# nested for loop

Discussion in 'Python' started by Wolfgang Buechel, May 10, 2004.

1. ### Wolfgang BuechelGuest

Hi,

I want to iterate over all 2x2 matrices with elements in range 0..25
(crypto-stuff).

To produce them, first I wrote a fourfold nested for loop:

M=26
for a in range(M):
for b in range (M):
for c in range (M):
for d in range (M):
matr = [[a,b],[c,d]]
(dosomething)

Then I had a look in comp.lang.python and found:

for (a,b,c,d) in [(x,y,z,t) for x in range(M)
for y in range(M)
for z in range(M)
for t in range(M)] :
matr = [[a,b],[c,d]]

Is there a shorter (and probably, with respect to exec time, faster)
way to write such a 4for loop?
(I want to scan 3x3, 4x4 matrices too (;-)

-- Wolfgang

Wolfgang Buechel, May 10, 2004

2. ### Tor Iver WilhelmsenGuest

(Wolfgang Buechel) writes:

> for (a,b,c,d) in [(x,y,z,t) for x in range(M)
> for y in range(M)
> for z in range(M)
> for t in range(M)] :
> matr = [[a,b],[c,d]]

Try assigning range(M) to a variable instead of creating four copies
of it. For larger ranges, experiment using xrange instead.

Tor Iver Wilhelmsen, May 10, 2004

3. ### Peter HansenGuest

Wolfgang Buechel wrote:

> I want to iterate over all 2x2 matrices with elements in range 0..25
> (crypto-stuff).
>
> To produce them, first I wrote a fourfold nested for loop:
>
> M=26
> for a in range(M):
> for b in range (M):
> for c in range (M):
> for d in range (M):
> matr = [[a,b],[c,d]]
> (dosomething)
>
> Then I had a look in comp.lang.python and found:
>
>
> for (a,b,c,d) in [(x,y,z,t) for x in range(M)
> for y in range(M)
> for z in range(M)
> for t in range(M)] :
> matr = [[a,b],[c,d]]
>
> Is there a shorter (and probably, with respect to exec time, faster)
> way to write such a 4for loop?

Hmm... why would you want something shorter, as it would
probably be less readable? Also, how fast do you need it
to run, or how many times faster than the above would you
like it to run?

(The second one is a facetious question... the first is
more serious. In effect: what's wrong with what you have?)

-Peter

Peter Hansen, May 11, 2004
4. ### Terry ReedyGuest

"Wolfgang Buechel" <> wrote in message
news:...
> Hi,
>
> I want to iterate over all 2x2 matrices with elements in range 0..25
> (crypto-stuff).
>
> To produce them, first I wrote a fourfold nested for loop:
>
> M=26
> for a in range(M):
> for b in range (M):
> for c in range (M):
> for d in range (M):
> matr = [[a,b],[c,d]]
> (dosomething)

This is completely clear and space efficient.

> Then I had a look in comp.lang.python and found:

Oh dear...

> for (a,b,c,d) in [(x,y,z,t) for x in range(M)
> for y in range(M)
> for z in range(M)
> for t in range(M)] :
> matr = [[a,b],[c,d]]

This is less clear. It took me about 10 seconds (versus 1) to see as
equivalent. It produces a list with M**4 elements that you don't actually
need and soon throw away.

> Is there a shorter (and probably, with respect to exec time, faster)
> way to write such a 4for loop?

Write a C extension, maybe with Pyrex. However, an hour of your time is
worth lots of hours of PC time. But for a million runs, it might be worth
it.

Terry J. Reedy

Terry Reedy, May 11, 2004
5. ### Sean RossGuest

"Wolfgang Buechel" <> wrote in message
news:...
> Hi,
>
> I want to iterate over all 2x2 matrices with elements in range 0..25
> (crypto-stuff).

[snip]
> Is there a shorter (and probably, with respect to exec time, faster)
> way to write such a 4for loop?
> (I want to scan 3x3, 4x4 matrices too (;-)
>
> -- Wolfgang

Hi.

