计算列表中的单词并将它们添加到字典中，以及出现次数 Python

Question

I'll type the question my code is in response to below for clarity, but I'm having 2 issues. It seems to be adding correctly up to a point, and the result for the count of the other words in the sentence seems to be accurate, but rabbit suddenly jumps from 1 to 4 and I'm not sure why.

I'm also getting this: Error: AttributeError: 'int' object has no attribute 'items'

Here is the problem followed by my code. Thanks!

Provided is a string saved to the variable name sentence. Split the string into a list of words, then create a dictionary that contains each word and the number of times it occurs. Save this dictionary to the variable name word_counts.

sentence = "The dog chased the rabbit into the forest but the rabbit was too quick."
sentence_case = sentence.lower()
sentence_list = sentence_case.split()
sentence_dictionary = {}
word_counts = 0

for item in sentence_list:
    if item in sentence_dictionary:
        word_counts += 1
        sentence_dictionary[item] = word_counts

    else:
        sentence_dictionary[item] = 1

Answer 1

If, i understand you right you might remove the word_count variable to count the frequency of words

sentence = "The dog chased the rabbit into the forest but the rabbit was too quick."
sentence_case = sentence.lower()
sentence_list = sentence_case.split()
sentence_dictionary = {}

for item in sentence_list:
    if item in sentence_dictionary:
        sentence_dictionary[item] += 1

    else:
        sentence_dictionary[item] = 1

print(sentence_dictionary)

if you want to save that in word_counts you can make it like that:

word_counts = sentence_dictionary

I hope i could help you

Answer 2

Try this

sentence = "The dog chased the rabbit into the forest but the rabbit was too quick."
sentence_case = sentence.lower()
sentence_list = sentence_case.split()
sentence_dictionary = {}

for item in sentence_list:
    if item in sentence_dictionary:
        sentence_dictionary[item] += 1

    else:
        sentence_dictionary[item] = 1

Answer 3

What is the purpose behind the variable word_counts ? The current usage of this variable conflates the counts of different words. Your problem now reduces to having different counters for each word. Fortunately, you don't need to do this explicitly since sentence_dictionary is a set of counters on its own:) Just increment sentence_dictionary[item] under your if block and get rid of word_counts .

As a sidenote, this is a good example to leverage the defaultdict class provided by Python. defaultdict is initialized along with a default_factory as its argument. This allows you to initialize dictionaries of counts, lists, sets, etc. without explicitly handling edge cases.

Consider the code sample below:

dic = defaultdict(int)
dic[0] += 1
dic[0] # prints 1

If you notice, I didn't have to explicitly check if the key was present in the dictionary since the defaultdict class handles this under the hood by automatically creating the key with value 0 (value 0 because the default_factory is int()). You can explore the Python docs to understand how defaultdict works. It will save you some time and debugging in the future!

Answer 4

You can try this,it worked for me.

sentence = "The dog chased the rabbit into the forest but the rabbit was too quick."
sentence_list = sentence.split()
sentence_dictionary = {}

for item in sentence_list:
    if item not in sentence_dictionary:
        sentence_dictionary[item] = 0
    sentence_dictionary[item] += 1
word_counts = sentence_dictionary
print(word_counts)