vary number of loops

[nullgraph]

Hi everyone,

I'm new to Python and the notion of lambda, and I'm trying to write a
function that would have a varying number of nested for loops
depending on parameter n. This just smells like a job for lambda for
me, but I can't figure out how to do it. Any hint?

For example, for n=2, I want the function to look something like:

def foo(2)
generate 2 sets of elements A, B
# mix elements by:
for a_elt in A
for b_elt in B
form all combinations of them


If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

Any help is greatly appreciated,

nullgraph

 

[Marc 'BlackJack' Rintsch]

I'm new to Python and the notion of lambda, and I'm trying to write a
function that would have a varying number of nested for loops
depending on parameter n. This just smells like a job for lambda for
me, but I can't figure out how to do it. Any hint?

That has nothing to do with ``lambda``. If you don't think "Hey, that's
smells like a job for a function." then it's no job for ``lambda``, which
is just a way to define a function without automatically binding it to a
name like ``def`` does.

One solution to your problem is recursion. Untested:

def foo(xs):
if xs:
for elt in xs[0]:
for ys in foo(xs[1:]):
yield [elt] + ys
else:
yield []

Called as ``foo([A, B])``.

Ciao,
Marc 'BlackJack' Rintsch

 

[Tim Chase]

I'm new to Python and the notion of lambda, and I'm trying to write a

function that would have a varying number of nested for loops
depending on parameter n. This just smells like a job for lambda for
me, but I can't figure out how to do it. Any hint?

I'm not sure lambda is the tool to use here. Doable, perhaps,
but improbable in my book.

For example, for n=2, I want the function to look something like:

def foo(2)
generate 2 sets of elements A, B
# mix elements by:
for a_elt in A
for b_elt in B
form all combinations of them

If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

You might be ineterested in this thread:

http://mail.python.org/pipermail/python-list/2008-January/473650.html

where various solutions were proposed and their various merits
evaluated.

-tkc

 

[colas.francis]

Hi everyone,

I'm new to Python and the notion of lambda, and I'm trying to write a
function that would have a varying number of nested for loops
depending on parameter n. This just smells like a job for lambda for
me, but I can't figure out how to do it. Any hint?

For example, for n=2, I want the function to look something like:

def foo(2)
generate 2 sets of elements A, B
# mix elements by:
for a_elt in A
for b_elt in B
form all combinations of them

If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

Any help is greatly appreciated,

nullgraph

You can try recursion in a more classic manner:

In [283]: def foo(n):
.....: def bar(n):
.....: my_elts = xrange(2)
.....: if n<=0:
.....: raise StopIteration
.....: elif n<=1:
.....: for elt in my_elts:
.....: yield (elt,)
.....: else:
.....: for elt in my_elts:
.....: for o_elt in bar(n-1):
.....: yield (elt,)+o_elt
.....: for elt in bar(n):
.....: print elt
.....:

In [284]: foo(2)
(0, 0)
(0, 1)
(1, 0)
(1, 1)

In [285]: foo(3)
(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)

In this case, I have an inner function to generate the whole set of
elements and then an outer loop to process them.
Note that you can have the generation of my_elts depend on rank n of
recursion (that is the index of the set in your list).

 

[Reedick, Andrew]

-----Original Message-----
From: [email protected] [mailtoython-
[email protected]] On Behalf Of Tim Chase
Sent: Wednesday, April 16, 2008 9:53 AM
To: (e-mail address removed)
Cc: (e-mail address removed)
Subject: Re: vary number of loops

If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

You might be ineterested in this thread:

http://mail.python.org/pipermail/python-list/2008-January/473650.html

where various solutions were proposed and their various merits
evaluated.

I second that. The thread compared building loops on the fly, building
comprehensions nested to arbitrarily levels, recursion (sloooow!), a
slick cookbook recipe using iterators, etc. and provided timings for
each method. Definitely worth bookmarking.



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[Mensanator]

Hi everyone,

I'm new to Python and the notion of lambda, and I'm trying to write a
function that would have a varying number of nested for loops
depending on parameter n. This just smells like a job for lambda for
me, but I can't figure out how to do it. Any hint?

For example, for n=2, I want the function to look something like:

def foo(2)
   generate 2 sets of elements A, B
   # mix elements by:
   for a_elt in A
      for b_elt in B
         form all combinations of them

If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

Any help is greatly appreciated,

nullgraph

There's always the stupid way:

def ooloop6(a, n, perm=True, repl=True):
if (not repl) and (n>len(a)): return
r0 = range(n)
r1 = r0[1:]
if perm and repl: # permutations with
replacement
v = ','.join(['c%s' % i for i in r0])
f = ' '.join(['for c%s in a' % i for i in r0])
e = ''.join(["p = [''.join((",v,")) ",f,"]"])
exec e
return p
if (not perm) and repl: # combinations with
replacement
v = ','.join(['c%s' % i for i in r0])
f = ' '.join(['for c%s in a' % i for i in r0])
i = ' and '.join(['(c%s>=c%s)' % (j,j-1) for j in r1])
e = ''.join(["p = [''.join((",v,")) ",f," if ",i,"]"])
exec e
return p
if perm and (not repl): # permutaions without
replacement
v = ','.join(['c%s' % i for i in r0])
f = ' '.join(['for c%s in a' % i for i in r0])
i = ' and '.join([' and '.join(['(c%s!=c%s)' % (j,k) for k in
range(j)]) for j in r1])
e = ''.join(["p = [''.join((",v,")) ",f," if ",i,"]"])
exec e
return p
if (not perm) and (not repl): # combinations without
replacement
v = ','.join(['c%s' % i for i in r0])
f = ' '.join(['for c%s in a' % i for i in r0])
i = ' and '.join(['(c%s>c%s)' % (j,j-1) for j in r1])
e = ''.join(["p = [''.join((",v,")) ",f," if ",i,"]"])
exec e
print '\n\n',e,'\n\n'
return p

a = 'abcdefghij'
n = 6

# for lotto use Combinations without Replacement
p = ooloop6(a,n,False, False)
##################################################################

Here's the code that gets executed:

## p = [''.join((c0,c1,c2,c3,c4,c5)) for c0 in a
## for c1 in a for c2 in a for c3 in a for c4 in a
## for c5 in a if (c1>c0) and (c2>c1) and (c3>c2)
## and (c4>c3) and (c5>c4)]

 

[nullgraph]

-----Original Message-----
From: [email protected] [mailtoython-
[email protected]] On Behalf Of Tim Chase
Sent: Wednesday, April 16, 2008 9:53 AM
To: (e-mail address removed)
Cc: (e-mail address removed)
Subject: Re: vary number of loops

If n=3, I want to have 3 sets of elements and mix them up using 3 for
loops.

You might be ineterested in this thread:

where various solutions were proposed and their various merits
evaluated.

I second that. The thread compared building loops on the fly, building
comprehensions nested to arbitrarily levels, recursion (sloooow!), a
slick cookbook recipe using iterators, etc. and provided timings for
each method. Definitely worth bookmarking.

Yes, I second that second Very nice thread, I'm leaning toward the
"pythonic method" from there, but thanks for all the other solutions
suggested here. Guess I need to go play with lambda more...

nullgraph