Odtnhj

I have a table below:

I want to add a column (Evaluation) that returns one of the elements in the cars columns (separated by comma). The element to return will be based on the Ferrari and Toyota columns. The Evaluation column returns the element that the individual does not have. So take the first row, for example, John has one Ferrari and no Toyota. Since John has no Toyota, the evaluation column returns Toyota.

I would prefer to have the decision made using the cars column, separating the text by comma and looking up the text against the values under Ferrari and Toyota

You can use:

df = pd.DataFrame('Names':['John'] * 2 + ['Peter'] * 2 + ['Sam'] * 2,
 'Cars':['Ferrari, Toyota','Ferrari','Ferrari, Toyota','Ferrari',
 'Ferrari, Toyota','Ferrari, Toyota'],
 'Ferrari': [1,1,0,0,1,1], 
 'Toyota': [0,1,1,0,1,0])

df['Evaluation1'] = df.iloc[:, 2:].ne(1).dot(df.columns[2:] + ', ').str.rstrip(', ')
print (df)
 Names Cars Ferrari Toyota Evaluation1
0 John Ferrari, Toyota 1 0 Toyota
1 John Ferrari 1 1 
2 Peter Ferrari, Toyota 0 1 Ferrari
3 Peter Ferrari 0 0 Ferrari, Toyota
4 Sam Ferrari, Toyota 1 1 
5 Sam Ferrari, Toyota 1 0 Toyota

Details:

First seelct all columns without first 2 by iloc and create boolean mask - compare by ne (!=):

print (df.iloc[:, 2:].ne(1))
 Ferrari Toyota
0 False True
1 False False
2 True False
3 True True
4 False False
5 False True

Then use matrix multiplication by dot with columns names with separator:

print (df.iloc[:, 2:].ne(1).dot(df.columns[2:] + ', '))
0 Toyota, 
1 
2 Ferrari, 
3 Ferrari, Toyota, 
4 
5 Toyota, 
dtype: object

And remove last separator by rstrip:

print (df.iloc[:, 2:].ne(1).dot(df.columns[2:] + ', ').str.rstrip(', '))
0 Toyota
1 
2 Ferrari
3 Ferrari, Toyota
4 
5 Toyota
dtype: object

If not possible select by position because positions should be changed of columns without 0,1 is possible use drop for remove unecessary columns:

df1 = df.drop(['Names','Ferrari'], axis=1).ne(1)
df['Evaluation2'] = df1.dot(df1.columns + ', ').str.rstrip(', ')

df = pd.DataFrame('a': [0,0,1,1], 'b': [0,1,0,1])

Creates the following DataFrame:

 a b
0 0 0
1 0 1
2 1 0
3 1 1

You can add a new column with a list of column names equal to zero with:

df['evaluated'] = df.apply(lambda x: ','.join(df.columns[x == 0]), axis=1)

Output:

 a b evaluated
0 0 0 a,b
1 0 1 a
2 1 0 b
3 1 1