List Combinations

[Gerdus van Zyl]

I have a list that looks like this:
[['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]

how can I get all the combinations thereof that looks like as follows:
3,9,5,4,2
3,1,5,4,2
3,9,5,4,5
3,1,5,4,5
etc.

Thank You,
Gerdus

 

[Shane Geiger]

2008/3/12, Gerdus van Zyl <[email protected]
<mailto:[email protected]>>:

I have a list that looks like this:
[['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]

how can I get all the combinations thereof that looks like as follows:
3,9,5,4,2
3,1,5,4,2
3,9,5,4,5
3,1,5,4,5
etc.

Thank You,
Gerdus

--


list = [['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]
newList = []
for l in list:
newList.extend(l)

#newList = [3,9,1,5,4,2,5,8]
list=[]
for l in newList:
if l not in list:
list.append(l)

#now list is [3,9,1,5,4,2,8]

list.sort()

#now your list is in sorted order

Then, you can just write a series of for loops to spin off your data
however you like.

def gen(lists):
out = '[' + ','.join(["v%s" % i for i in range(len(lists))]) + ']'
comp = ''.join([ " for v%d in lists[%d]" % (i, i) for i in
range(len(lists))])
return eval('[ ' + out + comp + ' ]')

a,b,c = [1,2,3],[4,5,6],[7,8,9]

print gen([a, b, c])





--
Shane Geiger
IT Director
National Council on Economic Education
(e-mail address removed) | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

 

[George Sakkis]

I have a list that looks like this:
[['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]

how can I get all the combinations thereof that looks like as follows:
3,9,5,4,2
3,1,5,4,2
3,9,5,4,5
3,1,5,4,5
etc.

Thank You,
Gerdus

Search for "cartesian product" recipes.

George

 

[Mark Dickinson]

I have a list that looks like this:
[['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]

how can I get all the combinations thereof that looks like as follows:

You could wait for Python 2.6, or download the current alpha:

Python 2.6a1+ (trunk:61353M, Mar 12 2008, 11:21:13)
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.

from itertools import product
for c in product(['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']): print c

...
('3', '9', '5', '4', '2')
('3', '9', '5', '4', '5')
('3', '9', '5', '4', '8')
('3', '1', '5', '4', '2')
('3', '1', '5', '4', '5')
('3', '1', '5', '4', '8')

If you can't wait, look at http://docs.python.org/dev/library/itertools.html
where equivalent code is given.

Mark

 

[Mel]

I have a list that looks like this:
[['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]

how can I get all the combinations thereof that looks like as follows:
3,9,5,4,2
3,1,5,4,2
3,9,5,4,5
3,1,5,4,5
etc.

Thank You,
Gerdus

What they said, or, if you want to see it done:


def combi (s):
if s:
for a in s[0]:
for b in combi (s[1:]):
yield [a] + b
else:
yield []

for y in combi ([['3'], ['9', '1'], ['5'], ['4'], ['2', '5', '8']]):
print y

 

[Reedick, Andrew]

-----Original Message-----

From: [email protected] [mailtoython-
[email protected]] On Behalf Of Shane Geiger
Sent: Wednesday, March 12, 2008 10:33 AM
To: Michael Wieher
Cc: (e-mail address removed)
Subject: Re: List Combinations


def gen(lists):
out = '[' + ','.join(["v%s" % i for i in range(len(lists))]) + ']'
comp = ''.join([ " for v%d in lists[%d]" % (i, i) for i in
range(len(lists))])
return eval('[ ' + out + comp + ' ]')

a,b,c = [1,2,3],[4,5,6],[7,8,9]

print gen([a, b, c])





--
Shane Geiger
IT Director
National Council on Economic Education
(e-mail address removed) | 402-438-8958 | http://www.ncee.net

It helps to quote your source...
http://www.mail-archive.com/[email protected]/msg178457.html


Start here

http://www.mail-archive.com/[email protected]/msg178356.html
and go through the thread. There are several ways to solve the problem
and we evaluated the performance and 'pythonicity' of each.




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[Arnaud Delobelle]

Start here

http://www.mail-archive.com/[email protected]/msg178356.html
and go through the thread.  There are several ways to solve the problem
and we evaluated the performance and 'pythonicity' of each.  

I used a kind of extended cartesian product function a while ago while
writing a parser. If I simplify it down to a simple product function
it goes like this:

def product(*sequences):
i, n = 0, len(sequences)
vals = [None]*n
iters = map(iter, sequences)
while i >= 0:
if i == n:
yield tuple(vals)
i -= 1
else:
for vals in iters:
i += 1
break
else:
iters = iter(sequences)
i -= 1

It's neither recursive nor a hack, I haven't tried to measure it
against other approaches (obviously it wouldn't beat the eval(...)
hack). I haven't optimised it for readability (by others) either