Selecting specific columns with specific values pandas

You can filter columns first with str.endswith, select columns by [] and compare by eq. Last add any for at least one 1 per row

cols = df.columns[df.columns.str.endswith('good')]
df1 = df[df[cols].eq(1).any(axis=1)]

Sample:

df = pd.DataFrame({'A':list('abcdef'),
                   'B':[1,1,4,5,5,1],
                   'C good':[7,8,9,4,2,3],
                   'D good':[1,3,5,7,1,0],
                   'E good':[5,3,6,9,2,1],
                   'F':list('aaabbb')})

print (df)
   A  B  C good  D good  E good  F
0  a  1       7       1       5  a
1  b  1       8       3       3  a
2  c  4       9       5       6  a
3  d  5       4       7       9  b
4  e  5       2       1       2  b
5  f  1       3       0       1  b

cols = df.columns[df.columns.str.endswith('good')]

print (df[cols].eq(1))
   C good  D good  E good
0   False    True   False
1   False   False   False
2   False   False   False
3   False   False   False
4   False    True   False
5   False   False    True

df1 = df[df[cols].eq(1).any(1)]
print (df1)
   A  B  C good  D good  E good  F
0  a  1       7       1       5  a
4  e  5       2       1       2  b
5  f  1       3       0       1  b

You solution was really close, only add any:

df1 = df[df[[i for i in df.columns if i.endswith('good')]].isin([1]).any(axis=1)]
print (df1)
   A  B  C good  D good  E good  F
0  a  1       7       1       5  a
4  e  5       2       1       2  b
5  f  1       3       0       1  b

EDIT:

If need only 1 and all another rows and columns remove:

df1 = df.loc[:, df.columns.str.endswith('good')]
df2 = df1.loc[df1.eq(1).any(1), df1.eq(1).any(0)]
print (df2)
   D good  E good
0       1       5
4       1       2
5       0       1