Girish Sahani wrote:
> hello ppl,
>
>   Consider a list like ['a.1','b.3','b.4','c.2']. Here 'a','b','c' are
> objects and 1,3,4,2 are their instance ids and they are unique e.g. a.1
> and b.1 cannot exist together. From this list i want to generate
> multiple lists such that each list must have one and only one instance of
> every object.
>  Thus, for the above list, my output should be:
>   [['a.1','b.3','c.2'],['a.1','b.4','c.2']]
>  Another example: Let l = ['a.1','b.3','b.4','c.2','c.6','d.3']. Then
> output should be [['a.1','b.3','c.2','d.3'],['a.1','b.3','c.6','d.3'],
> ['a.1','b.4','c.2','d.3'],[['a.1','b.4','c.6','d.3']
>
>  Can anyone suggest me a time-efficient method for doing this??
>


I don't understand what you mean by "'a','b','c' are objects and
1,3,4,2 are their instance ids", but I think the solution to whatever
your problem is, will involve the Cartesian Product of sets (lists) -
(also called the cross product).

If your data is just the strings that you give in your example, then
the following should work. If your 'object strings' are longer than one
character, you will have to adapt it.


print
import itertools as it

data1 = ['a.1','b.3','b.4','c.2']
data2 = ['a.1','b.3','b.4','c.2','c.6','d.3']

want1 = [['a.1', 'b.3', 'c.2'],
         ['a.1', 'b.4', 'c.2']]
want2 = [['a.1','b.3','c.2','d.3'],
         ['a.1','b.3','c.6','d.3'],
         ['a.1','b.4','c.2','d.3'],
         ['a.1','b.4','c.6','d.3'] ]

def split_data(data):
    ret = []
    for k, g in it.groupby(sorted(data), lambda x: x[0]):
        ret.append( list(g) )
    return ret

#following function from ASPN Cookbook by David Klaffenbach
#http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/302478
def combine(*seqin):
    '''returns a list of all combinations of argument sequences.
for example: combine((1,2),(3,4)) returns
[[1, 3], [1, 4], [2, 3], [2, 4]]'''
    def rloop(seqin,listout,comb):
        '''recursive looping function'''
        if seqin:                       # any more sequences to
process?
            for item in seqin[0]:
                newcomb=comb+[item]     # add next item to current comb
                # call rloop w/ rem seqs, newcomb
                rloop(seqin[1:],listout,newcomb)
        else:                           # processing last sequence
            listout.append(comb)        # comb finished, add to list
    listout=[]                      # listout initialization
    rloop(seqin,listout,[])         # start recursive process
    return listout


assert combine(*split_data(data1)) == want1
assert combine(*split_data(data2)) == want2

 --------------------------------

Gerard

-- 
http://mail.python.org/mailman/listinfo/python-list

Reply via email to