Since your keys are not unique, I would think that you would want a list
of values for the object corresponding to each key. Something like
Mydict = {}
Mydict.setdefault(mykey, []).append(avalue)
-----Original Message-----
From: python-list-bounces+frsells=adventistcare....@python.org
[mailto:python-list-bounces+frsells=adventistcare....@python.org] On
Behalf Of Peter Otten
Sent: Sunday, November 07, 2010 4:01 PM
To: python-list@python.org
Subject: Re: { '0':'c->c->a' ,'1':'a->b->a' .........}
chris wrote:
> have anybody a hint , how i get a dict from non unique id's and their
> different related values.
>
> Thanks for advance
> Chris
>
> ###random data #
> a=range(10)*3
> def seqelem():
> i=random.randint(0,2)
> elem=['a','b','c'][i]
> return elem
>
> s=[seqelem() for t in range(30)]
> print zip(a,s)
>
> ## favored result:
> { '0':'c->c->a' ,'1':'a->b->a' .........}
>>> import random
>>> from collections import defaultdict
>>> a = range(10)*3
>>> s = [random.choice("abc") for _ in a]
>>> d = defaultdict(list)
>>> for k, v in zip(a, s):
... d[k].append(v)
...
>>> d
defaultdict(<type 'list'>, {0: ['b', 'a', 'a'], 1: ['c', 'a', 'c'], 2:
['c',
'c', 'c'], 3: ['c', 'a', 'a'], 4: ['b', 'c', 'a'], 5: ['b', 'c', 'c'],
6:
['c', 'a', 'b'], 7: ['b', 'b', 'a'], 8: ['a', 'c', 'c'], 9: ['b', 'a',
'b']})
>>> dict((k, "->".join(v)) for k, v in d.iteritems())
{0: 'b->a->a', 1: 'c->a->c', 2: 'c->c->c', 3: 'c->a->a', 4: 'b->c->a',
5:
'b->c->c', 6: 'c->a->b', 7: 'b->b->a', 8: 'a->c->c', 9: 'b->a->b'}
Peter
--
http://mail.python.org/mailman/listinfo/python-list
--
http://mail.python.org/mailman/listinfo/python-list
