Rearrange python list to avoid any position matches-CodePudding

I have 2 lists list1 and list2. How would I rearrange list1 (without rearranging list2) so that there are no matches on any positions eg:

UNDESIRED: list1 = [‘A’, ‘B’, ‘C’] list2 = [‘X’, ‘B’, ‘Z’] as you can see the B is on the same position, 1, in both lists…so I would then like to rearrange list1 to list1 = [‘B’, ‘A’, ‘C’] or any other order where there are no positional matches with list2 WITHOUT rearranging list2

CodePudding user response：

You could do this with the help of the itertools module. There may be better / more efficient mechanisms. Note that the process() function will return None when there is no possible solution.

import itertools

def process(L1, L2):
    for s in set(itertools.permutations(L1, len(L1))):
        if all([a != b for a, b in zip(s, L2)]):
            return s
    return None

list1 = ['A', 'B', 'C']
list2 = ['X', 'B', 'Z']

print(process(list1, list2))

CodePudding user response：

from itertools import permutations

def diff(list1, list2):
    for item in permutations(list1, len(list1)):
        if all(map(lambda a, b: a!=b, item, list2)):
            return list(item)
    return None

list1 = ['A', 'B', 'C']
list2 = ['X', 'B', 'Z']

result = diff(list1, list2)
if result:
    print(result, 'vs', list2)

['A', 'C', 'B'] vs ['X', 'B', 'Z']

CodePudding user response：

I solved it using this piece of code - however, as noted above, there are many possible cases where this very solution may not work. In this very case, you can try this:

import random
list1 = ['A', 'B', 'C']
list2 = ['X', 'B', 'Z']
for i in list1:
    for j in list2:
        if i == j in list2:
            while list1.index(i) == list2.index(j):
                list1.remove(i)
                z = len(list1)
                rand = random.choice(range(z))
                list1.insert(rand, i)
print(list1, list2)

Also, note that you use quite weird apostrophes inside the list: ‘ instead of ' or ". AFAIK, Python won't be able to comprehend them correctly.