From 39430daa3b82a121b80633436170d0e0757f0fac Mon Sep 17 00:00:00 2001 From: Eli Bendersky Date: Sat, 14 Mar 2015 20:14:23 -0700 Subject: [PATCH] Issue #23549: Clarify confusion in heapq doc - accessing the mininmal element The current documentation only mentions heap[0] as the smallest element in the beginning, and not in any of the methods' docs. There's no method to access the minimal element without popping it, and the documentation of nsmallest is confusing because it may suggest that min() is the way to go for n==1. --- Doc/library/heapq.rst | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) diff --git a/Doc/library/heapq.rst b/Doc/library/heapq.rst index 43088ad9e37..f8970bed735 100644 --- a/Doc/library/heapq.rst +++ b/Doc/library/heapq.rst @@ -47,7 +47,8 @@ The following functions are provided: .. function:: heappop(heap) Pop and return the smallest item from the *heap*, maintaining the heap - invariant. If the heap is empty, :exc:`IndexError` is raised. + invariant. If the heap is empty, :exc:`IndexError` is raised. To access the + smallest item without popping it, use ``heap[0]``. .. function:: heappushpop(heap, item) @@ -112,7 +113,8 @@ The module also offers three general purpose functions based on heaps. The latter two functions perform best for smaller values of *n*. For larger values, it is more efficient to use the :func:`sorted` function. Also, when ``n==1``, it is more efficient to use the built-in :func:`min` and :func:`max` -functions. +functions. If repeated usage of these functions is required, consider turning +the iterable into an actual heap. Basic Examples