Column of lists, convert list to string as a new column

One way you could do it is to use list comprehension, str, and join:

df['liststring'] = df.lists.apply(lambda x: ', '.join([str(i) for i in x]))

Output:

                lists        liststring
1  [1, 2, 12, 6, ABC]  1, 2, 12, 6, ABC
2     [1000, 4, z, a]     1000, 4, z, a

List Comprehension

If performance is important, I strongly recommend this solution and I can explain why.

df['liststring'] = [','.join(map(str, l)) for l in df['lists']]
df

                lists    liststring
0  [1, 2, 12, 6, ABC]  1,2,12,6,ABC
1     [1000, 4, z, a]    1000,4,z,a

You can extend this to more complicated use cases using a function.

def try_join(l):
    try:
        return ','.join(map(str, l))
    except TypeError:
        return np.nan

df['liststring'] = [try_join(l) for l in df['lists']]

---

Series.apply/Series.agg with ','.join

You need to convert your list items to strings first, that's where the map comes in handy.

df['liststring'] = df['lists'].apply(lambda x: ','.join(map(str, x)))

Or,

df['liststring'] = df['lists'].agg(lambda x: ','.join(map(str, x)))

<!- >

df
                lists    liststring
0  [1, 2, 12, 6, ABC]  1,2,12,6,ABC
1     [1000, 4, z, a]    1000,4,z,a

---

pd.DataFrame constructor with DataFrame.agg

A non-loopy/non-lambda solution.

df['liststring'] = (pd.DataFrame(df.lists.tolist())
                      .fillna('')
                      .astype(str)
                      .agg(','.join, 1)
                      .str.strip(','))

df
                lists    liststring
0  [1, 2, 12, 6, ABC]  1,2,12,6,ABC
1     [1000, 4, z, a]    1000,4,z,a

All of these didn't work for me (dealing with text data) what worked for me is this:

    df['liststring'] = df['lists'].apply(lambda x: x[1:-1])