How to map integer to string value in pandas dataframe-CodePudding

I have this python dictionary:

dictionary = {
        '1':'A',
        '2':'B',
        '3':'C',
        '4':'D',
        '5':'E',
        '6':'F',
        '7':'G',
        '8':'H',
        '8':'I',
        '9':'J',
        '0':'L'
        }

The I have created this simple pandas dataframe:

import pandas as pd
ds = {'col1' : [12345,67890], 'col2' : [12364,78910]}

df = pd.DataFrame(data=ds)
print(df)

Which looks like this:

    col1   col2
0  12345  12364
1  67890  78910

I would like to transform each and every digit in col1 (which is an int field) to the correspondent letter as per dictionary indicated above. So, basically I'd like the resulting dataframe to look like this:

    col1   col2 col1_transformed
0  12345  12364            ABCDE
1  67890  78910            FGHIJ

Is there a quick, pythonic way to do so by any chance?

CodePudding user response：

A possible solution (notice that 8 is repeated in your dictionary -- a typo? -- and, therefore, my result does not match yours):

def f(x):
    return ''.join([dictionary[y] for y in str(x)])

df['col3'] = df['col1'].map(f)

Output:

    col1   col2   col3
0  12345  12364  ABCDE
1  67890  78910  FGIJL

CodePudding user response：

Try:

df[df.columns   "_transformed"] = df.apply(
    lambda x: [
        "".join(dictionary.get(ch, "") for ch in s) for s in map(str, x)
    ],
    axis=1,
    result_type="expand",
)
print(df)

Prints:

    col1   col2 col1_transformed col2_transformed
0  12345  12364            ABCDE            ABCFD
1  67890  78910            FGIJL            GIJAL