Question

我对为修改作为输入的列表而编写的两个函数的行为感到有点困惑。在我看来，两者都应该做同样的事情，但事实并非如此。看看：

def headerFormat1(list1):
    char_to_remove = ['(', ')']
    list2 = list1[:]
    list1 = []
    for headers in list2:
        headers = headers.replace(' ', '_')
        for character in char_to_remove:
            headers = headers.replace(character, '')
        list1.append(headers)
    del list2
    return list1

def headerFormat2(list1):
    char_to_remove = ['(', ')']
    for i, headers in enumerate(list1):
        list1[i] = headers.replace(' ', '_')
        for character in char_to_remove:
            list1[i] = headers.replace(character, '')
    return list1

theList = ['I want Underscores Instead', 'R(e(m(o(v(e()P)a)r)e)n)t)h)e)s)e)s', 'LetMeBe']
print(headerFormat1(theList)) #prints: ['I_want_Underscores_Instead', 'RemoveParentheses', 'LetMeBe']
print(headerFormat2(theList)) #prints: ['I want Underscores Instead', 'R(e(m(o(v(e(Parentheses', 'LetMeBe']

我的意思是好的，第一个工作，因为它应该是altough我发现列表重复冗余，可能没有必要\ overkill（？）

但是第二个......除去括号后的奇怪之处是什么？此外，即使它不能正常工作，它也表明可以就地修改列表元素（删除＆＃39;）然后为什么不替换空格？

感谢。

Answer 1

您每次使用原始标题更改list1 [i] 检查以下代码是否有效

def headerFormat1(list1):
    char_to_remove = ['(', ')']
    list2 = list1[:]
    list1 = []
    for headers in list2:
        headers = headers.replace(' ', '_')
        for character in char_to_remove:
            headers = headers.replace(character, '')
        list1.append(headers)
    del list2
    return list1

def headerFormat2(list1):
    char_to_remove = ['(', ')']
    for i, headers in enumerate(list1):
        headers = headers.replace(' ', '_')
        for character in char_to_remove:
            headers = headers.replace(character, '')
        list1[i]=headers
    return list1

theList = ['I want Underscores Instead', 'R(e(m(o(v(e()P)a)r)e)n)t)h)e)s)e)s', 'LetMeBe']
print(headerFormat1(theList))
print(headerFormat2(theList))

修改列表值

1 个答案: