Creating loop in Pandas DataFrame with conditional value in cell

Question

In the script below, I assign the values 0 or 1 if the DataFrame cell has No or Yes in them.

answer= {'account': ['Adam', 'Ben', 'Tom', 'Isabel'],
     'a1': ['No', 'Yes', 'Yes', 'No'],
     'a2': ['No', 'Yes', 'No', 'No'],
     'a3': ['No', 'Yes', 'No', 'No'],
     'a4': ['Yes', 'No', 'Yes', 'Yes']}
RPI = pd.DataFrame.from_dict(answer)

I'm trying to create a loop or a function for the RPI.loc statements below so I don't have to repeat the 50 columns that I have. Is there a way to do that?

RPI.loc[RPI['a1'] == 'No', 'a1'] = 0
RPI.loc[RPI['a1'] == 'Yes', 'a1'] = 1
RPI.loc[RPI['a2'] == 'No', 'a2'] = 0
RPI.loc[RPI['a2'] == 'Yes', 'a2'] = 1
RPI.loc[RPI['a3'] == 'No', 'a3'] = 0
RPI.loc[RPI['a3'] == 'Yes', 'a3'] = 1
RPI.loc[RPI['a4'] == 'No', 'a4'] = 0
RPI.loc[RPI['a4'] == 'Yes', 'a4'] = 1

    a1  a2  a3  a4  account
0   0   0   0   1   Adam
1   1   1   1   0   Ben
2   1   0   0   1   Tom
3   0   0   0   1   Isabel

jezrael · Accepted Answer · 2017-05-29 05:21:40Z

6

Need replace by dict:

RPI = RPI.replace({'No':0, 'Yes':1})
print (RPI)
   a1  a2  a3  a4 account
0   0   0   0   1    Adam
1   1   1   1   0     Ben
2   1   0   0   1     Tom
3   0   0   0   1  Isabel

If need specify columns for replace by positions add iloc:

print (RPI.iloc[:, 0:4])
    a1   a2   a3   a4
0   No   No   No  Yes
1  Yes  Yes  Yes   No
2  Yes   No   No  Yes
3   No   No   No  Yes

RPI.iloc[:, 0:4] = RPI.iloc[:, 0:4].replace({'No':0, 'Yes':1})
print (RPI)
  a1 a2 a3 a4 account
0  0  0  0  1    Adam
1  1  1  1  0     Ben
2  1  0  0  1     Tom
3  0  0  0  1  Isabel

edited May 29, 2017 at 5:21

answered May 29, 2017 at 5:16

jezrael

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1 Comment

Adam Schroeder Over a year ago

Thank you. This worked very well for me. I appreciate it very much.

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