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 o
> On 7/6/06, Girish Sahani <[EMAIL PROTECTED]> wrote:
> 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'],
> ['
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
> p_code = ''
> for i,k in enumerate(ks):
> p_code += i*' ' + "for item%s in elts['%s']:\n" %(i,k)
> p_code += len(ks)*' '+'print ['+','.join([ "item%s" %i
> for i,k in enumerate(ks) ])+']'
>
> # print the code
> print p_code
>
> >for item0 in elts['a']:
> > for item1 in elts['b']:
> > for
# first step : build a dictionary mapping the objects
# to all possible ids
alist = ['a.1','b.3','b.4','c.2','c.6','d.3']
elts = {}
for item in alist:
obj=item.split('.')[0]
if elts.has_key(obj):
elts[obj].append(item)
else:
elts[obj] = [item]
# then build the Python c
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
