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Cost of len() function

Posted by: admin November 1, 2017 Leave a comment

Questions:

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

Answers:

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.

Questions:
Answers:

Calling len() on those data types is O(1) in CPython, the 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

Questions:
Answers:

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

Questions:
Answers:

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