Reserve memory for list in Python?

Question

When programming in Python, is it possible to reserve memory for a list that will be populated with a known number of items, so that the list will not be reallocated several times while building it? I've looked through the docs for a Python list type, and have not found anything that seems to do this. However, this type of list building shows up in a few hotspots of my code, so I want to make it as efficient as possible.

Edit: Also, does it even make sense to do something like this in a language like Python? I'm a fairly experienced programmer, but new to Python and still getting a feel for its way of doing things. Does Python internally allocate all objects in separate heap spaces, defeating the purpose of trying to minimize allocations, or are primitives like ints, floats, etc. stored directly in lists?

Answer 1

Here's four variants:

• an incremental list creation
• "pre-allocated" list
• array.array()
• numpy.zeros()

 

python -mtimeit -s"N=10**6" "a = []; app = a.append;"\     "for i in xrange(N):  app(i);" 10 loops, best of 3: 390 msec per loop  python -mtimeit -s"N=10**6" "a = [None]*N; app = a.append;"\     "for i in xrange(N):  a[i] = i" 10 loops, best of 3: 245 msec per loop  python -mtimeit -s"from array import array; N=10**6" "a = array('i', [0]*N)"\     "for i in xrange(N):" "  a[i] = i" 10 loops, best of 3: 541 msec per loop  python -mtimeit -s"from numpy import zeros; N=10**6" "a = zeros(N,dtype='i')"\     "for i in xrange(N):" "  a[i] = i" 10 loops, best of 3: 353 msec per loop 

It shows that [None]*N is the fastest and array.array is the slowest in this case.

Answer 2

you can create list of the known length like this:

>>> [None] * known_number