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pythondictionarynestedduplicates
thxclown
thxclown

Reputation: 37

Getting duplicates from nested dictionary

I'm fairly new to python and have the following problem. I have a nested dictionary in the form of

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

and would like to find all the keys that have the same values. The output should look similar to this. 

1 : [a,b]
2 : [a,c]

..

Many thanks in Advance for any help!

Upvotes: 1

Views: 1086

Answers (6)

Soheil__K
Soheil__K

Reputation: 642

This can be easily achieved with two inner for loops:

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
out = {}
for key in dict:
    for value in dict[key]:
        if value not in out:
            out[value]= [key]
        else:
            out[value]+= [key]

print out  # {'1': ['a', 'b'], '3': ['c'], '2': ['a', 'c'], '5': ['b']}

Upvotes: 0

han solo
han solo

Reputation: 6590

This can be easily done using defaultdict from collections,

>>> d = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
>>> from collections import defaultdict
>>> dd = defaultdict(list)
>>> for key,vals in d.items():
...   for val in vals:
...     dd[val].append(key)
... 
>>>>>> dict(dd)
{'1': ['a', 'b'], '3': ['c'], '2': ['a', 'c'], '5': ['b']}

Upvotes: 0

Samagra Singh Tomar
Samagra Singh Tomar

Reputation: 52

before we go to the solution, lemme tell you something. What you've got there is not a nested dictionary but rather sets within the dictionary.

Some python terminologies to clear that up:

  • Array: [ 1 , 2 ]

    Arrays are enclosed in square braces & separated by commas.

  • Dictionary: { "a":1 , "b":2 }

    Dictionaries are enclosed in curly braces & separate "key":value pairs with comma. Here, "a" & "b" are keys & 1 & 2 would be their respective values.

  • Set: { 1 , 2 }

    Sets are enclosed in curly braces & separated by commas.

    dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

Here, {'1', '2'} is a set in a dictionary with key 'a'. Thus, what you've got is actually set in a dictionary & not a nested dictionary.

Solution

Moving on to the solution, sets are not iterable meaning you can't go through them one by one. So, you gotta turn them into lists & then iterate them.

# Initialize the dictionary to be processed
data = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

# Create dictionary to store solution
sol = {} # dictionary to store element as a key & sets containing that element as an array
# Eg., sol = { "1" : [ "a" , "b" ] }
# This shows that the value 1 is present in the sets contained in keys a & b.

# Record all elements & list every set containing those elements
for key in data. keys (): # iterate all keys in the dictionary
    l = list ( data [ key ] ) # convert set to list
    for elem in l: # iterate every element in the list
        if elem in sol. keys (): # check if elem already exists in solution as a key
            sol [ elem ]. append ( key ) # record that key contains elem
        else:
            sol [ elem ] = [ key ] # create a new list with elem as key & store that key contains elem

# At this time, sol would be
# {
#     "1" : [ "a" , "b" ] ,
#     "2" : [ "a" , "C" ] ,
#     "3" : [ "c" ] ,
#     "5" : [ "b" ]
# }

# Since, you want only the ones that are present in more than 1 sets, let's remove them
for key in sol : # iterate all keys in sol
    if sol [ key ]. length < 2 : # Only keys in at least 2 sets will be retained
        del sol [ key ] # remove the unrequired element

# Now, you have your required output in sol
print ( sol )
# Prints:
# {
#     "1" : [ "a" , "b" ] ,
#     "2" : [ "a" , "c" ]
# }

I hope that helps you...

Upvotes: 1

Thierry Lathuille
Thierry Lathuille

Reputation: 24232

You can use a defaultdict to build the output easily (and sort it if you want the keys in sorted order):

from collections import defaultdict

d = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

out  = defaultdict(list)

for key, values in d.items():
    for value in values:
        out[value].append(key)

# for a sorted output (dicts are ordered since Python 3.7):        
sorted_out = dict((k, out[k]) for k in sorted(out))
print(sorted_out)

#{'1': ['a', 'b'], '2': ['a', 'c'], '3': ['c'], '5': ['b']}

Upvotes: 1

recnac
recnac

Reputation: 3744

you can reverse the key-value in dict, create a value-key dict, if you only want duplicated values(find all the keys that have the same values), you can filter it:

from collections import defaultdict
def get_duplicates(dict1):
    dict2 = defaultdict(list)
    for k, v in dict1.items():
        for c in v:
            dict2[c].append(k)
    # if you want to all values, just return dict2
    # return dict2
    return dict(filter(lambda x: len(x[1]) > 1, dict2.items()))

output:

{'1': ['a', 'b'], '2': ['a', 'c']}

Upvotes: 0

Giorgos Myrianthous
Giorgos Myrianthous

Reputation: 39800

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
output = {}

for key, value in dict.items():
    for v in value:
        if v in output.keys():
            output[v].append(key)
        else:
            output[v] = [ key ]

print(output)

And the output will be 

{'2': ['a', 'c'], '1': ['a', 'b'], '5': ['b'], '3': ['c']}

Upvotes: 4

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