"""Unittests for heapq.""" from test.test_support import verify, vereq, verbose, TestFailed from heapq import heappush, heappop, heapify import random def check_invariant(heap): # Check the heap invariant. for pos, item in enumerate(heap): if pos: # pos 0 has no parent parentpos = (pos-1) >> 1 verify(heap[parentpos] <= item) # An iterator returning a heap's elements, smallest-first. class heapiter(object): def __init__(self, heap): self.heap = heap def next(self): try: return heappop(self.heap) except IndexError: raise StopIteration def __iter__(self): return self def test_main(): # 1) Push 100 random numbers and pop them off, verifying all's OK. heap = [] data = [] check_invariant(heap) for i in range(256): item = random.random() data.append(item) heappush(heap, item) check_invariant(heap) results = [] while heap: item = heappop(heap) check_invariant(heap) results.append(item) data_sorted = data[:] data_sorted.sort() vereq(data_sorted, results) # 2) Check that the invariant holds for a sorted array check_invariant(results) # 3) Naive "N-best" algorithm heap = [] for item in data: heappush(heap, item) if len(heap) > 10: heappop(heap) heap.sort() vereq(heap, data_sorted[-10:]) # 4) Test heapify. for size in range(30): heap = [random.random() for dummy in range(size)] heapify(heap) check_invariant(heap) # 5) Less-naive "N-best" algorithm, much faster (if len(data) is big # enough ) than sorting all of data. However, if we had a max # heap instead of a min heap, it could go faster still via # heapify'ing all of data (linear time), then doing 10 heappops # (10 log-time steps). heap = data[:10] heapify(heap) for item in data[10:]: if item > heap[0]: # this gets rarer the longer we run heappop(heap) # we know heap[0] isn't in best 10 anymore heappush(heap, item) vereq(list(heapiter(heap)), data_sorted[-10:]) # Make user happy if verbose: print "All OK" if __name__ == "__main__": test_main()