CODEWITHSUNDEEP
pythonpython-3.xdictionarykeyrename
nfmcclure
nfmcclure

Reputation: 3141

Aggregating and Renaming Keys in Dictionary

I have a word occurrence dictionary, and a synonym dictionary.

Word occurrence dictionary example:

word_count = {'grizzly': 2, 'panda': 4, 'beer': 3, 'ale': 5}

Synonym dictionary example:

synonyms = {
            'bear': ['grizzly', 'bear', 'panda', 'kodiak'],
            'beer': ['beer', 'ale', 'lager']
           }

I would like to comibine/rename aggregate the word count dictionary as

new_word_count = {'bear': 6, 'beer': 8}

I thought I would try this:

new_dict = {}
for word_key, word_value in word_count.items():           # Loop through word count dict
    for syn_key, syn_value in synonyms.items():           # Loop through synonym dict
        if word_key in [x for y in syn_value for x in y]: # Check if word in synonyms
            if syn_key in new_dict:                       # If so:
                new_dict[syn_key] += word_value           #   Increment count
            else:                                         # If not:
                new_dict[syn_key] = word_value            #   Create key

But this isn't working, new_dict ends up empty. Also, is there an easier way to do this? Maybe using dictionary comprehension?

Upvotes: 3

Views: 156

Answers (2)

vaultah
vaultah

Reputation: 46533

Using dict comprehension, sum and dict.get:

In [11]: {w: sum(word_count.get(x, 0) for x in ws) for w, ws in synonyms.items()}
Out[11]: {'bear': 6, 'beer': 8}

Using collections.Counter and dict.get:

from collections import Counter
ec = Counter()
for x, vs in synonyms.items():
    for v in vs:
        ec[x] += word_count.get(v, 0)
print(ec) # Counter({'bear': 6, 'beer': 8})

Upvotes: 4

inspectorG4dget
inspectorG4dget

Reputation: 114005

Let's change your synonym dictionary a little. Instead of mapping from a word to a list of all its synonyms, let's map from a word to its parent synonym (i.e. ale to beer). This should speed up lookups

synonyms = {
            'bear': ['grizzly', 'bear', 'panda', 'kodiak'],
            'beer': ['beer', 'ale', 'lager']
           }
synonyms = {syn:word for word,syns in synonyms.items() for syn in syns}

Now, let's make your aggregate dictionary:

word_count = {'grizzly': 2, 'panda': 4, 'beer': 3, 'ale': 5}
new_word_count = {}
for word,count in word_count:
    word = synonyms[word]
    if word not in new_word_count:
        new_word_count[word] = 0
    new_word_count[word] += count

Upvotes: 1

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