Python solution present column values to list

Question

I have data something as below:

df = pd.DataFrame({'column1': ['Y', 'Y', 'Y'],
                   'value_5': ['N', 'Y', 'Y'],
                   'value_6': ['N', 'Y', 'N'],
                   'value_10': ['Y', 'N', 'N'],
                   'value_20': ['N', 'N', 'Y']},
                  index=['key1','key2','key4'])
print(df)
     column1 value_5 value_6 value_10 value_20
key1       Y       N       N        Y        N
key2       Y       Y       Y        N        N
key4       Y       Y       N        N        Y   

From that data I would like to create the last column. But the number of columns and values may be different in each run.

Answer 1

First extract integers from columns with substrings value_ selected by DataFrame.filter , then compare values by Y and if match convert columns names to lists:

f = lambda x: int(x.split('_')[-1])
df1 = df.filter(like='value_').rename(columns=f)

df['new'] = df1.eq('Y').agg(lambda x: x.index[x].tolist(), axis=1)

Another idea with regex for integers from columns names and for list is used list comprehension:

import re
f = lambda x: next(map(int,re.findall(r'\d+',x)))
df1 = df.filter(like='value_').rename(columns=f)

df['new'] = [df1.columns[x].tolist() for x in df1.eq('Y').to_numpy()]

Or use Series.str.extract :

df1 = df.filter(like='value_')
df1.columns = df1.columns.str.extract('(\d+)', expand=False).astype(int)

df['new'] = [df1.columns[x].tolist() for x in df1.eq('Y').to_numpy()]
print (df)
     column1 value_5 value_6 value_10 value_20      new
key1       Y       N       N        Y        N     [10]
key2       Y       Y       Y        N        N   [5, 6]
key4       Y       Y       N        N        Y  [5, 20]

Answer 2

Assuming df is this:

import pandas as pd

df = pd.DataFrame({'column1': ['Y', 'Y', 'Y'],
                   'value_5': ['N', 'Y', 'Y'],
                   'value_6': ['N', 'Y', 'N'],
                   'value_10': ['Y', 'N', 'N'],
                   'value_20': ['N', 'N', 'Y']  })
print(df)
  column1 value_5 value_6 value_10 value_20
0       Y       N       N        Y        N
1       Y       Y       Y        N        N
2       Y       Y       N        N        Y

I have extracted the column index where Y is present

expected = []
df1 = df[df.columns[['value' in c for c in df.columns]]]
for i in range(len(df1)):
    idx = df1.iloc[i, :][df1.iloc[i, :]=='Y'].index
    expected.append([int(e.split('_')[-1]) for e in idx])
df['Expected'] = expected
print(df)

  column1 value_5 value_6 value_10 value_20 Expected
0       Y       N       N        Y        N     [10]
1       Y       Y       Y        N        N   [5, 6]
2       Y       Y       N        N        Y  [5, 20]

Answer 3

Here's a way to define a function to check for the value 'Y' :

def check_y(row):
    checklis = [k for k,v in zip(row.index, row.values) if v=='Y']
    return [int(k.split('_')[1]) for k in checklis]            

df1 = df.filter(like='value_')
df['Expected'] = df1.apply(lambda row: check_y(row), axis=1)
print(df)

     column1 value_5 value_6 value_10 value_20 Expected
key1       Y       N       N        Y        N     [10]
key2       Y       Y       Y        N        N   [5, 6]
key4       Y       Y       N        N        Y  [5, 20]