How to use collections.deque most effectively (popleft vs. appendleft)

Question

I have been learning queues as I study data structures in Python and wanted to ask a question regarding its use.

I suppose there are two ways append/pop from a queue. First way would be to use deque.append() and deque.popleft(). Another way would be to use deque.appendleft() and deque.pop(). Is there a performance difference between the two? If not, which one is used more often from your experience? Do you recommend one over another for some other reason?

From my point of view, they essentially do the same thing as they both implement first-in-first-out. Your input would be much appreciated!

kaya3 · Accepted Answer · 2019-11-14T02:58:02.453

According to the docs:

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.

So there is no asymptotic difference between the two options; either way, your "enqueue" and "poll" operations are both done in constant time. This is to be expected, since the whole point of a deque (double-ended queue) is to have efficient add and remove operations on both sides.

Empirical results using timeit confirm there is basically no difference:

from collections import deque

def append_popleft():
    d = deque()
    for i in range(10000):
        d.append(i)
    for j in range(10000):
        d.popleft()

def appendleft_pop():
    d = deque()
    for i in range(10000):
        d.appendleft(i)
    for j in range(10000):
        d.pop()

import timeit
t = timeit.timeit(append_popleft, number=10000)
print('append / popleft:', t)
t = timeit.timeit(appendleft_pop, number=10000)
print('appendleft / pop:', t)

Output:

append / popleft: 12.000681700999849
appendleft / pop: 11.937629571999423

ggorlen · Answer 2 · 2019-11-14T03:07:01.907

It doesn't matter which end you use. Both will be equally optimal. Unlike a list, which only offers O(1) append and pop operations on the right/top of the list, append and pop operations are guaranteed to be O(1) on both ends of the deque. If they weren't, there'd be no point to its existence and we might as well use a list.

Looking at the source code for CPython 3.8, the methods pop and popleft are almost identical:

static PyObject *
deque_pop(dequeobject *deque, PyObject *unused)
{
    PyObject *item;
    block *prevblock;

    if (Py_SIZE(deque) == 0) {
        PyErr_SetString(PyExc_IndexError, "pop from an empty deque");
        return NULL;
    }
    item = deque->rightblock->data[deque->rightindex];
    deque->rightindex--;
    Py_SIZE(deque)--;
    deque->state++;

    if (deque->rightindex < 0) {
        if (Py_SIZE(deque)) {
            prevblock = deque->rightblock->leftlink;
            assert(deque->leftblock != deque->rightblock);
            freeblock(deque->rightblock);
            CHECK_NOT_END(prevblock);
            MARK_END(prevblock->rightlink);
            deque->rightblock = prevblock;
            deque->rightindex = BLOCKLEN - 1;
        } else {
            assert(deque->leftblock == deque->rightblock);
            assert(deque->leftindex == deque->rightindex+1);
            /* re-center instead of freeing a block */
            deque->leftindex = CENTER + 1;
            deque->rightindex = CENTER;
        }
    }
    return item;
}

static PyObject *
deque_popleft(dequeobject *deque, PyObject *unused)
{
    PyObject *item;
    block *prevblock;

    if (Py_SIZE(deque) == 0) {
        PyErr_SetString(PyExc_IndexError, "pop from an empty deque");
        return NULL;
    }
    assert(deque->leftblock != NULL);
    item = deque->leftblock->data[deque->leftindex];
    deque->leftindex++;
    Py_SIZE(deque)--;
    deque->state++;

    if (deque->leftindex == BLOCKLEN) {
        if (Py_SIZE(deque)) {
            assert(deque->leftblock != deque->rightblock);
            prevblock = deque->leftblock->rightlink;
            freeblock(deque->leftblock);
            CHECK_NOT_END(prevblock);
            MARK_END(prevblock->leftlink);
            deque->leftblock = prevblock;
            deque->leftindex = 0;
        } else {
            assert(deque->leftblock == deque->rightblock);
            assert(deque->leftindex == deque->rightindex+1);
            /* re-center instead of freeing a block */
            deque->leftindex = CENTER + 1;
            deque->rightindex = CENTER;
        }
    }
    return item;
}

And the same is true for appends:

static inline int
deque_append_internal(dequeobject *deque, PyObject *item, Py_ssize_t maxlen)
{
    if (deque->rightindex == BLOCKLEN - 1) {
        block *b = newblock();
        if (b == NULL)
            return -1;
        b->leftlink = deque->rightblock;
        CHECK_END(deque->rightblock->rightlink);
        deque->rightblock->rightlink = b;
        deque->rightblock = b;
        MARK_END(b->rightlink);
        deque->rightindex = -1;
    }
    Py_SIZE(deque)++;
    deque->rightindex++;
    deque->rightblock->data[deque->rightindex] = item;
    if (NEEDS_TRIM(deque, maxlen)) {
        PyObject *olditem = deque_popleft(deque, NULL);
        Py_DECREF(olditem);
    } else {
        deque->state++;
    }
    return 0;
}

static inline int
deque_appendleft_internal(dequeobject *deque, PyObject *item, Py_ssize_t maxlen)
{
    if (deque->leftindex == 0) {
        block *b = newblock();
        if (b == NULL)
            return -1;
        b->rightlink = deque->leftblock;
        CHECK_END(deque->leftblock->leftlink);
        deque->leftblock->leftlink = b;
        deque->leftblock = b;
        MARK_END(b->leftlink);
        deque->leftindex = BLOCKLEN;
    }
    Py_SIZE(deque)++;
    deque->leftindex--;
    deque->leftblock->data[deque->leftindex] = item;
    if (NEEDS_TRIM(deque, deque->maxlen)) {
        PyObject *olditem = deque_pop(deque, NULL);
        Py_DECREF(olditem);
    } else {
        deque->state++;
    }
    return 0;
}

Deques in CPython allocate and deallocate memory in blocks to minimize heap access and improve cache locality (sequential elements in an array are likely to belong to the same cache block). These methods all allocate or deallocate memory as necessary and increment/decrement a few indices. All constant time.

How to use collections.deque most effectively (popleft vs. appendleft)

2 Answers2