在python中，将一个列表的每个相同元素分配给另一列表的相同索引的最佳方法是什么？

Question

I thought about having a word as a string, making it into a "regularList" of strings, generating a "dummyList" which contains a string '-'for each letter in the word , then looping through the "regularList", simultaneously removing each instance of my guessed letter from "regularList" and reassigning it to the same index of "dummyList". Basically, I need to make:

regularList = [['a', 'a', 'r', 'd', 'v', 'a', 'r', 'k']]
dummyList = ['_','_','_','_','_','_','_']

Into:

regularList = [['r', 'd', 'v', 'r', 'k']]
dummyList = ['a','a','_','_','_','a','_','_']

Here is my attempt:

word = 'aardvark'

def changeLetter(word):
    guess = raw_input('Guess a letter:') # When called, guess:a
    print word
    dummyList = []
    for i in word:
        dummyList.append('_ ')
    print dummyList
    regularList = [list(i) for i in word.split('\n')]
    print regularList
    numIters = 0
    while guess in regularList[0]:
        numIters += 1
        index = regularList[0].index(guess)
        dummyList[index] = guess
        del regularList[0][index]
    print regularList
    print dummyList
    print numIters


changeLetter(word)

This code produces:

Samuels-MacBook:python amonette$ python gametest.py
Guess a letter:a
aardvark
['_ ', '_ ', '_ ', '_ ', '_ ', '_ ', '_ ', '_ ']
[['a', 'a', 'r', 'd', 'v', 'a', 'r', 'k']]
[['r', 'd', 'v', 'r', 'k']]
['a', '_ ', '_ ', 'a', '_ ', '_ ', '_ ', '_ ']
3

As you can see, the proper indices aren't being reassigned.

Answer 1

word = 'aardvark'

def changeLetter(word):
    guess = raw_input('Guess a letter:') # When called, guess:a
    print word
    dummyList = []
    for i in word:
        dummyList.append('_ ')
    print dummyList
    regularList = [list(i) for i in word.split('\n')]
    print regularList
    numIters = 0
    position = 0
    length = len(regularList[0])
    while numIters < len(regularList[0]):
        if regularList[0][numIters] == guess:
            dummyList[position] = guess
            del regularList[0][numIters]
            numIters -=1
        position +=1
        numIters +=1
    print regularList
    print dummyList
    print numIters


changeLetter(word)

Your program has one mistake when you delete an element the size of array becomes small and the element which should be next becomes previous.

regularList[0] = ['a', 'a', 'r', 'd', 'v', 'a', 'r', 'k']

while guess in regularList[0]:

In this loop, when you remove first a , list becomes ['a', 'r', 'd', 'v', 'a', 'r', 'k']

Now when your loop continues, guess becomes 'r' that is next element in previous list. Hence a which was previously at position 1 is neglected (0 based indexing).

Answer 2

Maybe with enumerate and then remove the elements:

for index, element in enumerate(regularList):
    if element == "a":
        dummyList[index] = element
regularList.remove("a")

Answer 3

The accepted answer is way too long, you can just use a clean list comprehension

currentWord = [letter if letter in guessedLetters else "_" for letter in solution]

A fully functioning guess the word program without win logic would be 6 lines

solution = "aardvark"
guessedLetters = []
while True:
    guessedLetters.append(input("Guess a letter "))
    currentWord = [letter if letter in guessedLetters else "_" for letter in solution]
    print(" ".join(currentWord))

This outputs:

Guess a letter a
a a _ _ _ a _ _
Guess a letter k
a a _ _ _ a _ k
Guess a letter v
a a _ _ v a _ k
Guess a letter r
a a r _ v a r k
Guess a letter d
a a r d v a r k
Guess a letter