Question

我正在研究如何从列表中删除特定项目。

"peppers", "cheese", "mushrooms", "bananas", "peppers"

我可以找到项目“辣椒”，并将其更改为“已消失！”，但我真的想删除该项目，使用

del blist[idx]

但这会导致错误，我不知道为什么。

myList = ["peppers", "cheese", "mushrooms", "bananas", "peppers"]

def findInList (needle, haystack):
    needle = needle.lower()
    findy = []
    # loops over list elements
    for i in range(0, len(haystack)):
        temp = haystack[i].lower()
        idx = temp.find(needle)
        if (idx != -1): findy.append(i)
    return findy

def deleteItemInList(alist, blist):
    for i in range(0, len(alist)):
        idx = alist[i]
        blist[idx] = "Gone!"
        # del blist[idx]

# find items in list
mySearch = findInList("Peppers", myList)

# remove item from list
deleteItemInList(mySearch, myList)

print myList

追溯：如下

Traceback (most recent call last):
  File "delete_in_list.py", line 23, in <module>
    deleteItemInList(mySearch, myList)
  File "delete_in_list.py", line 16, in deleteItemInList
    blist[idx] = "Gone!"
IndexError: list assignment index out of range

有人可以查看上面的代码并指出我出错的地方。

Answer 1

您可以使用列表推导。

def removeWord(needle, haystack):
    return [word for word in haystack if word.lower() != needle.lower()]

Answer 2

要查找元素，请使用此功能。或者只是像往常一样定义它：

>>> find = lambda _list, query: [item.lower() for item in _list].index(query.lower())

>>> l = ['red', 'pepper']
>>> q = 'Pepper'
>>> find(l, q)
1

要按索引删除，只需使用del：

>>> del l[find(l, q)]
>>> l
['red']

Answer 3

我终于明白了！在列表中迭代删除列表中的项目时，我实际上是在锯掉我正坐在的分支上。您需要反向循环遍历列表：

def deleteItemInList(alist, blist):
    for i in range(len(alist) -1, -1, -1):
        idx = alist[i]
        del blist[idx]

从列表中删除某些项目

3 个答案: