Try to find unique numbers in a long list, and return them in order. (prev. version did not work)-CodePudding

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?

CodePudding user response：

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

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