Create a list with initial capacity in Python

Code like this often happens:

l = []
while foo:
    # baz
    l.append(bar)
    # qux

This is really slow if you're about to append thousands of elements to your list, as the list will have to be constantly resized to fit the new elements.

In Java, you can create an ArrayList with an initial capacity. If you have some idea how big your list will be, this will be a lot more efficient.

I understand that code like this can often be refactored into a list comprehension. If the for/while loop is very complicated, though, this is unfeasible. Is there an equivalent for us Python programmers?


Asked by: Chloe649 | Posted: 05-10-2021






Answer 1

Warning: This answer is contested. See comments.

def doAppend( size=10000 ):
    result = []
    for i in range(size):
        message= "some unique object %d" % ( i, )
        result.append(message)
    return result

def doAllocate( size=10000 ):
    result=size*[None]
    for i in range(size):
        message= "some unique object %d" % ( i, )
        result[i]= message
    return result

Results. (evaluate each function 144 times and average the duration)

simple append 0.0102
pre-allocate  0.0098

Conclusion. It barely matters.

Premature optimization is the root of all evil.

Answered by: Adelaide311 | Posted: 06-11-2021



Answer 2

Python lists have no built-in pre-allocation. If you really need to make a list, and need to avoid the overhead of appending (and you should verify that you do), you can do this:

l = [None] * 1000 # Make a list of 1000 None's
for i in xrange(1000):
    # baz
    l[i] = bar
    # qux

Perhaps you could avoid the list by using a generator instead:

def my_things():
    while foo:
        #baz
        yield bar
        #qux

for thing in my_things():
    # do something with thing

This way, the list isn't every stored all in memory at all, merely generated as needed.

Answered by: Paul137 | Posted: 06-11-2021



Answer 3

Short version: use

pre_allocated_list = [None] * size

to preallocate a list (that is, to be able to address 'size' elements of the list instead of gradually forming the list by appending). This operation is very fast, even on big lists. Allocating new objects that will be later assigned to list elements will take much longer and will be the bottleneck in your program, performance-wise.

Long version:

I think that initialization time should be taken into account.

Since in Python everything is a reference, it doesn't matter whether you set each element into None or some string - either way it's only a reference. Though it will take longer if you want to create a new object for each element to reference.

For Python 3.2:

import time
import copy

def print_timing (func):
  def wrapper (*arg):
    t1 = time.time()
    res = func (*arg)
    t2 = time.time ()
    print ("{} took {} ms".format (func.__name__, (t2 - t1) * 1000.0))
    return res

  return wrapper

@print_timing
def prealloc_array (size, init = None, cp = True, cpmethod = copy.deepcopy, cpargs = (), use_num = False):
  result = [None] * size
  if init is not None:
    if cp:
      for i in range (size):
          result[i] = init
    else:
      if use_num:
        for i in range (size):
            result[i] = cpmethod (i)
      else:
        for i in range (size):
            result[i] = cpmethod (cpargs)
  return result

@print_timing
def prealloc_array_by_appending (size):
  result = []
  for i in range (size):
    result.append (None)
  return result

@print_timing
def prealloc_array_by_extending (size):
  result = []
  none_list = [None]
  for i in range (size):
    result.extend (none_list)
  return result

def main ():
  n = 1000000
  x = prealloc_array_by_appending(n)
  y = prealloc_array_by_extending(n)
  a = prealloc_array(n, None)
  b = prealloc_array(n, "content", True)
  c = prealloc_array(n, "content", False, "some object {}".format, ("blah"), False)
  d = prealloc_array(n, "content", False, "some object {}".format, None, True)
  e = prealloc_array(n, "content", False, copy.deepcopy, "a", False)
  f = prealloc_array(n, "content", False, copy.deepcopy, (), False)
  g = prealloc_array(n, "content", False, copy.deepcopy, [], False)

  print ("x[5] = {}".format (x[5]))
  print ("y[5] = {}".format (y[5]))
  print ("a[5] = {}".format (a[5]))
  print ("b[5] = {}".format (b[5]))
  print ("c[5] = {}".format (c[5]))
  print ("d[5] = {}".format (d[5]))
  print ("e[5] = {}".format (e[5]))
  print ("f[5] = {}".format (f[5]))
  print ("g[5] = {}".format (g[5]))

if __name__ == '__main__':
  main()

