How do I iterate through a dictionary of lists and pair each iteration with an item from all the keys?

Question

I need to iterate through a dictionary of lists, not knowing how many lists the dictionary will have, but still pair off each list value with any other list value produced from another key in the dictionary (if another one exists). I have the following code:

def loop_rec(codes, currentcode={}):
    if len(codes.keys())>1:
        for key in sorted(codes):
            codespop = dict(codes)
            loop = codespop.pop(key)
            for x in loop:
                currentcode[key]=x
                loop_rec(codespop,currentcode)
            break
    else:
        for key in codes.keys():
            loop = codes[key]
            for x in loop:
                currentcode[key]=x
                print currentcode

So if I have the following dictionary:

codes = {"coarse":range(4),"fine":range(2)}

I get this result:

>>> loop_rec(codes)
{'fine': 0, 'coarse': 0}
{'fine': 1, 'coarse': 0}
{'fine': 0, 'coarse': 1}
{'fine': 1, 'coarse': 1}
{'fine': 0, 'coarse': 2}
{'fine': 1, 'coarse': 2}
{'fine': 0, 'coarse': 3}
{'fine': 1, 'coarse': 3}

This is a brute force method and would like a more "Pythonic" way of doing this. I searched around a lot for something equivalent, but most methods don't result in the coarse and fine values being together for each iteration. Also would prefer it loop through coarse first, but the sorted command isn't working.

EDIT: Just realized the sorted command is working, the printout just isn't sorted. I don't care if it is printed in order.

I don't think this is immediately relevant to your problem, but be careful using a dictionary as a default argument, as it can have surprising behavior when you call the function more than once. Related reading: [“Least Astonishment” and the Mutable Default Argument](https://stackoverflow.com/q/1132941/953482) — Kevin, Sep 08 '17 at 17:45
@Kevin: I'm trying to be. That's why a copy of the dictionary is made and making sure the original dictionary doesn't get modified as a result of the original call. — Joe Davis, Sep 08 '17 at 18:07

Kevin · Accepted Answer · 2017-09-08T18:17:09.167

If I understand your question correctly, you want to take the cartesian product of all of the lists that are the values of a dict. You can use itertools.product to accomplish this.

import itertools
def dict_product(d):
    list_of_dicts = []
    for values in itertools.product(*d.values()):
        item = dict(zip(d.keys(),values))
        list_of_dicts.append(item)
    return list_of_dicts


codes = {"coarse":range(4),"fine":range(2),"zesty":range(3)}
for item in dict_product(codes):
    print(item)

Result:

{'zesty': 0, 'fine': 0, 'coarse': 0}
{'zesty': 0, 'fine': 0, 'coarse': 1}
{'zesty': 0, 'fine': 0, 'coarse': 2}
{'zesty': 0, 'fine': 0, 'coarse': 3}
{'zesty': 0, 'fine': 1, 'coarse': 0}
{'zesty': 0, 'fine': 1, 'coarse': 1}
{'zesty': 0, 'fine': 1, 'coarse': 2}
{'zesty': 0, 'fine': 1, 'coarse': 3}
{'zesty': 1, 'fine': 0, 'coarse': 0}
{'zesty': 1, 'fine': 0, 'coarse': 1}
{'zesty': 1, 'fine': 0, 'coarse': 2}
{'zesty': 1, 'fine': 0, 'coarse': 3}
{'zesty': 1, 'fine': 1, 'coarse': 0}
{'zesty': 1, 'fine': 1, 'coarse': 1}
{'zesty': 1, 'fine': 1, 'coarse': 2}
{'zesty': 1, 'fine': 1, 'coarse': 3}
{'zesty': 2, 'fine': 0, 'coarse': 0}
{'zesty': 2, 'fine': 0, 'coarse': 1}
{'zesty': 2, 'fine': 0, 'coarse': 2}
{'zesty': 2, 'fine': 0, 'coarse': 3}
{'zesty': 2, 'fine': 1, 'coarse': 0}
{'zesty': 2, 'fine': 1, 'coarse': 1}
{'zesty': 2, 'fine': 1, 'coarse': 2}
{'zesty': 2, 'fine': 1, 'coarse': 3}

In this example, the iteration order is coarse-fine-zesty, but this behavior is not guaranteed. In CPython 3.6 and higher, dictionaries are ordered, but this is an implementation detail and may change in the future.

I'm going to use: item = dict(zip(sorted(codes),values)) to get coarse to iterate before fine. — Joe Davis, Sep 08 '17 at 18:17

How do I iterate through a dictionary of lists and pair each iteration with an item from all the keys?

1 Answers1