CODEWITHSUNDEEP
pythondictionarynested
apol96
apol96

Reputation: 210

How to fill in keys of nested dictionary

I have a nested dictionary whose values are all 0. e.g.

dic = {0:{'year': 0, 'drive': 0, 'local': 0,},
       1:{'day': 0, 'element': 0, 'party': 0},
       2:{'carry': 0}}

I also have a list of words:

wordlist = ['light', 'day', 'year', 'day', 'party', 'care', 'local']

I want to check for each word in the wordlist if it appears in the dictionary, and if it does, append the relative key by 1.

So for the above example I want the following output:

dic = {0:{'year': 1, 'drive': 0, 'local': 1,},
       1:{'day': 2, 'element': 0, 'party': 1},
       2:{'carry': 0}}

Upvotes: 0

Views: 333

Answers (3)

Thierry Lathuille
Thierry Lathuille

Reputation: 24232

Here is a more efficient solution that iterates only once on the list of names, and once on the nested dict, so it will be O(nb of items of dict + size of list):

from collections import Counter

dic = {0:{'year': 0, 'drive': 0, 'local': 0,},
       1:{'day': 0, 'element': 0, 'party': 0},
       2:{'carry': 0}}

wordlist = ['light', 'day', 'year', 'day', 'party', 'care', 'local']

word_counts = Counter(wordlist)

for subdict in dic.values():
    for key in subdict:
        if key in word_counts:
            subdict[key] += word_counts[key]
            
print(dic)
# {0: {'year': 1, 'drive': 0, 'local': 1}, 1: {'day': 2, 'element': 0, 'party': 1}, 2: {'carry': 0}}

(note that key in word_counts is an O(1) operation)

Upvotes: 1

tomo_iris427
tomo_iris427

Reputation: 158

It is simple to store into dict after counting.

dic = {0:{'year': 0, 'drive': 0, 'local': 0,},
       1:{'day': 0, 'element': 0, 'party': 0},
       2:{'carry': 0}}

wordlist = ['light', 'day', 'year', 'day', 'party', 'care', 'local']

keys=[k for v in dic.values() for k in v.items() ]

dic_counts={k:wordlist.count(k) for k in keys}
dic={k1:{k2:dic_counts[k2] for k2 in v1.items()} for k1, v1 in dic.items()}

You don't need to ready keys as dict.

dic_keys = {0:['year', 'drive', 'local' ],...}
wordlist = ['light', 'day', 'year', 'day', 'party', 'care', 'local']

keys=[k for v in dic.values() for k in v.items() ]

dic_counts={k:wordlist.count(k) for k in keys}
#dic={k1:{k2:dic_counts[k2] for k2 in v1.items()} for k1, v1 in dic.items()}
dic={k1:{k2:dic_counts[k2] for k2 in v1} for k1, v1 in dic.items()}

Upvotes: 1

Red
Red

Reputation: 27557

Here is how you can use a nested for loop:

dic = {0:{'year': 0, 'drive': 0, 'local': 0,},
       1:{'day': 0, 'element': 0, 'party': 0},
       2:{'carry': 0}}

wordlist = ['light', 'day', 'year', 'day', 'party', 'care', 'local']

for word in wordlist:
    for index in dic:
        if word in dic[index]:
            dic[index][word] += 1

Output:

{0: {'year': 1, 'drive': 0, 'local': 1}, 1: {'day': 2, 'element': 0, 'party': 1}, 2: {'carry': 0}}

Upvotes: 2

Related Questions