Evaluation:

prealloc_array_by_appending took 118.00003051757812 ms
prealloc_array_by_extending took 102.99992561340332 ms
prealloc_array took 3.000020980834961 ms
prealloc_array took 49.00002479553223 ms
prealloc_array took 316.9999122619629 ms
prealloc_array took 473.00004959106445 ms
prealloc_array took 1677.9999732971191 ms
prealloc_array took 2729.999780654907 ms
prealloc_array took 3001.999855041504 ms
x[5] = None
y[5] = None
a[5] = None
b[5] = content
c[5] = some object blah
d[5] = some object 5
e[5] = a
f[5] = []
g[5] = ()

As you can see, just making a big list of references to the same None object takes very little time.

Prepending or extending takes longer (I didn't average anything, but after running this a few times I can tell you that extending and appending take roughly the same time).

Allocating new object for each element - that is what takes the most time. And S.Lott's answer does that - formats a new string every time. Which is not strictly required - if you want to preallocate some space, just make a list of None, then assign data to list elements at will. Either way it takes more time to generate data than to append/extend a list, whether you generate it while creating the list, or after that. But if you want a sparsely-populated list, then starting with a list of None is definitely faster.

Answered by: Lyndon971 | Posted: 06-11-2021



Answer 4

The Pythonic way for this is:

x = [None] * numElements

Or whatever default value you wish to prepopulate with, e.g.

bottles = [Beer()] * 99
sea = [Fish()] * many
vegetarianPizzas = [None] * peopleOrderingPizzaNotQuiche

(Caveat Emptor: The [Beer()] * 99 syntax creates one Beer and then populates an array with 99 references to the same single instance)

Python's default approach can be pretty efficient, although that efficiency decays as you increase the number of elements.

Compare

import time

class Timer(object):
    def __enter__(self):
        self.start = time.time()
        return self

    def __exit__(self, *args):
        end = time.time()
        secs = end - self.start
        msecs = secs * 1000  # Millisecs
        print('%fms' % msecs)

Elements   = 100000
Iterations = 144

print('Elements: %d, Iterations: %d' % (Elements, Iterations))


def doAppend():
    result = []
    i = 0
    while i < Elements:
        result.append(i)
        i += 1

def doAllocate():
    result = [None] * Elements
    i = 0
    while i < Elements:
        result[i] = i
        i += 1

def doGenerator():
    return list(i for i in range(Elements))


def test(name, fn):
    print("%s: " % name, end="")
    with Timer() as t:
        x = 0
        while x < Iterations:
            fn()
            x += 1


test('doAppend', doAppend)
test('doAllocate', doAllocate)
test('doGenerator', doGenerator)

with

#include <vector>
typedef std::vector<unsigned int> Vec;

static const unsigned int Elements = 100000;
static const unsigned int Iterations = 144;

void doAppend()
{
    Vec v;
    for (unsigned int i = 0; i < Elements; ++i) {
        v.push_back(i);
    }
}

void doReserve()
{
    Vec v;
    v.reserve(Elements);
    for (unsigned int i = 0; i < Elements; ++i) {
        v.push_back(i);
    }
}

void doAllocate()
{
    Vec v;
    v.resize(Elements);
    for (unsigned int i = 0; i < Elements; ++i) {
        v[i] = i;
    }
}

