Cost of len() function

Question

What is the cost of len() function for Python built-ins? (list/tuple/string/dictionary)

Answer 1

It's O(1) (constant time, not depending of actual length of the element - very fast) on every type you've mentioned, plus set and others such as array.array.

Answer 2

Calling len() on those data types is O(1) in CPython, the official and most common implementation of the Python language. Here's a link to a table that provides the algorithmic complexity of many different functions in CPython:

TimeComplexity Python Wiki Page

Answer 3

All those objects keep track of their own length. The time to extract the length is small (O(1) in big-O notation) and mostly consists of [rough description, written in Python terms, not C terms]: look up "len" in a dictionary and dispatch it to the built_in len function which will look up the object's __len__ method and call that ... all it has to do is return self.length

Answer 4

The below measurements provide evidence that len() is O(1) for oft-used data structures.

A note regarding timeit: When the -s flag is used and two strings are passed to timeit the first string is executed only once and is not timed.

List:

$ python -m timeit -s "l = range(10);" "len(l)"
10000000 loops, best of 3: 0.0677 usec per loop

$ python -m timeit -s "l = range(1000000);" "len(l)"
10000000 loops, best of 3: 0.0688 usec per loop

Tuple:

$ python -m timeit -s "t = (1,)*10;" "len(t)"
10000000 loops, best of 3: 0.0712 usec per loop

$ python -m timeit -s "t = (1,)*1000000;" "len(t)"
10000000 loops, best of 3: 0.0699 usec per loop

String:

$ python -m timeit -s "s = '1'*10;" "len(s)"
10000000 loops, best of 3: 0.0713 usec per loop

$ python -m timeit -s "s = '1'*1000000;" "len(s)"
10000000 loops, best of 3: 0.0686 usec per loop

Dictionary (dictionary-comprehension available in 2.7+):

$ python -mtimeit -s"d = {i:j for i,j in enumerate(range(10))};" "len(d)"
10000000 loops, best of 3: 0.0711 usec per loop

$ python -mtimeit -s"d = {i:j for i,j in enumerate(range(1000000))};" "len(d)"
10000000 loops, best of 3: 0.0727 usec per loop

Array:

$ python -mtimeit -s"import array;a=array.array('i',range(10));" "len(a)"
10000000 loops, best of 3: 0.0682 usec per loop

$ python -mtimeit -s"import array;a=array.array('i',range(1000000));" "len(a)"
10000000 loops, best of 3: 0.0753 usec per loop

Set (set-comprehension available in 2.7+):

$ python -mtimeit -s"s = {i for i in range(10)};" "len(s)"
10000000 loops, best of 3: 0.0754 usec per loop

$ python -mtimeit -s"s = {i for i in range(1000000)};" "len(s)"
10000000 loops, best of 3: 0.0713 usec per loop

Deque:

$ python -mtimeit -s"from collections import deque;d=deque(range(10));" "len(d)"
100000000 loops, best of 3: 0.0163 usec per loop

$ python -mtimeit -s"from collections import deque;d=deque(range(1000000));" "len(d)"
100000000 loops, best of 3: 0.0163 usec per loop

Answer 5

len is an O(1) because in your RAM, lists are stored as tables (series of contiguous addresses). To know when the table stops the computer needs two things : length and start point. That is why len() is a O(1), the computer stores the value, so it just needs to look it up.

Answer 6

It is O(1) in CPython because length is derived from the size attribute on the Pyobject representing the list. See [1], [2] and [3] in that order:

[1]:

static PyObject *
listiter_len(_PyListIterObject *it, PyObject *Py_UNUSED(ignored))
{
    Py_ssize_t len;
    if (it->it_seq) {
        len = PyList_GET_SIZE(it->it_seq) - it->it_index;
        if (len >= 0)
            return PyLong_FromSsize_t(len);
    }
    return PyLong_FromLong(0);
}

[2]:

static inline Py_ssize_t PyList_GET_SIZE(PyObject *op) {
    PyListObject *list = _PyList_CAST(op);
    return Py_SIZE(list);
}

[3]

static inline Py_ssize_t Py_SIZE(PyObject *ob) {
    assert(ob->ob_type != &PyLong_Type);
    assert(ob->ob_type != &PyBool_Type);
    PyVarObject *var_ob = _PyVarObject_CAST(ob);
    return var_ob->ob_size;
}

[1] listiter_len

[2] PyList_GET_SIZE

[3] Py_SIZE