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.

diff --git a/Doc/library/heapq.rst b/Doc/library/heapq.rst
index 43088ad9e37881e244b4882a764be25766dd5529..f8970bed735e395e8d32c91e9f6a16d0cf140081 100644 (file)
--- 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