Flexible List Cleaner

Question

I have a list:

sol = ["U", "U'","U","R","R'", "R", "R", "R", "R","U","U'"]

I also created a function to clean this list:

def cleanSol(sol):
    res = '-'.join(sol)
    moveList = ['U', "U'", 'U2', 'D', "D'", 'D2', 'R', "R'", 'R2', 'L', "L'", 'L2', 'F', "F'", 'F2', 'B', "B'", 'B2', 'Y', "Y'", 'X', "X'", 'Z', "Z'", 'M', "M'", 'M2', 'E', "E'", 'E2', 'S', "S'", 'S2']
    for move in moveList:
        if move in res: # save time
            if (literal:= (f"{move}-"*4)[:-1]) and literal in res:
                res = res.replace(literal,"")
            if (literal:= f"{move}-{move}'") and literal in res:
                res = res.replace(literal,"")
            if (literal:= f"{move}'-{move}") and literal in res: 
                res = res.replace(literal,"")
            if (literal:= f"{move}-{move}") and literal in res:
                res = res.replace(literal,move+"2")
    return [r for r in res.split('-') if r] # remove emptry strings

My basic requirements are:

If 4 continuous instances of same move is in sol, remove them. For example, if R, R, R, R is in sol remove all the instances.
If move and it’s inverse is in sol, remove them. For example, if U, U' is in sol remove all the instances.
Same as above, just U' U inverse.
If only 2 moves are same, add a 2.

Samples:-

Input:

sol = ["U", "U'","U","R","R'", "R", "R", "R", "R","U'","U"]

Output:

>>> ['U']

Expected Output:

>>> ['U']

Tweek: I am using - as a simple separator and is not related with the list.

My solution works pretty well, but I think it’s too length and I’m looking for efficient solutions 🙂

Asked By: The Myth

||

Source

Answer 1

Mine is also lengthy but easy to follow, I just wanted to share as a second approach.

Here I check the list in a while True many times, every time from the beginning. If first items are removable, I remove them and start the loop again, if not I check next few items. I’ve put comment for each section.

def cleanSol(lst):
    while True:
        flag = False
        for i, item in enumerate(lst):
            inverse = item.replace("'", "") if "'" in item else item[0] + "'" + item[1:]

            # To remove "U" and "U'"
            if i < len(lst) - 1 and lst[i + 1] == inverse:
                del lst[i : i + 2]
                flag = True
                break

            # To remove "U", "U", "U", "U"
            if len(lst) >= 4 and set(lst[i : i + 4]) == {item}:
                del lst[i : i + 4]
                flag = True
                break

            # The two "U2", "U2", makes "U22" this should Also be removed.
            if "22" in item:
                del lst[i]
                flag = True
                break

            # To remove "U", "U2", "U"
            if i < len(lst) - 2 and item == lst[i + 2] and item == f"{lst[i+1]}2":
                del lst[i : i + 3]
                flag = True
                break

            # To convert "U", "U" to "U2"
            if i < len(lst) - 1 and item == lst[i + 1]:
                lst[i] = f"{item}2"
                del lst[i + 1]
                flag = True
                break

        if not flag:
            break
    return lst


for test in (
    ["U", "U'", "U", "R", "R'", "R", "R", "R", "R", "U'", "U"],
    ["U2", "U", "U", "R2", "R'", "R", "R"],
    ["U2", "U", "R", "R'", "U", "R", "R"],
):
    print(cleanSol(test))

This is now more flexible compared to the previous answer. For example it can now handle ["U2", "U", "U"] and ["U2", "U2"].

Answered By: S.B

Flexible List Cleaner

Question:

Answers: