Find columns where at least one row contains an alphabetical letter

Question

Let's say I have the following data set:

import pandas as pd

df = pd.DataFrame(
        {'A': [1, 2, 3],
         'B': ['one', 2, 3],
         'C': [4, 5, '6Y']
         })

I would like to find out - without any cumbersome for loop - which columns contain at least one case with an alphabetical letter (here: B and C). I guess the result should either be a list of booleans or indices.

Thank you for your help!

Answer 1

As a quick and simple solution, you can use replace and filter:

df.replace('(?i)[a-z]', '', regex=True).ne(df).any()

A    False
B     True
C     True
dtype: bool

df.columns[df.replace('(?i)[a-z]', '', regex=True).ne(df).any()]
# Index(['B', 'C'], dtype='object')

---

Another option is applying str.contains column-wise:

mask = df.astype(str).apply(
    lambda x: x.str.contains(r'[a-z]', flags=re.IGNORECASE)).any()
mask

A    False
B     True
C     True
dtype: bool

df.columns[mask]
# Index(['B', 'C'], dtype='object')

Answer 2

We could use pd.to_numeric:

df.apply(pd.to_numeric, errors='coerce').isna().any().tolist()
# [False, True, True]

Answer 3

In that case you can do with to_numeric

df.apply(pd.to_numeric,errors='coerce').isnull().any()
Out[37]: 
A    False
B     True
C     True
dtype: bool

---

Update

df.stack().str.contains('[a-zA-Z]').groupby(level=1).any()
Out[62]: 
A    False
B     True
C     True
dtype: bool