Also available in python: import heapq n = 2 heapq.nlargest(n, vals)[n-1] Edit: and out of morbid curiosity, here is the same implemented by a scan. It

Also available in python: import heapq n = 2 heapq.nlargest(n, vals)[n-1] Edit: and out of morbid curiosity, here is the same implemented by a scan. It could be arranged as a single statement. swapIfGt = lambda xs, q: xs if xs[0] >= q else sorted([q]+xs[1:]) n = 2 reduce(swapIfGt, vals, [None]*n)[0] Edit2: rewrote swapIfGt