Python For Loop Slowing With Time

Question

So I'm having a little trouble dealing with for loops in Python - as far as I can tell, they're getting slower with time. I'm looping over a range inside of a range, and as time passes, the loop noticeably slows. This is done inside of a game engine, if it matters. Could anyone tell me what the issue is?

Here's a quick example.

for x in range(xs): # xs, ys, and zs are all pre-determined size values

     for z in range(zs):

          for y in range(ys):

              vp = [x * vs, y * vs, z * vs]

              v = Cube(vp)

The initial speed of this process is fine, but with time the loop slows. I know it's not anything else like the Rasterizer of the game engine because when the loop is done, the rest of the engine runs at 60 FPS. So what could be the problem?

EDIT: I'm using Python 3, so there is no xrange.

EDIT 2: For this example, vs is 1.0, and the predetermined size values of xs, ys, and zs are all 20.

What if you comment out either of the two lines inside the loops? — jtbandes
– jtbandes, Commented Aug 21, 2011 at 3:50
If you are using python 2.x xrange will be faster than range — GWW
– GWW, Commented Aug 21, 2011 at 3:55
How are you measuring "initial speed" vs. "slows with time"? Are we talking about iterations within a single run of the shown code, or are we talking about successive runs of the shown code? If the former, how are you measuring the percentage completion of the process? — Karl Knechtel
– Karl Knechtel, Commented Aug 21, 2011 at 8:07

agf · Accepted Answer · 2011-08-21 03:55:55Z

16

This is another case of "need more information". However, Python has a standard way of constructing nested loops like this efficiently, itertools.product:

from itertools import product

for x, y, z in product(xrange(xs), xrange(zs), xrange(ys)):
    vp = [x * vs, y * vs, z * vs]
    v = Cube(vp)

It doesn't require the construction of ranges every time in the inner loop. I also switched your use of range to xrange, as it's better for large ranges, although this is really irrelevant with product.

@JohnZ's question is good -- if your "predetermined size values" are very large, and especially if vs is also large, you could be constructing some large values, and it could be taking a long time for Cube to process them.

I doubt the loop itself is slowing down, but the numbers are getting larger, so your calculations might be.

answered Aug 21, 2011 at 3:55

agf

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6 Comments

SolarLune Over a year ago

Also, my 'predetermined size values' are, for this example, 20 each.

SolarLune Over a year ago

But that's the thing that's weird - the for loop should stay at a stable speed, shouldn't it? It's definitely getting slower with time to a point...

agf Over a year ago

If it's getting slower, it's because of what you're doing in Cube, because the values are getting very large, or because of something else going on in parallel with the loop. There isn't any other explanation. Loops just don't "get slower" in Python.

agf Over a year ago

When you say "for the purposes of this example", do you mean you actually tested this loop with xs etc. set to 20 and saw a slowdown? Because that doesn't seem to make any sense -- with values that small (8000 iterations) the loop will take no time at all.

SolarLune Over a year ago

That seemed to help some. Thanks for that! It's still slowing down with time, but I'm accessing a global dictionary that I'm appending onto, so maybe that slowdown is to be expected.

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JSideris · Accepted Answer · 2011-08-21 04:23:21Z

1

Three things I can think of:

Memory - if you're storing all of the generated values somewhere, the loop might be slowing down because the of all the memory being used. Python has it's own memory manager, so using up lots of memory could eventually make the program slower.

Complexity of calculations - Python uses arbitrary precision numeric data types. If you are multiplying extremely large numbers together (especially floats) the program will slow down. It really depends on how large these values get.

Cube - It could be a bug in the Cube code (although I'm sure it's probably just as simple as it sounds).

answered Aug 21, 2011 at 4:23

JSideris

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