CODEWITHSUNDEEP
pythonlistpriority-queue
user1050619
user1050619

Reputation: 20856

heapq module python

I'm using the heapq module to determine the smallest item in a list.

I have this below code, but the heapq.heapify() return value is None.

How do I get the result in a new list?

>>> a=heapq.heapify(lista)
>>> a
>>> lista=[1,2,3,4,5]
>>> a=heapq.heapify(lista)
>>> print(a)
None

Upvotes: 6

Views: 2674

Answers (2)

Mayank Maheshwari
Mayank Maheshwari

Reputation: 303

You have to use heapq.nsmallest(k,arr) function. //k here is 1 as you want one minimum value.

>>> a=heapq.heapify(lista)
>>> a
>>> lista=[1,2,3,4,5]
>>> a=heapq.nsmallest(1,lista)
>>> print(a)
1

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121814

heapq.heapify doesn't return anything, it heapifies the list in place; it's far more efficient to do it that way:

>>> import heapq
>>> lista = [44, 42, 3, 89, 10]
>>> heapq.heapify(lista)
>>> lista
[3, 10, 44, 89, 42]

If you need a new list, create a copy first:

>>> lista = [44, 42, 3, 89, 10]
>>> newlist = lista[:]
>>> heapq.heapify(newlist)
>>> lista
[44, 42, 3, 89, 10]
>>> newlist
[3, 10, 44, 89, 42]

That defeats the purpose somewhat, of course, as copying the list has a (linear) cost too.

If all you need is the smallest item in a list, the min() function will be just as fast when locating just the one smallest element (both heapify() and min() scan the input list once, so O(n) cost):

>>> min(lista)
3

If you need more than one smallest value, by all means use a heapq, especially if you add items later on. If you cannot alter the original list, need several smallest items, see looking for an inverted heap in python for an efficient nsmallest implementation that creates a new heap from an input heap with only a fixed number of smallest values.

Upvotes: 10

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