Pandas mathematical operation, conditional on column value

Question

I need to make a mathematical operation which is conditional on the value in a second column. Here is the setup.

Given a simple dataframe (df):

Answer 1

It was throwing ValueError since you were not using the loc correctly. Here is the solution using loc:

df.loc[:,'math'] = 10 + df.loc[df['col1'] == "B", 'col3']

Output:

 col1 col2 col3 math
0    A   2   0    NaN
1    A   1   1    NaN
2    B   9   9    19.0
3    NaN 8   4    NaN
4    D   7   2    NaN
5    C   4   3    NaN

Answer 2

where

I perform the math then mask it using pandas.Series.where by passing the boolean series df.col1.eq('B')

df.assign(math=df.col3.add(10).where(df.col1.eq('B')))

  col1  col2  col3  math
0    A     2     0   NaN
1    A     1     1   NaN
2    B     9     9  19.0
3  NaN     8     4   NaN
4    D     7     2   NaN
5    C     4     3   NaN

Answer 3

Using loc

df['math'] = df.loc[df.col1.eq('B'), 'col3'].add(10)

  col1  col2  col3  math
0    A     2     0   NaN
1    A     1     1   NaN
2    B     9     9  19.0
3  NaN     8     4   NaN
4    D     7     2   NaN
5    C     4     3   NaN

Answer 4

I was also able to do the following...

df['math'] = 10 + df.loc[df['col1'] == 'B']['col3']  

Which is a variation on @user3483203 answer above. Ultimately, my 'B' is a variable, so I modified for @RafaelC 's comments.

Answer 5

Use:(Not a safe way to achieve it see the comment below )

df['New']=df.col3[df.col1=='B']+10
df
Out[11]: 
  col1  col2  col3   New
0    A     2     0   NaN
1    A     1     1   NaN
2    B     9     9  19.0
3  NaN     8     4   NaN
4    D     7     2   NaN
5    C     4     3   NaN

Update

pd.concat([df,(df.col3[df.col1=='B']+10).to_frame('New')],1)
Out[51]: 
  col1  col2  col3   New
0    A     2     0   NaN
1    A     1     1   NaN
2    B     9     9  19.0
3  NaN     8     4   NaN
4    D     7     2   NaN
5    C     4     3   NaN