如何在不更改第一个列表顺序的情况下找到n个最小的数字？

Question

I intend to get the n smallest numbers in a list but keep the numbers in the same order they appear in the list. For example:

This is my list:

A = [1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]

I like to get the first three lowest numbers as it has been ordered in the first list:

[1, 2, 0]

I do not want to sort the result as:

[0, 1, 2]

I have tried:

heapq.nsmallest(3,A)

but i wonder if it is possible to retain this list as:[1, 2, 0]

By the way, I'm not a Python coder so thanks for the help in advance.

Answer 1

You can try this:

new_a = []
A=[1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]
for a in A:
   if a not in new_a:
      new_a.append(a)
new_a = [i for i in new_a if i in sorted(new_a)[:3]]

Output:

[1, 2, 0]

Answer 2

You could use heapq.nsmallest() to get the n smallest elements from the list. Then use collections.Counter to create a multiset from that list which you can use to check which elements from the original list to include in the result, eg

>>> from heapq import nsmallest
>>> from collections import Counter
>>> A = [1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]
>>> n = 3    
>>> c = Counter(nsmallest(n, A))
>>> result = []
>>> for elem in A:
...     if c.get(elem, 0):
...         result.append(elem)
...         c[elem] -= 1
... 
>>> result
[1, 2, 0]