#include <iostream>
#include <chrono>
using namespace std;

void test(const char* name, void(*fn)(void))
{
    cout << name << ": ";

    auto start = chrono::high_resolution_clock::now();
    for (unsigned int i = 0; i < Iterations; ++i) {
        fn();
    }
    auto end = chrono::high_resolution_clock::now();

    auto elapsed = end - start;
    cout << chrono::duration<double, milli>(elapsed).count() << "ms\n";
}

int main()
{
    cout << "Elements: " << Elements << ", Iterations: " << Iterations << '\n';

    test("doAppend", doAppend);
    test("doReserve", doReserve);
    test("doAllocate", doAllocate);
}

On my Windows 7 Core i7, 64-bit Python gives

Elements: 100000, Iterations: 144
doAppend: 3587.204933ms
doAllocate: 2701.154947ms
doGenerator: 1721.098185ms

While C++ gives (built with Microsoft Visual C++, 64-bit, optimizations enabled)

Elements: 100000, Iterations: 144
doAppend: 74.0042ms
doReserve: 27.0015ms
doAllocate: 5.0003ms

C++ debug build produces:

Elements: 100000, Iterations: 144
doAppend: 2166.12ms
doReserve: 2082.12ms
doAllocate: 273.016ms

The point here is that with Python you can achieve a 7-8% performance improvement, and if you think you're writing a high-performance application (or if you're writing something that is used in a web service or something) then that isn't to be sniffed at, but you may need to rethink your choice of language.

Also, the Python code here isn't really Python code. Switching to truly Pythonesque code here gives better performance:

import time

class Timer(object):
    def __enter__(self):
        self.start = time.time()
        return self

    def __exit__(self, *args):
        end = time.time()
        secs = end - self.start
        msecs = secs * 1000  # millisecs
        print('%fms' % msecs)

Elements   = 100000
Iterations = 144

print('Elements: %d, Iterations: %d' % (Elements, Iterations))


def doAppend():
    for x in range(Iterations):
        result = []
        for i in range(Elements):
            result.append(i)

def doAllocate():
    for x in range(Iterations):
        result = [None] * Elements
        for i in range(Elements):
            result[i] = i

def doGenerator():
    for x in range(Iterations):
        result = list(i for i in range(Elements))


def test(name, fn):
    print("%s: " % name, end="")
    with Timer() as t:
        fn()


test('doAppend', doAppend)
test('doAllocate', doAllocate)
test('doGenerator', doGenerator)

Which gives

Elements: 100000, Iterations: 144
doAppend: 2153.122902ms
doAllocate: 1346.076965ms
doGenerator: 1614.092112ms

(in 32-bit, doGenerator does better than doAllocate).

Here the gap between doAppend and doAllocate is significantly larger.

Obviously, the differences here really only apply if you are doing this more than a handful of times or if you are doing this on a heavily loaded system where those numbers are going to get scaled out by orders of magnitude, or if you are dealing with considerably larger lists.

The point here: Do it the Pythonic way for the best performance.

But if you are worrying about general, high-level performance, Python is the wrong language. The most fundamental problem being that Python function calls has traditionally been up to 300x slower than other languages due to Python features like decorators, etc. (PythonSpeed/PerformanceTips, Data Aggregation).

Answered by: Blake979 | Posted: 06-11-2021



Answer 5

As others have mentioned, the simplest way to preseed a list is with NoneType objects.

That being said, you should understand the way Python lists actually work before deciding this is necessary.

In the CPython implementation of a list, the underlying array is always created with overhead room, in progressively larger sizes ( 4, 8, 16, 25, 35, 46, 58, 72, 88, 106, 126, 148, 173, 201, 233, 269, 309, 354, 405, 462, 526, 598, 679, 771, 874, 990, 1120, etc), so that resizing the list does not happen nearly so often.

Because of this behavior, most list.append() functions are O(1) complexity for appends, only having increased complexity when crossing one of these boundaries, at which point the complexity will be O(n). This behavior is what leads to the minimal increase in execution time in S.Lott's answer.

