基于dicts内部值的所有可能的dicts组合

Question

I want to generate all possible ways of using dicts, based on the values in them. To explain in code, I have:

a = {'name' : 'a', 'items': 3}
b = {'name' : 'b', 'items': 4}
c = {'name' : 'c', 'items': 5}

I want to be able to pick (say) exactly 7 items from these dicts, and all the possible ways I could do it in.

So:

x = itertools.product(range(a['items']), range(b['items']), range(c['items']))
y = itertools.ifilter(lambda i: sum(i)==7, x)

would give me:

(0, 3, 4)
(1, 2, 4)
(1, 3, 3)
...

What I'd really like is:

({'name' : 'a', 'picked': 0}, {'name': 'b', 'picked': 3}, {'name': 'c', 'picked': 4})
({'name' : 'a', 'picked': 1}, {'name': 'b', 'picked': 2}, {'name': 'c', 'picked': 4})
({'name' : 'a', 'picked': 1}, {'name': 'b', 'picked': 3}, {'name': 'c', 'picked': 3})
....

Any ideas on how to do this, cleanly?

Answer 1

Here it is

import itertools
import operator

a = {'name' : 'a', 'items': 3}
b = {'name' : 'b', 'items': 4}
c = {'name' : 'c', 'items': 5}

dcts = [a,b,c]

x = itertools.product(range(a['items']), range(b['items']), range(c['items']))
y = itertools.ifilter(lambda i: sum(i)==7, x)
z = (tuple([[dct, operator.setitem(dct, 'picked', vval)][0] \
       for dct,vval in zip(dcts, val)]) for val in y)
for zz in z:
    print zz

You can modify it to create copies of dictionaries. If you need a new dict instance on every iteration, you can change z line to

z = (tuple([[dct, operator.setitem(dct, 'picked', vval)][0] \
      for dct,vval in zip(map(dict,dcts), val)]) for val in y)

Answer 2

easy way is to generate new dicts:

names = [x['name'] for x in  [a,b,c]]
ziped = map(lambda x: zip(names, x), y)
maped = map(lambda el: [{'name': name, 'picked': count} for name, count in el],
            ziped)