Reputation: 207
my_dict = {'a':[1,2], 'b':[3], 'c':{'d':[4,5], 'e':[6,7]}}
I need to derive all the combinations out of it as below.
{'a':1, 'b':3, 'c':{'d':4, 'e':6}}
{'a':1, 'b':3, 'c':{'d':4, 'e':7}}
{'a':1, 'b':3, 'c':{'d':5, 'e':6}}
{'a':1, 'b':3, 'c':{'d':5, 'e':7}}
{'a':2, 'b':3, 'c':{'d':4, 'e':6}}
and so on. There could be any level of nesting here
Please let me know how to achieve this
Something that I tried is pasted below but definitely was reaching nowhere
def gen_combinations(data):
my_list =[]
if isinstance(data, dict):
for k, v in data.iteritems():
if isinstance(v, dict):
gen_combinations(v)
elif isinstance(v, list):
for i in range(len(v)):
temp_dict = data.copy()
temp_dict[k] = v[i]
print temp_dict
my_dict = {'a':[1,2], 'b':[3], 'c':{'d':[4,5], 'e':[6,7]}}
gen_combinations(my_dict)
Which resulted in
{'a': 1, 'c': {'e': [6, 7], 'd': [4, 5]}, 'b': [3]}
{'a': 2, 'c': {'e': [6, 7], 'd': [4, 5]}, 'b': [3]}
{'e': 6, 'd': [4, 5]}
{'e': 7, 'd': [4, 5]}
{'e': [6, 7], 'd': 4}
{'e': [6, 7], 'd': 5}
{'a': [1, 2], 'c': {'e': [6, 7], 'd': [4, 5]}, 'b': 3}
Upvotes: 7
Views: 386
Reputation: 39546
from itertools import product
my_dict = {'a':[1,2], 'b':[3], 'c':{'d':[4,5], 'e':[6,7]}}
def process(d):
to_product = [] # [[('a', 1), ('a', 2)], [('b', 3),], ...]
for k, v in d.items():
if isinstance(v, list):
to_product.append([(k, i) for i in v])
elif isinstance(v, dict):
to_product.append([(k, i) for i in process(v)])
else:
to_product.append([(k, v)])
return [dict(l) for l in product(*to_product)]
for i in process(my_dict):
print(i)
Output:
{'a': 1, 'b': 3, 'c': {'e': 6, 'd': 4}}
{'a': 2, 'b': 3, 'c': {'e': 6, 'd': 4}}
{'a': 1, 'b': 3, 'c': {'e': 6, 'd': 5}}
{'a': 2, 'b': 3, 'c': {'e': 6, 'd': 5}}
{'a': 1, 'b': 3, 'c': {'e': 7, 'd': 4}}
{'a': 2, 'b': 3, 'c': {'e': 7, 'd': 4}}
{'a': 1, 'b': 3, 'c': {'e': 7, 'd': 5}}
{'a': 2, 'b': 3, 'c': {'e': 7, 'd': 5}}
Upd:
Code that works as asked here:
from itertools import product
my_dict = {'a':[1,2], 'e':[7], 'f':{'x':[{'a':[3,5]}, {'a':[4]}] } }
def process(d):
to_product = [] # [[('a', 1), ('a', 2)], [('b', 3),], ...]
for k, v in d.items():
if isinstance(v, list) and all(isinstance(i, dict) for i in v):
# specific case, when list of dicts process differently...
c = product(*list(process(i) for i in v))
to_product.append([(k, list(l)) for l in c])
elif isinstance(v, list):
to_product.append([(k, i) for i in v])
elif isinstance(v, dict):
to_product.append([(k, i) for i in process(v)])
else:
to_product.append([(k, v)])
return [dict(l) for l in product(*to_product)]
for i in process(my_dict):
print(i)
Output:
{'f': {'x': [{'a': 3}, {'a': 4}]}, 'a': 1, 'e': 7}
{'f': {'x': [{'a': 3}, {'a': 4}]}, 'a': 2, 'e': 7}
{'f': {'x': [{'a': 5}, {'a': 4}]}, 'a': 1, 'e': 7}
{'f': {'x': [{'a': 5}, {'a': 4}]}, 'a': 2, 'e': 7}
Upvotes: 11
Reputation: 4855
Solve it with two steps.
First replace each dict with a list of dicts generated by gen_combinations, called recursively.
Second, make the inner join between all keys. Each key has a flat list now.
Upvotes: -1