Source: Python list implementation

Answered by: Michelle784 | Posted: 06-11-2021



Answer 6

Concerns about preallocation in Python arise if you're working with NumPy, which has more C-like arrays. In this instance, preallocation concerns are about the shape of the data and the default value.

Consider NumPy if you're doing numerical computation on massive lists and want performance.

Answered by: Samantha954 | Posted: 06-11-2021



Answer 7

I ran S.Lott's code and produced the same 10% performance increase by preallocating. I tried Ned Batchelder's idea using a generator and was able to see the performance of the generator better than that of the doAllocate. For my project the 10% improvement matters, so thanks to everyone as this helps a bunch.

def doAppend(size=10000):
    result = []
    for i in range(size):
        message = "some unique object %d" % ( i, )
        result.append(message)
    return result

def doAllocate(size=10000):
    result = size*[None]
    for i in range(size):
        message = "some unique object %d" % ( i, )
        result[i] = message
    return result

def doGen(size=10000):
    return list("some unique object %d" % ( i, ) for i in xrange(size))

size = 1000
@print_timing
def testAppend():
    for i in xrange(size):
        doAppend()

@print_timing
def testAlloc():
    for i in xrange(size):
        doAllocate()

@print_timing
def testGen():
    for i in xrange(size):
        doGen()


testAppend()
testAlloc()
testGen()

Output

testAppend took 14440.000ms
testAlloc took 13580.000ms
testGen took 13430.000ms

Answered by: Anna151 | Posted: 06-11-2021



Answer 8

Python's list doesn't support preallocation. Numpy allows you to preallocate memory, but in practice it doesn't seem to be worth it if your goal is to speed up the program.

This test simply writes an integer into the list, but in a real application you'd likely do more complicated things per iteration, which further reduces the importance of the memory allocation.

import timeit
import numpy as np

def list_append(size=1_000_000):
    result = []
    for i in range(size):
        result.append(i)
    return result

def list_prealloc(size=1_000_000):
    result = [None] * size
    for i in range(size):
        result[i] = i
    return result

def numpy_prealloc(size=1_000_000):
    result = np.empty(size, np.int32)
    for i in range(size):
        result[i] = i
    return result

setup = 'from __main__ import list_append, list_prealloc, numpy_prealloc'
print(timeit.timeit('list_append()', setup=setup, number=10))     # 0.79
print(timeit.timeit('list_prealloc()', setup=setup, number=10))   # 0.62
print(timeit.timeit('numpy_prealloc()', setup=setup, number=10))  # 0.73

Answered by: Emma605 | Posted: 06-11-2021



Answer 9

For some applications, a dictionary may be what you are looking for. For example, in the find_totient method, I found it more convenient to use a dictionary since I didn't have a zero index.

def totient(n):
    totient = 0

    if n == 1:
        totient = 1
    else:
        for i in range(1, n):
            if math.gcd(i, n) == 1:
                totient += 1
    return totient

def find_totients(max):
    totients = dict()
    for i in range(1,max+1):
        totients[i] = totient(i)

    print('Totients:')
    for i in range(1,max+1):
        print(i,totients[i])

This problem could also be solved with a preallocated list:

def find_totients(max):
    totients = None*(max+1)
    for i in range(1,max+1):
        totients[i] = totient(i)

    print('Totients:')
    for i in range(1,max+1):
        print(i,totients[i])

I feel that this is not as elegant and prone to bugs because I'm storing None which could throw an exception if I accidentally use them wrong, and because I need to think about edge cases that the map lets me avoid.

It's true the dictionary won't be as efficient, but as others have commented, small differences in speed are not always worth significant maintenance hazards.

Answered by: Victoria183 | Posted: 06-11-2021



Answer 10

From what I understand, Python lists are already quite similar to ArrayLists. But if you want to tweak those parameters I found this post on the Internet that may be interesting (basically, just create your own ScalableList extension):

http://mail.python.org/pipermail/python-list/2000-May/035082.html

Answered by: Sophia102 | Posted: 06-11-2021



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