Get column name based on condition in pandas

Firstly

Your question is very ambiguous and I recommend reading this link in @sammywemmy’s comment. If I understand your problem correctly… we’ll talk about this mask first:

df.columns[      
    (df == 1)        # mask 
    .any(axis=0)     # mask
]

What’s happening? Lets work our way outward starting from within df.columns[**HERE**] :

1. (df == 1) makes a boolean mask of the df with True/False(1/0)
2. .any() as per the docs:

"Returns False unless there is at least one element within a series or along a Dataframe axis that is True or equivalent".

This gives us a handy Series to mask the column names with.

We will use this example to automate for your solution below

---

Next:

Automate to get an output of (<row index> ,[<col name>, <col name>,..]) where there is 1 in the row values. Although this will be slower on large datasets, it should do the trick:

import pandas as pd

data = {'foo':[0,0,0,0], 'bar':[0, 1, 0, 0], 'baz':[0,0,0,0], 'spam':[0,1,0,1]}
df = pd.DataFrame(data, index=['a','b','c','d'])

print(df)

   foo  bar  baz  spam
a    0    0    0     0
b    0    1    0     1
c    0    0    0     0
d    0    0    0     1

# group our df by index and creates a dict with lists of df's as values
df_dict = dict(
    list(
        df.groupby(df.index)
    )
)

Next step is a for loop that iterates the contents of each df in df_dict, checks them with the mask we created earlier, and prints the intended results:

for k, v in df_dict.items():               # k: name of index, v: is a df
    check = v.columns[(v == 1).any()]
    if len(check) > 0:
        print((k, check.to_list()))

('b', ['bar', 'spam'])
('d', ['spam'])

Side note:

You see how I generated sample data that can be easily reproduced? In the future, please try to ask questions with posted sample data that can be reproduced. This way it helps you understand your problem better and it is easier for us to answer it for you.