CODEWITHSUNDEEP
pythondictionarykeydefaultdict
alvas
alvas

Reputation: 122148

Getting the values of many defaultdicts that contains a certain key

I have several defaultdict that looks like this:

en1 = {1:['handler', 'coach', 'manager'],2:['a','c']}
es2 = {1:['entrenador', 'míster', 'entrenadora'],2:['h','i'],4:['j','k','m']}
it3 = {1:['mister', 'allenatore' 'coach'],3:['p','q'],5:['r','q']}
...

And I need get the string from each defaultdict sharing the same index. Something like this:

for i in en1:
  for j in en1[i]:
    for k in es2[i]:
      for l in it3[i]:
        print j, k, l

but i have to do it for like 30+ defaultdict, so how can that be done without write that horrendous loop above?

Is there a way to get all the values of the defaultdict with the same keys and then output the values easily?

It should output combinations of tuples made up of values sharing the same key for all dicts, e.g.:

handler entrenador mister
handler entrenador allenatore
handler entrenador coach
handler mister mister
handler mister allenator
...

Upvotes: 0

Views: 118

Answers (2)

Pawel Miech
Pawel Miech

Reputation: 7822

I am not sure if I understand your expectations regarding output, ("string from each defaultdict sharing the same index"? Your sample output contains items that do not share the same index, e.g. a g a, g has second position in dd2 key1) but this will give you the dictionary containing fused values of items with the same key within all three dicts :

d1 = {1:['a','b'],2:['a','c'],3:['d','e']}
d2 = {1:['f','g','l'],2:['h','i'],4:['j','k','m']}
d3 = {1:['n','a'],3:['p','q'],5:['r','q']}

d4 = [d1,d2,d3]
new = {}
for i in d4:
    for m in i:
        try:
            new[m].extend(i[m])
        except:
            new[m] = []
            new[m].extend(i[m])
print new
{1: ['a', 'b', 'f', 'g', 'l', 'n', 'a'], 
 2: ['a', 'c', 'h', 'i'], 3: ['d', 'e', 'p', 'q'], 
 4: ['j', 'k', 'm'], 5: ['r', 'q']}

And you can then make another list containing combinations of all items in those lists if you need them. 

from itertools import combinations
combs = []
#this is just for one key, but you can loop over list of dicts to do this for each dict
for m in combinations(new[1],3):
    perms.append(m)
print combs
[('a', 'b', 'f'), ('a', 'b', 'g'), ('a', 'b', 'l'), ('a', 'b', 'n'), ('a', 'b', 'a'), ('a', 'f', 'g'), ('a', 'f', 'l'), ('a', 'f', 'n'), ('a', 'f', 'a'), ('a', 'g', 'l'), ('a', 'g', 'n'), ('a', 'g', 'a'), ('a', 'l', 'n'), ('a', 'l', 'a'), ('a', 'n', 'a'), ('b', 'f', 'g'), ('b', 'f', 'l'), ('b', 'f', 'n'), ('b', 'f', 'a'), ('b', 'g', 'l'), ('b', 'g', 'n'), ('b', 'g', 'a'), ('b', 'l', 'n'), ('b', 'l', 'a'), ('b', 'n', 'a'), ('f', 'g', 'l'), ('f', 'g', 'n'), ('f', 'g', 'a'), ('f', 'l', 'n'), ('f', 'l', 'a'), ('f', 'n', 'a'), ('g', 'l', 'n'), ('g', 'l', 'a'), ('g', 'n', 'a'), ('l', 'n', 'a')]

the list is long.

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1123480

Start with using a list of dictionaries instead of 30 separate variables:

dd = [{1: ['a', 'b'], 2: ['a', 'c'], 3: ['d', 'e']}, {1: ['f', 'g', 'l'], 2: ['h', 'i'], 4: ['j', 'k', 'm']}, ...]

And use itertools.product() to generate all combinations between matching contained lists:

from itertools import product

for key in dd[0]:
    for combination in product(*(d[key] for d in dd)):
        print ' '.join(combination)

Here the generator expression retrieves the same key from all dictionaries in one statement.

Output for your sample input; note, there is no dd2[3] and no dd3[2], so for keys 2 and 3 no output is generated.

>>> for key in dd[0]:
...     for combination in product(*(d[key] for d in dd)):
...         print ' '.join(combination)
... 
handler entrenador mister
handler entrenador allenatorecoach
handler míster mister
handler míster allenatorecoach
handler entrenadora mister
handler entrenadora allenatorecoach
coach entrenador mister
coach entrenador allenatorecoach
coach míster mister
coach míster allenatorecoach
coach entrenadora mister
coach entrenadora allenatorecoach
manager entrenador mister
manager entrenador allenatorecoach
manager míster mister
manager míster allenatorecoach
manager entrenadora mister
manager entrenadora allenatorecoach

Looping over many keys then producing products from 30 different lists is going to take a while. You may have to rethink what you are trying to do and avoid doing too much work.

Upvotes: 3

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