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).

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

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