How to reduce a List of element following a Logic in python?-CodePudding

I have Two list:

L = ['A','B','C']
L2 = ['A', 'B', ('B', 'A'), 'C']

I would create a single list with the following element:

L3 = ['A', ('B', 'A'), 'C']

Every time an element in the list L is present in more element in the L2 list I would pick the longest one. Important, Only if B Is at the first place of the tuple

I tried the following code: the following code

temp_length = 0
for d in L:
    for d2 in L2:
        if d in d2 and temp_length<len(d2):
           temp_op = d
    L3.append(temp_op)

But is not adding ('B','A') instead of 'B'

CodePudding user response：

Well this is one way to do it, but this question may have cleaner solutions as well:

L = ['A','B','C']
L2 = ['A', 'B', ('B', 'A'), 'C']    
answer = []
    for l2 in L2:
        if len(l2) == 1: 
            if l2 in L:
                answer.append(l2)
            
            
        elif l2[0] in L:
            answer.append(l2)
            
            if l2[0] in answer:
                answer.remove(l2[0])
    
            
    print(answer)

['A', ('B', 'A'), 'C']

CodePudding user response：

First, find the longest elements in L2 (according to the first sub-element match) and keep them in a dictionary using the matching sub-element as a key.

It's much faster than looking for elements in lists many times and repeat checking the shorter ones needlessly.

from typing import Dict, Iterable

l = ['A','B','C']
l2 = ['A', 'B', ('B', 'A'), 'C']


def keep_longest_elements(seq: Iterable) -> Dict:
    res = {}
    for el in seq:
        if (exist_el := res.get(el)) is not None:
            if exist_el[0] == el[0] and len(el) > len(exist_el):
                res[exist_el[0]] = el
        else:
            res[el[0]] = el
    return res

longest = keep_longest_elements(l2)
print(longest)
l3 = [longest[el] for el in l]
print(l3)

produces

{'A': 'A', 'B': ('B', 'A'), 'C': 'C'}
['A', ('B', 'A'), 'C']