Question

第n条评论被删除，第n个用户名在其他评论中被检查，如果第n个用户名不在所有评论中（第n条评论已被删除），它将在warn_usernames中传递/附加，之后，第n条评论获得再次在评论中添加。循环继续

usernames = ["a", "b", "c"]
warn_usernames = []
comments = [["b"], ["a", "c"], ["a", "b"]]

n = 0
while n < len(usernames):
    commente = comments[n]
    print(commente)
    usernamee = usernames[n]
    print(usernamee)
    comments.pop(n)
    if str(usernamee) not in str(comments):
       warn_usernames.append(usernamee)
    comments.insert(n, commente)
    n = n + 1
print(warn_usernames)

结果我得到了

[]

结果我希望得到

[c]

提前致谢，我希望你能得到我的问题。

Answer 1

我认为你可以这样做：

usernames = ["a", "b", "c"]
warn_usernames = []
comments = [["b"], ["a", "c"], ["a", "b"]]
# use a loop for with enumerate on usernames to get a username and its index
for n, username in enumerate(usernames):
    #loop on all others comments than the nth
    for comments_to_check in comments[:n]+comments[n+1:]:
        #check the username is in comments_to_check and 
        if username not in comments_to_check:
            # append to warn_usernames if not
            warn_usernames.append(username)
            # stop the loop on comments to go to the next loop on usernames
            # to prevent multiple time the same username to be appened
            break 
print warn_usernames

您将获得预期的输出

附加和删除以检查文本是否在列表中

1 个答案: