在列表中找到第n个最小的数字？

Question

i need a efficient way of getting the nth smallest number AND its index in a list containing up to 15000 enties (so speed is not super crucial).

I sadly can't use numpy or any other non-standard library.

Im using Python 2.7

Answer 1

use heapq.nsmallest (and enumerate to get the index):

nums = [random.randint(1,1000000) for _ in range(10000)]

import heapq
import operator

heapq.nsmallest(10,enumerate(nums),key=operator.itemgetter(1))
Out[26]: 
[(5544, 35),
 (1702, 43),
 (6547, 227),
 (1540, 253),
 (4919, 360),
 (7993, 445),
 (1608, 495),
 (5832, 505),
 (1388, 716),
 (5103, 814)]

Answer 2

Use enumerate to store the original position, then sort by the value:

original_list = [36, 44, 23, 24, 47, 19, 49, 36, 30, 3]
sorted_lookup = sorted(enumerate(original_list), key=lambda i:i[1])
position, value = sorted_lookup[4] # 4 is my "n"

sorted_lookup in my example is:

[(9, 3), (5, 19), (2, 23), (3, 24), (8, 30), (0, 36), (7, 36), (1, 44), (4, 47), (6, 49)]

And position is 8 , value is 30