of players that played against eachother in the respective round.
I now want to cumulatively count the tournament participations for, say, player A. In pseudocode this means: If the player with name A comes up in either the player1 or player2 column per tournament ID, increment the counter by 1.
The result should look like this (note: in my example player A did participate in tournaments with the IDs 1 and 3):
ID  round   player1     player2     playerAparticipated
1   1       A           B           1
1   2       A           C           1
1   3       B           D           1
2   1       B           C           0
2   2       C           D           0
2   3       C           E           0
3   1       B           C           2
3   2       C           D           2
3   3       C           A           2
My current status is, that I added a "helper" column containing the values 1 or 0 denoting, if the respective player participated in the tournament:
ID  round   player1     player2     helper
1   1       A           B           1
1   2       A           C           1
1   3       B           D           1
2   1       B           C           0
2   2       C           D           0
2   3       C           E           0
3   1       B           C           1
3   2       C           D           1
3   3       C           A           1
I think that I just need one final step, e.g., a smart use of cumsum() that counts the helper column in the desired way. However, I could not come up with the solution yet.