The following code isn't necessarily shorter or faster (or more readable),
but it's a bit more general:

# slightly modified code from
http://twistedmatrix.com/wiki/python/PostYourCode
def sequences(n, things):
"generates sequences of n items from a set of things"
if n == 0:
yield []
else:
for x in things:
for y in sequences(n-1, things):
yield [x] + y

def nXn_matrices(n, elements):
"generates nXn matrices from elements"
for s in sequences(n*n, elements):
yield [s[i*ni+1)*n] for i in xrange(n)]

# we'll try it over a small range ...
M = 3
for m in nXn_matrices(2, range(M)):
print m

Output:
[[0, 0], [0, 0]]
[[0, 0], [0, 1]]
[[0, 0], [0, 2]]
[[0, 0], [1, 0]]
[[0, 0], [1, 1]]
[[0, 0], [1, 2]]
[[0, 0], [2, 0]]
....
....
[[2, 2], [2, 0]]
[[2, 2], [2, 1]]
[[2, 2], [2, 2]]

# now 3X3 ... this takes a _l_o_n_g_ time ...
M = 3
for m in nXn_matrices(3,range(M)):
print m

Output:
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
[[0, 0, 0], [0, 0, 0], [0, 0, 1]]
[[0, 0, 0], [0, 0, 0], [0, 0, 2]]
[[0, 0, 0], [0, 0, 0], [0, 1, 0]]
[[0, 0, 0], [0, 0, 0], [0, 1, 1]]
[[0, 0, 0], [0, 0, 0], [0, 1, 2]]
....
....
[[2, 2, 2], [2, 2, 2], [2, 1, 0]]
[[2, 2, 2], [2, 2, 2], [2, 1, 1]]
[[2, 2, 2], [2, 2, 2], [2, 1, 2]]
[[2, 2, 2], [2, 2, 2], [2, 2, 0]]
[[2, 2, 2], [2, 2, 2], [2, 2, 1]]
[[2, 2, 2], [2, 2, 2], [2, 2, 2]]

I'm believe there are several opportunities for optimization both in the
code and in the algorithm (for instance, it may be possible to take
advantage of repetition in the sub-matrices), but I won't be trying that
now.

Good luck with what you're doing,
Sean

Sean Ross, May 11, 2004
6. ### Sean RossGuest

"Sean Ross" <> wrote in message
news:VDWnc.21679\$...
[snip]
> # slightly modified code from
> http://twistedmatrix.com/wiki/python/PostYourCode
> def sequences(n, things):
> "generates sequences of n items from a set of things"
> if n == 0:
> yield []
> else:
> for x in things:
> for y in sequences(n-1, things):
> yield [x] + y
>
> def nXn_matrices(n, elements):
> "generates nXn matrices from elements"
> for s in sequences(n*n, elements):
> yield [s[i*ni+1)*n] for i in xrange(n)]

[snip]
> believe there are several opportunities for optimization both in the
> code and in the algorithm (for instance, it may be possible to take
> advantage of repetition in the sub-matrices), but I won't be trying that
> now.

Looks like I'll be trying it now after all:

def nXn_matrices2(n, elements):
for m in sequences(n, list(sequences(n, elements))):
yield m

This is slightly faster than the first version, but it has more overhead
since it builds a list of the submatrices. That could be problem. It can
probably be avoided, but I haven't figured out how to do it just yet. We'll
see what happens.

Sean

Sean Ross, May 11, 2004
7. ### Anton VredegoorGuest

(Wolfgang Buechel) wrote:

>I want to iterate over all 2x2 matrices with elements in range 0..25
>(crypto-stuff).
>
>To produce them, first I wrote a fourfold nested for loop:
>
>M=26
> for a in range(M):
> for b in range (M):
> for c in range (M):
> for d in range (M):
> matr = [[a,b],[c,d]]
> (dosomething)

[snip]

>Is there a shorter (and probably, with respect to exec time, faster)
>way to write such a 4for loop?
>(I want to scan 3x3, 4x4 matrices too (;-)

This question is very much related to the one in a thread below here
(titled "Inverse of int(s, base)?"). In fact with a small adaptation
that code can produce this kind of matrices:

from itertools import islice

def tobase(i,base,digits):
R = []
for j in xrange(digits):
i,k = divmod(i,base)
R.append(k)
R.reverse()
return R

def as_matrix(R,rows,cols):
it = iter(R)
return [list(islice(it,cols)) for i in xrange(rows)]

def test():
for i in range(10):
R = tobase(i,25,6)
print as_matrix(R,3,2)

if __name__=='__main__':
test()

Anton

Anton Vredegoor, May 11, 2004
8. ### Wolfgang BuechelGuest

(Wolfgang Buechel) wrote in message news:<>...
> Hi,
>
> I want to iterate over all 2x2 matrices with elements in range 0..25
> (crypto-stuff).

