CODEWITHSUNDEEP
pythondictionary
user136819
user136819

Reputation: 239

Python Dictionary: Replace values based on key occurrence in other values

My question might involve comprehension if I'm not mistaken.
So I have a dictionary as: 

data = {
 1: [2, 4], 
 2: [3], 
 3: [1, 2, 6], 
 4: [1, 3, 6], 
 6: [3, 4]
}

I wish to rearrange the dictionary values (not keys) such that with every occurrence of a "key" in any of the other values, the current key's values should now contain the "key" in who's value it originally occurred.
For example the dictionary should now look like this: 

new_data = {
 1: [3, 4], 
 2: [1, 3], 
 3: [2, 4, 6], 
 4: [1, 6], 
 6: [3, 4]
}

I know for sure that I must use another empty dictionary to solve this, and I must use looping. I have a starting point, but it's not efficient: 

for pk,pv in zip(data.keys(), data.values()):
    #print(pk,pv)
    for pvs in pv:
        #print(pvs)
        if pk == pvs:
            new_data.setdefault(pk, [])
            new_data[pk].append(pvs)

print(new_data)

Any help is appreciated. Thanks in advance :)
(Using Ubuntu 14.04 32-Bit VM and Python 2.7)

Upvotes: 2

Views: 1547

Answers (2)

bow
bow

Reputation: 2563

Perhaps this is not the most efficient way to do this, but I would argue it's one of the nicest to read :).

One thing I should say before, if it's possible to change the initial dictionary values to set instead of list, that could result in some performance gains since set membership checks can be faster than listss.

Here goes:

def present_in(k, d):
    return {idx for idx in d if k in d[idx]}

new_data = {k: present_in(k, data) for k in data}

Basically the present_in function creates a set of values corresponding to keys where k is present as a member. We simply then use this to create the new_data dictionary using a dictionary comprehension. Note again the values of this dictionary is a set, instead of a list. You can easily wrap them in list() if you do need lists.

EDIT:

If you would like new_data to have keys that may be missing as keys in data but present as one of the values, you can change the code to create a set of all possible keys first:

all_keys = {v for vs in data.values() for v in vs}

def present_in(k, d):
    return {idx for idx in d if k in d[idx]}

new_data = {k: present_in(k, data) for k in all_keys}

Upvotes: 6

Transhuman
Transhuman

Reputation: 3547

{k:[i for i,j in data.items() if k in j] for k,_ in data.items()}
#Output:
#{1: [3, 4], 2: [1, 3], 3: [2, 4, 6], 4: [1, 6], 6: [3, 4]}

The below part essentially checks if the key(k from outer dict) is present in any of the values in dict, if present, it'll get the corresponding key into the list

[i for i,j in data.items() if k in j]

Upvotes: 1

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