What is the fastest way to initialize an integer array in python?

Question

Say I wanted to create an array (NOT list) of 1,000,000 twos in python, like this:

array = [2, 2, 2, ...... , 2]

What would be a fast but simple way of doing it?

Answer 1

The currently-accepted answer is NOT the fastest way using array.array ; at least it's not the slowest -- compare these:

[source: johncatfish (quoting chauncey), Bartek]
python -m timeit -s"import array" "arr = array.array('i', (2 for i in range(0,1000000)))"
10 loops, best of 3: 543 msec per loop

[source: g.d.d.c]
python -m timeit -s"import array" "arr = array.array('i', [2] * 1000000)"
10 loops, best of 3: 141 msec per loop

python -m timeit -s"import array" "arr = array.array('i', [2]) * 1000000"
100 loops, best of 3: 15.7 msec per loop

That's a ratio of about 9 to 1 ...

Answer 2

Is this what you're after?

# slower.
twosArr = array.array('i', [2] * 1000000)

# faster.
twosArr = array.array('i', [2]) * 1000000

You can get just a list with this:

twosList = [2] * 1000000

-- EDITED --

I updated this to reflect information in another answer. It would appear that you can increase the speed by a ratio of ~ 9 : 1 by adjusting the syntax slightly. Full credit belongs to @john-machin. I wasn't aware you could multiple the array object the same way you could do to a list.

Answer 3

A hybrid approach works fastest for me

$ python -m timeit -s"import array" "arr = array.array('i', [2]*100) * 10000"
100 loops, best of 3: 5.38 msec per loop

---

$ python -m timeit -s"import array" "arr = array.array('i', [2]) * 1000000"
10 loops, best of 3: 20.3 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*10) * 100000"
100 loops, best of 3: 6.69 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*100) * 10000"
100 loops, best of 3: 5.38 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*1000) * 1000"
100 loops, best of 3: 5.47 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*10000) * 100"
100 loops, best of 3: 6.13 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*100000) * 10"
10 loops, best of 3: 14.9 msec per loop
$ python -m timeit -s"import array" "arr = array.array('i', [2]*1000000)"
10 loops, best of 3: 77.7 msec per loop

Answer 4

Using the timeit module you can kind of figure out what the fastest of doing this is:

First off, putting that many digits in a list will kill your machine most likely as it will store it in memory.

However, you can test the execution using something like so. It ran on my computer for a long time before I just gave up, but I'm on an older PC:

timeit.Timer('[2] * 1000000').timeit()

Ther other option you can look into is using the array module which is as stated, efficient arrays of numeric values

array.array('i', (2 for i in range(0, 1000000)))

I did not test the completion time of both but I'm sure the array module, which is designed for number sets will be faster.

Edit: Even more fun, you could take a look at numpy which actually seems to have the fastest execution:

from numpy import *
array( [2 for i in range(0, 1000000)])

Even faster from the comments:

a = 2 * ones(10000000)

Awesome!

Answer 5

aList = [2 for x in range(1000000)]

或基于昌西链接

anArray =array.array('i', (2 for i in range(0,1000000)))

Answer 6

If the initial value doesn't have to be non-zero and if you have /dev/zero available on your platform, the following is about 4.7 times faster than the array('L',[0])*size solution:

myarray = array.array('L')
f = open('/dev/zero', 'rb')
myarray.fromfile(f, size)
f.close()

In question How to initialise an integer array.array object with zeros in Python I'm looking for a better way.