尝试在长列表中找到唯一的数字，然后按顺序返回。（上一个版本不起作用）

发布于 2025-02-12 08:52:18 字数 501 浏览 1 评论 0原文

您可能会认为这个问题是上述的。回答，但最后一个版本与2位数字无效给定数字列表，查找并打印仅出现一次的元素。这些元素应按照原始列表中出现的顺序打印。

因此，这是我的代码（由其他pythoneer修复），

a = [int(s) for s in input().split()]
sortedLst = sorted(a)
unique = []
uniqueOrd = []

for i in range(len(a) - 2):
     if sortedLst[i + 1] != sortedLst[i] and sortedLst[i + 1] != sortedLst[i + 2]:
            unique.append(sortedLst[i + 1])

for num in a:
    if num in unique:
        uniqueOrd.append(num)

print(*uniqueOrd)

但是输出不适合2位数字编号，我该如何修复此错误？

原文

You may consider this question is prev. answered, but the last version doesn't work with 2 digits number
Given a list of numbers, find and print the elements that appear in it only once. Such elements should be printed in the order in which they occur in the original list.

so here's my code (being fixed by other pythoneers)

a = [int(s) for s in input().split()]
sortedLst = sorted(a)
unique = []
uniqueOrd = []

for i in range(len(a) - 2):
     if sortedLst[i + 1] != sortedLst[i] and sortedLst[i + 1] != sortedLst[i + 2]:
            unique.append(sortedLst[i + 1])

for num in a:
    if num in unique:
        uniqueOrd.append(num)

print(*uniqueOrd)

but the output does not work with 2 digits number, how can I fix this bug?

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青春如此纠结 2025-02-19 08:52:18

您不必为从长列表中找到唯一而努力工作，只需考虑collections - 计数器的nice lib即可。它更快，更柔软。

A = [int(s) for s in input().split()]
sortedLst = sorted(A)         # this is just to confirm your inputs..

from collections import Counter

counts = Counter(A)
print(counts)

uniques = [k for k, v in counts.items() if v == 1] # just check the frequency (v) 

print(uniques)

尝试一些输入，例如：2 3 1 4 22 33 44 55 33 22

[2, 3, 1, 4, 44, 55]

You don't have to labor yourself on finding the uniques from a long list, just consider the nice lib from collections - Counter. It's faster and more Pythonic.

A = [int(s) for s in input().split()]
sortedLst = sorted(A)         # this is just to confirm your inputs..

from collections import Counter

counts = Counter(A)
print(counts)

uniques = [k for k, v in counts.items() if v == 1] # just check the frequency (v) 

print(uniques)

Try some inputs like: 2 3 1 4 22 33 44 55 33 22