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?
6 Answers
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 ...
2 Comments
g.d.d.c
+1 and I've updated my answer with the other syntax and a comment. Thank you for pointing it out.
John La Rooy
slooow :) ... hybrid of those two version is better
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.
Comments
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
Comments
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!
3 Comments
Philipp
Numpy also has dedicated factory functions:
a = 2 * ones(1000000)
Bartek
@Philipp: That's awesome! This is why I love SO. Curiosity to answer a question leads to many learnings for myself. Cheers :-)
John Machin
If you can't fit a million-element list or array into your machine's memory, it's dead already. Also, I don't understand "It ran on my computer for a long time" ... see my answer for (a) how to do simple timing using
timeit at the command prompt (b) how small the measured times (milliseconds!) are (4-year-old laptop running Win XP SP2)aList = [2 for x in range(1000000)]
or base on chauncey link
anArray =array.array('i', (2 for i in range(0,1000000)))
Comments
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.
array = [2 for x in 1..1000000]?[2 for x in xrange(1000000)];[2] * 1000000would be faster and simpler; however these produce alist--arrayandlistmean different things in Python.