Re: vary number of loops
On Wed, 16 Apr 2008 06:31:04 -0700, nullgraph wrote: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: vary number of loops
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 -- http://mail.python.org/mailman/listinfo/python-list
Re: vary number of loops
On 16 avr, 15:31, [EMAIL PROTECTED] wrote: 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). -- http://mail.python.org/mailman/listinfo/python-list
RE: vary number of loops
-Original Message- From: [EMAIL PROTECTED] [mailto:python- [EMAIL PROTECTED] On Behalf Of Tim Chase Sent: Wednesday, April 16, 2008 9:53 AM To: [EMAIL PROTECTED] Cc: python-list@python.org 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 (slw!), a slick cookbook recipe using iterators, etc. and provided timings for each method. Definitely worth bookmarking. * The information transmitted is intended only for the person or entity to which it is addressed and may contain confidential, proprietary, and/or privileged material. Any review, retransmission, dissemination or other use of, or taking of any action in reliance upon this information by persons or entities other than the intended recipient is prohibited. If you received this in error, please contact the sender and delete the material from all computers. GA625 -- http://mail.python.org/mailman/listinfo/python-list
Re: vary number of loops
On Apr 16, 8:31 am, [EMAIL PROTECTED] wrote: 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 (nlen(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%sc%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 (c1c0) and (c2c1) and (c3c2) ## and (c4c3) and (c5c4)] -- http://mail.python.org/mailman/listinfo/python-list
Re: vary number of loops
On Apr 16, 10:12 am, Reedick, Andrew [EMAIL PROTECTED] wrote: -Original Message- From: [EMAIL PROTECTED] [mailto:python- [EMAIL PROTECTED] On Behalf Of Tim Chase Sent: Wednesday, April 16, 2008 9:53 AM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] 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 (slw!), 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 -- http://mail.python.org/mailman/listinfo/python-list