Try generators (new in Python 2.3.3).
There is an example you can/must adapt:

(PythonPath)/Lib/test/test_generators.py
def conjoin()

and in the whatsnew-documentation:

http://www.python.org/doc/2.3.3/whatsnew/section-generators.html

--W.

Wolfgang Buechel, May 23, 2004
9. ### Dan BishopGuest

(Wolfgang Buechel) wrote in message news:<>...
> Hi,
>
> I want to iterate over all 2x2 matrices with elements in range 0..25
> (crypto-stuff).
>
> To produce them, first I wrote a fourfold nested for loop:

....
> Then I had a look in comp.lang.python and found:
>
>
> for (a,b,c,d) in [(x,y,z,t) for x in range(M)
> for y in range(M)
> for z in range(M)
> for t in range(M)] :
> matr = [[a,b],[c,d]]

This may be shorter, but it's slower. Your old code took "only" 1.5
seconds to run on this computer, but the new way takes 3.5 seconds.

What you *can* do to make your code faster (if you don't change matr
once it's created) is to precompute the 676 possible matrix rows.

ELEMENT_RANGE = range(26)
MATRIX_ROWS = [[x, y] for x in ELEMENT_RANGE
for y in ELEMENT_RANGE]
for row1 in MATRIX_ROWS:
for row2 in MATRIX_ROWS:
matr = [row1, row2]

That takes only 532 ms -- almost 3 times faster than the original.

Dan Bishop, May 23, 2004
10. ### Terry ReedyGuest

"Dan Bishop" <> wrote in message
news:...
> What you *can* do to make your code faster (if you don't change matr
> once it's created) is to precompute the 676 possible matrix rows.
>
> ELEMENT_RANGE = range(26)
> MATRIX_ROWS = [[x, y] for x in ELEMENT_RANGE
> for y in ELEMENT_RANGE]
> for row1 in MATRIX_ROWS:
> for row2 in MATRIX_ROWS:
> matr = [row1, row2]
>
> That takes only 532 ms -- almost 3 times faster than the original.

Cute. While I am quite familiar with and have experience applying the
principle of computing once instead of multiple times, I missed its
application here.

tjr

Terry Reedy, May 24, 2004
11. ### Peter OttenGuest

Dan Bishop wrote:

> What you can do to make your code faster (if you don't change matr
> once it's created) is to precompute the 676 possible matrix rows.
>
> ELEMENT_RANGE = range(26)
> MATRIX_ROWS = [[x, y] for x in ELEMENT_RANGE
> for y in ELEMENT_RANGE]
> for row1 in MATRIX_ROWS:
> for row2 in MATRIX_ROWS:
> matr = [row1, row2]
>
> That takes only 532 ms -- almost 3 times faster than the original.

Nice. Another speed gain (from 435 to 246ms on my machine) is in for you if
you use tuples instead of lists. And if you allow for somewhat less elegant
code that builds on your recipe,

from itertools import izip, repeat
ELEMENT_RANGE = range(26)
MATRIX_ROWS = [(x, y) for x in ELEMENT_RANGE
for y in ELEMENT_RANGE]
for row in MATRIX_ROWS:
for matr in izip(repeat(row), MATRIX_ROWS):
pass

you can bring that down to 138ms.

For the record: the straightforward solution (the original with tuples and
range() factored out)

r = range(26)
for a in r:
for b in r:
for c in r:
for d in r:
matr = ((a,b),(c,d))

takes 478ms. The "improved" variant is evil performance-wise (1598ms):

r = range(26)
for (a,b,c,d) in [(x,y,z,t) for x in r
for y in r
for z in r
for t in r] :
matr = ((a,b),(c,d))

It might be interesting how much this can be improved with 2.4's generator
expressions.

Peter

Peter Otten, May 24, 2004