Make column rows the same value with matching value of another column when condition is met-CodePudding

I have a dataframe like below

df = pd.DataFrame({'col1': ['A', 'A', 'B', 'C', 'D', 'D'],
                   'col2': [1,0,1,0,0,1]})

col1	col2
A	1
A	0
B	1
C	0
D	0
D	1

If there's one row from col2 = 1, I want to make all col2 rows = 1 for the same value col1 value. The table I'd want is as below:

col1	col2
A	1
A	1
B	1
C	0
D	1
D	1

Any help is much appreciated!

CodePudding user response：

You can chunk it in two steps:

Get the rows where col2 is 1:

filters = df.loc[df.col2.eq(1), 'col1']

Assign the new values to rows, where col1 is in filters:

df.loc[df.col1.isin(filters), 'col2'] = 1

df
 
  col1  col2
0    A     1
1    A     1
2    B     1
3    C     0
4    D     1
5    D     1

CodePudding user response：

If you only have 0 and 1 in the column, you can use groupby transform:

df['col2'] = df.col2.groupby(df.col1).transform('max')

df
  col1  col2
0    A     1
1    A     1
2    B     1
3    C     0
4    D     1
5    D     1

In general you can pass a lambda to return 1 if any value is 1:

df['col2'] = df.col2.groupby(df.col1).transform(lambda g: 1 if (g == 1).any() else g)

df    
  col1  col2
0    A     1
1    A     1
2    B     1
3    C     0
4    D     1
5    D     1