Как кодировать и декодировать столбец в пандах Python?

Question

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

Как кодировать и декодировать столбец F.

Ожидаемый результат:

Должно быть еще два столбца с закодированными и декодированными значениями

    A   B   C   D   E   F Encoded_Column           Decoded_Column
0   a   4   7   1   5   a   <!--Encoded Value-->    a
1   b   5   8   3   3   a   <!--Encoded Value-->    a
2   c   4   9   5   6   a   <!--Encoded Value-->    a 
3   d   5   4   4   9   b   <!--Encoded Value-->    b
4   e   5   2   2   2   b   <!--Encoded Value-->    b
5   f   4   0   0   4   b   <!--Encoded Value-->    b

Answer 1

Вы можете использовать это решение , реализованное для панд по Series.apply:

from Crypto.Cipher import XOR
import base64

def encrypt(key, plaintext):
  cipher = XOR.new(key)
  return base64.b64encode(cipher.encrypt(plaintext))

def decrypt(key, ciphertext):
  cipher = XOR.new(key)
  return cipher.decrypt(base64.b64decode(ciphertext))

load['Encoded_Column'] = load['F'].apply(lambda x: encrypt('password',x))
load['Decoded_Column'] = (load['Encoded_Column'].apply(lambda x: decrypt('password', x))
                                                .str.decode("utf-8"))
print (load)
   A  B  C  D  E  F Encoded_Column Decoded_Column
0  a  4  7  1  5  a        b'EQ=='              a
1  b  5  8  3  3  a        b'EQ=='              a
2  c  4  9  5  6  a        b'EQ=='              a
3  d  5  4  4  9  b        b'Eg=='              b
4  e  5  2  2  2  b        b'Eg=='              b
5  f  4  0  0  4  b        b'Eg=='              b

Другое решение :

import base64
def encode(key, clear):
    enc = []
    for i in range(len(clear)):
        key_c = key[i % len(key)]
        enc_c = chr((ord(clear[i]) + ord(key_c)) % 256)
        enc.append(enc_c)
    return base64.urlsafe_b64encode("".join(enc).encode()).decode()

def decode(key, enc):
    dec = []
    enc = base64.urlsafe_b64decode(enc).decode()
    for i in range(len(enc)):
        key_c = key[i % len(key)]
        dec_c = chr((256 + ord(enc[i]) - ord(key_c)) % 256)
        dec.append(dec_c)
    return "".join(dec)

load['Encoded_Column'] = load['F'].apply(lambda x: encode('password',x))
load['Decoded_Column'] = load['Encoded_Column'].apply(lambda x: decode('password', x))

Или используйте list comprehension:

load['Encoded_Column'] = [encode('password',x) for x in load['F']]
load['Decoded_Column'] = [decode('password', x) for x in load['Encoded_Column']]

print (load)
   A  B  C  D  E  F Encoded_Column Decoded_Column
0  a  4  7  1  5  a           w5E=              a
1  b  5  8  3  3  a           w5E=              a
2  c  4  9  5  6  a           w5E=              a
3  d  5  4  4  9  b           w5I=              b
4  e  5  2  2  2  b           w5I=              b
5  f  4  0  0  4  b           w5I=              b

Answer 2

import pandas as pd
import binascii

load = pd.DataFrame({'A':list('abcdef'),
                   'B':[4,5,4,5,5,4],
                   'C':[7,8,9,4,2,0],
                   'D':[1,3,5,4,2,0],
                   'E':[5,3,6,9,2,4],
                   'F':[binascii.hexlify(x.encode()) for x in 'aaabbb']
                    })

   A  B  C  D  E      F
0  a  4  7  1  5  b'61'
1  b  5  8  3  3  b'61'
2  c  4  9  5  6  b'61'
3  d  5  4  4  9  b'62'
4  e  5  2  2  2  b'62'
5  f  4  0  0  4  b'62'


# decode
binascii.unhexlify(load.loc[1]['F']).decode('utf-8') -->> 'a'

пример

print(binascii.hexlify('HelloWorld'.encode())) --> b'48656c6c6f576f726c64'

print(binascii.unhexlify('48656c6c6f576f726c64'.encode())) --> b'HelloWorld'