How to refactor this function to reduce its cognitive complexity issue for using of multiple if-elif-else statements in python?

Question

My question is about how to refactor if-elif-else conditions in the code below.

Piece of code has been changed here due to new implementation.Here V, X, Y are numbers. This code is for generating a random strings.

def get_uid(type, key):

    if type == 'orange_fruit':
        if key == 'juice':
            uid = 'zofj' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zofs' + uuidV().hex[:X]
        else:
            uid = 'zofx' + uuidV().hex[:X]
        return uid
    elif type == 'grape':
        if key == 'juice':
            uid = 'zzgj' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zzgs' + uuidV().hex[:X]
        else:
            uid = 'zzgx' + uuidV().hex[:X]
        return uid
    elif type == 'kiwi_healthy':
        if key == 'juice':
            uid = 'zkij' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zkis' + uuidV().hex[:X]
        else:
            uid = 'zkix' + uuidV().hex[:X]
        return uid
    elif type == 'anar_tasty':
        if key == 'juice':
            uid = 'zatj' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zats' + uuidV().hex[:X]
        else:
            uid = 'yx' + uuidV().hex[:X]
        return uid
    elif type == 'apple':
        if key == 'juice':
            uid = 'zppj' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zpps' + uuidV().hex[:X]
        else:
            uid = 'zppx' + uuidV().hex[:X]
        return uid
    elif type == 'cherry':
        if key == 'juice':
            uid = 'zchj' + uuidV().hex[:X]
        elif key == 'salad':
            uid = 'zchs' + uuidV().hex[:X]
        else:
            uid = 'zchx' + uuidV().hex[:X]
        return uid

    return uuidV().hex[:Y]

Thanks in advance.

Answer 1

Well, in terms of simplicity...

def get_uid(my_env, my_type):
    i = my_env[-1]
    t = type[0:2]
    uid = f'z{t}t{i}z' + uuid4().hex[:6]
    
    return uid

Btw I can't really understand the condition for the last return statement. You could Assign None to my_env and my_type and if anyone is None return it:

def get_uid(my_env=None, my_type=None):
    if my_env == None or my_type == None:
        return uuid4().hex[:18]
    else:
        i = my_env[-1]
        t = type[0:2]
        uid = f'z{t}t{i}z' + uuid4().hex[:6]
    
        return uid

Answer 2

I assume that you put 3 in zkit3z by mistake. here should be zkitnz. If I'm right then the following code doing the same functionality

def get_uid(type, env):
    if type in ["apple", "grape", "kiwi"]:
        if env in ["test1", "test2"]:
            uid = "z"+ type[:2]+ "t"+ env[-1]+"z" + uuid4().hex[:x]
        else:
            uid = "z" + type[:2] + "tnz" + uuid4().hex[:x]
        return uid
    return uuid4().hex[:xx]

if 3 must be in zkit3z then the following code should proceed

def get_uid(type, env):
    if type in ["apple", "grape", "kiwi"]:
        if env in ["test1", "test2"]:
            uid = "z"+ type[:2]+ "t"+ env[-1]+"z" + uuid4().hex[:x]
        else:
            if type != "kiwi":
                uid = "z" + type[:2] + "tnz" + uuid4().hex[:x]
            else:
                uid = "z" + type[:2] + "t3z" + uuid4().hex[:x]
        return uid
    return uuid4().hex[:xx]