又出现循环问题

发布于 2025-01-16 13:59:12 字数 294 浏览 0 评论 0原文

我刚刚编写了一堆简单的代码，必须删除列表中的重复项，但有一个问题。有人可以解释一下为什么它会跳过数字 3 吗？

lis = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]
for i in lis:
    while lis.count(i) > 1:
        print(f'removing {i} ... ')
        lis.remove(i)
        print(lis.count(i))
print(lis)

原文

I just wrote a simple bunch of code, which had to remove duplicates in list, but there was an issue.
Can somebody explain me why it is skipping number 3 ?

lis = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]
for i in lis:
    while lis.count(i) > 1:
        print(f'removing {i} ... ')
        lis.remove(i)
        print(lis.count(i))
print(lis)

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原来是傀儡 2025-01-23 13:59:12

只需将您的列表转换为集合，然后再转换回列表即可：

listofthings = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]

newlist=list(set(listofthings))

print(newlist)

Just take your list, convert it to a set and then back to a list:

listofthings = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]

newlist=list(set(listofthings))

print(newlist)

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蓝天 2025-01-23 13:59:12

如果您不想删除所有 3 或所有 1，请不要使用 list.remove()

listofthings = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]


for things in listofthings:
    exists = 1
    obj = things
    cnt = 0
    for things in listofthings:
       cnt = cnt + 1
       dup = things
       if dup == obj:
          exists = exists + 1 
       if exists == 2:
           del listofthings[cnt]

dont use list.remove() if u dont want to remove all 3s or all 1s

listofthings = [1, 1, 3, 4, 5, 5, 5, 6, 1, 1, 3, 6, 6, 6, 1, 4, 7, 6, 5]


for things in listofthings:
    exists = 1
    obj = things
    cnt = 0
    for things in listofthings:
       cnt = cnt + 1
       dup = things
       if dup == obj:
          exists = exists + 1 
       if exists == 2:
           del listofthings[cnt]

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