Return columns with a row containing substring in an array

Question

I have df with n columns (named as col'n') and 'name' column, I want to loc any row with col'n' have any keywords in 'searchkey' array.

It seems the list comprehension cannot apply in df column names list.

df[[c for c in df.columns if c[3:] = 'Col'].str.isin(searchkey))

Also, I tried to use apply function, but still face the problem.

df = pd.DataFrame({'name':['AA','BB','CC','DD','EE','FF'],
                   'col1':['mn','mxn','ca','sd','xa','ac'], 
                   'col2':['m','naa','x','ddn','q','y'],
                   'col3':['mn','mddn','csfd','sad','xxa','aad'], 
                   ... ... 
                   'coln':['sfn','mxc','cxa','sxxd','xada','axxc'],                    
})

searchkey = ['xx', 'aa', 'dd']

def func(x):
    return [x.columns] in searchkey  ## Error

df.apply(func, axis=1)

Any help is greatly appreciated!

What is the expected output, a list of columns or the rows? And is this a exact match, or a substring match? — ALollz
– ALollz, Commented Jul 17, 2019 at 15:56

tawab_shakeel · Accepted Answer · 2019-07-17 15:59:07Z

3

Try this

df = df[df.isin(search_key).any(1)]

answered Jul 17, 2019 at 15:59

tawab_shakeel

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1 Comment

tawab_shakeel Over a year ago

no actually it is mentioned in question I want to loc any row with col'n' have any keywords in 'searchkey' array.

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