Python：嵌套字典 - 如果键不存在则创建，否则求和 1

Question

ESCENARIO
I am trying to count the number of times a word appears in a sentence, for a list of sentences. Each sentence is a list of words.
I want the final dictionary to have a key for each word in the entire corpus, and a second key indicating the sentences in which they appear, with the value being the number of times it appears in it.

CURRENT SOLUTION
The following code works correctly:

dfm = dict()
for i,sentence in enumerate(setences):
    for word in sentence:
        if word not in df.keys():
            dfm[word] = dict()
        if i not in dfm[word].keys():
            dfm[word][i] = 1
        else:
            dfm[word][i] += 1

QUESTION
Is there any cleaner way to do it with python?
I have already gone through this and this where they suggest using:

dic.setdefault(key,[]).append(value)  

and,

d = defaultdict(lambda: defaultdict(dict))

I think they are good solution, but I can't figure out how to adapt that to my particular solution.

Thanks !

Answer 1

Say you have this input:

sentences = [['dog','is','big'],['cat', 'is', 'big'], ['cat', 'is', 'dark']]

Your solution:

dfm = dict()
for i,sentence in enumerate(sentences):
    for word in sentence:
        if word not in dfm.keys():
            dfm[word] = dict()
        if i not in dfm[word].keys():
            dfm[word][i] = 1
        else:
            dfm[word][i] += 1

Defaultdict int:

from collections import defaultdict

dfm2 = defaultdict(lambda: defaultdict(int))

for i,sentence in enumerate(sentences):
    for word in sentence:
        dfm2[word][i] += 1 

Test:

dfm2 == dfm  # True


#{'dog': {0: 1},
# 'is': {0: 1, 1: 1, 2: 1},
# 'big': {0: 1, 1: 1},
# 'cat': {1: 1, 2: 1},
# 'dark': {2: 1}}

Answer 2

for cleaner version use Counter

from collections import Counter

string = 'this is america this is america'
x=Counter(string.split())
print(x)

output

Counter({'this': 2, 'is': 2, 'america': 2})

if want some own code then

copying input data (sentence) from @rassar

def func(list_:list):      
    dic = {}
    for sub_list in list_:
        for word in sub_list:
            if word not in dic.keys():
                dic.update({word:1})
            else:
                dic[word]+=1
    return dic


sentences = [['dog','is','big'],['cat', 'is', 'big'], ['cat', 'is', 'dark']]


print(func(sentences))

output

{'dog': 1, 'is': 3, 'big': 2, 'cat': 2, 'dark': 1}

Answer 3

Use counters

from collections import Counter
    
sentences = ["This is Day", "Never say die", "Chat is a good bot", "Hello World", "Two plus two equals four","A quick brown fox jumps over the lazy dog", "Young chef, bring whisky with fifteen hydrogen ice cubes"]

sentenceWords = ( Counter(x.lower() for x in sentence.split()) for sentence in sentences)

#print result
print("\n".join(str(c) for c in sentenceWords))