创建文字中的单词的字典

Question

I want to create a dictionary of all unique words in the text. The key is the word and the value is the word's frequency

dtt = ['you want home at our peace', 'we went our home', 'our home is nice', 'we want peace at home']
word_listT = str(' '.join(dtt)).split()
wordsT = {v:k for (k, v) in enumerate(word_listT)}
print wordsT

I expect something like this:

{'we': 2, 'is': 1, 'peace': 2, 'at': 2, 'want': 2, 'our': 3, 'home': 4, 'you': 1, 'went': 1, 'nice': 1}

However, I receive this:

{'we': 14, 'is': 12, 'peace': 16, 'at': 17, 'want': 15, 'our': 10, 'home': 18, 'you': 0, 'went': 7, 'nice': 13}

Apparently, I am misusing the functionality or doing something wrong.

Please, help

Answer 1

The problem with what you are doing is you are storing the array index of where the word is instead of a count of those words.

To achieve this you can just use collections.Counter

from collections import Counter

dtt = ['you want home at our peace', 'we went our home', 'our home is nice', 'we want peace at home']
counted_words = Counter(' '.join(dtt).split())
# if you want to see what the counted words are you can print it
print counted_words

>>> Counter({'home': 4, 'our': 3, 'we': 2, 'peace': 2, 'at': 2, 'want': 2, 'is': 1, 'you': 1, 'went': 1, 'nice': 1})

SOME CLEANUP: as mentioned in the comments

str() is unnecessary for your ' '.join(dtt).split()

You can also remove the list assignment and do your counter on the same line

Counter(' '.join(dtt).split())

A little more detail about your list indices; first you have to understand what your code is doing.

dtt = [
    'you want home at our peace', 
    'we went our home', 
    'our home is nice', 
    'we want peace at home'
]

Notice you have 19 words here; print len(word_listT) returns 19. Now on the next line word_listT = str(' '.join(dtt)).split() you are making a list of all of the words, which looks like this

word_listT = [
    'you', 
    'want', 
    'home', 
    'at', 
    'our', 
    'peace', 
    'we', 
    'went', 
    'our', 
    'home', 
    'our', 
    'home', 
    'is', 
    'nice', 
    'we', 
    'want', 
    'peace', 
    'at', 
    'home'
] 

Count them again: 19 words. The very last word is 'home'. And list indices start at 0 so 0 to 18 = 19 elements. yourlist[18] is 'home'. This has nothing to do with the string location or anything, just the index of your new array. :)

Answer 2

Try this:

from collections import defaultdict

dtt = ['you want home at our peace', 'we went our home', 'our home is nice', 'we want peace at home']
word_list = str(' '.join(dtt)).split()
d = defaultdict(int)
for word in word_list:
    d[word] += 1

Answer 3

enumerate returns a list of words with their indices, not with their frequency. That is, when you create the wordsT dictionary, each v is actually the index in word_listT of the last instance of k . To do what you want, using a for-loop is probably the most straightforward.

wordsT = {}
for word in word_listT:
    try:
        wordsT[word]+=1
    except KeyError:
        wordsT[word] = 1