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

我考虑具有字作为一个字符串，使得它变成一个字符串“regularList”，产生一“dummyList”，其中包含一个字符串“-'for在单词的每个字母，然后通过‘regularList’循环，同时除去各我从“ regularList”猜到的字母的实例，并将其重新分配给“ dummyList”的相同索引。 基本上，我需要做：

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

成：

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

这是我的尝试：

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)

此代码产生：

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

如您所见，没有重新分配适当的索引。

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)

删除元素时，程序会犯一个错误，即数组的大小变小，而下一个要变的元素将成为前一个。

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

while guess in regularList[0]:

在此循环中，当您首先删除a时，列表将变为['a'，'r'，'d'，'v'，'a'，'r'，'k']

现在，当您的循环继续进行时，猜测将变为“ r”，即上一个列表中的下一个元素。 因此，先前位于位置1的a被忽略（基于0的索引）。

Answer 2

也许用枚举然后删除元素：

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

Answer 3

接受的答案太长了，您可以使用一个整洁的列表理解

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

一个没有猜测逻辑的全功能单词程序将是6行

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))

输出：

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