You can try using theheapqmodule.

import heapq numbers = [1, 3, 5, 2, 4, 1.1, 3.5, 4.8, 0.5, 2.4, -1.5] # 输出元祖第一个元素是index，第二元素是比较的数值 print(heapq.nsmallest(5, enumerate(numbers), key=lambda x: x[1])) # [(10, -1.5), (8, 0.5), (0, 1), (5, 1.1), (3, 2)] print(heapq.nlargest(5, enumerate(numbers), key=lambda x: x[1])) # [(2, 5), (7, 4.8), (4, 4), (6, 3.5), (1, 3)]

When the number of elements to be found is relatively small, functionsnlargest()和nsmallest()是很合适的。 如果你仅仅想查找唯一的最小或最大(N=1)的元素的话，那么使用min()和max()函数会更快些。 类似的，如果N的大小和集合大小接近的时候，通常先排序这个集合然后再使用切片操作会更快点(sorted(items)[:N]或者是sorted(items)[-N:] )。 需要在正确场合使用函数nlargest()和nsmallest()can take advantage of them (if N is close to the size of the set, it will be better to use sorting operations).

nums = [1,2,3,2,4,5,4,3,2,1,8,9,10,11,10,9,8] peaks = [] troughs = [] for idx in range(1, len(nums)-1): if nums[idx-1] < nums[idx] > nums[idx+1]: peaks.append((idx, nums[idx])) if nums[idx-1] > nums[idx] < nums[idx+1]: troughs.append((idx, nums[idx])) print(peaks) # [(2, 3), (5, 5), (13, 11)] print(troughs) # [(3, 2), (9, 1)]

First loop through to find all the peaks and troughs, and then find the top five