How to peek front of deque without popping?

Question

I want to check a condition against the front of a queue before deciding whether or not to pop. How can I achieve this in python with collections.deque?

list(my_deque)[0]

seems ugly and poor for performance.

Asked By: corinjg

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Source

Answer 1

TL;DR: assuming your deque is called d, just inspect d[0], since the "leftmost" element in a deque is the front (you might want to test before the length of the deque to make sure it’s not empty). Taking @asongtoruin’s suggestion, use if d: to test whether the deque is empty (it’s equivalent to if len(d) == 0:, but more pythonic)

###Why not converting to list?
Because deques are indexable and you’re testing the front. While a deque has an interface similar to a list, the implementation is optimized for front- and back- operations. Quoting the documentation:

Deques support thread-safe, memory efficient appends and pops from
either side of the deque with approximately the same O(1) performance
in either direction.

Though list objects support similar operations, they are optimized for
fast fixed-length operations and incur O(n) memory movement costs for
pop(0) and insert(0, v) operations which change both the size and
position of the underlying data representation.

Converting to list might be desirable if you have lots of operations accessing the "middle" of the queue. Again quoting the documentation:

Indexed access is O(1) at both ends but slows to O(n) in the middle.
For fast random access, use lists instead.

Conversion to list is O(n), but every subsequent access is O(1).

Answered By: GPhilo

Answer 2

You can simply find the last element using my_deque[-1] or my_deque[len(my_deque)-1] .

Answered By: tsukyonomi06

Answer 3

Here is a simple implementation that allowed me to check the front of the queue before popping (using while and q[0]):

Apply your own condition against q[0], before q.popleft(), below:

testLst = [100,200,-100,400,340]
q=deque(testLst)

while q:
    print(q)
    print('{}{}'.format("length of queue: ", len(q)))
    print('{}{}'.format("head: ", q[0]))
    print()

    q.popleft()

output:

deque([100, 200, -100, 400, 340])
length of queue: 5
head: 100

deque([200, -100, 400, 340])
length of queue: 4
head: 200

deque([-100, 400, 340])
length of queue: 3
head: -100

deque([400, 340])
length of queue: 2
head: 400

deque([340])
length of queue: 1
head: 340

Answered By: Grant Shannon

Answer 4

Assuming your deque is implemented from collections python

from collections import deque
deque = deque() //syntax

Deque too can be interpreted as a list in terms of accessing using indices.
You can peek front element by using deque[0] and peek last using deque[-1]
This works without popping elements from left or right and seems efficient too.

Answered By: vijay9908

How to peek front of deque without popping?

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