how to refactor nested for loop into smaller for loop assume each of them independent?

[BartC]

On 08/10/2016 11:54, Chris Angelico wrote:
> On Sat, Oct 8, 2016 at 9:12 PM, BartC <bc at freeuk.com> wrote:
>> On 08/10/2016 11:03, Chris Angelico wrote:
>>>
>>> On Sat, Oct 8, 2016 at 8:58 PM, meInvent bbird <jobmattcon at gmail.com>
>>> wrote:
>>>>
>>>> how to refactor nested for loop into smaller for loop assume each of them
>>>> independent?
>>>>
>>>> because memory is not enough
>>>>
>>>> for ii in range(1,2000):
>>>>  for jj in range(1,2000):
>>>>   for kk in range(1,2000):
>>>>     print run(ii,jj,kk)
>>>
>>>
>>> Since you're using Python 2, you could try xrange instead of range.
>>> But it won't make a lot of difference. You're trying to call your run
>>> function 8,000,000,000 times... that is a *lot* of function calls.
>>> Consider a faster algorithm instead!
>>
>>
>> Printing even one character 8 billion times is likely to take quite a few
>> hours too.
>>
>> The OP's code however is a good demonstration of how crazy Python's original
>> for-range loop was: you need to construct a list of N elements just to be
>> able to count to N. How many years was it until xrange was introduced?
>
> Hardly matters in this case. The memory usage of a list of 2000 ints is:
>
>>>> sys.getsizeof(range(2000))
> 16072
>>>> 2000 * sys.getsizeof(1999)
> 48000

> About 64KB. Not overly significant.

Well, there'll be three lots. But it's true there won't be 8 billion 
elements which is what I immediately thought when the OP said they'd run 
out of memory. So perhaps something is going on inside run() to account 
for it.

-- 
Bartc