为文件中的每个单词创建一个字典，并计算紧随其后的单词的出现频率

问题描述

限时删除！！

我正试图解决一个棘手的问题，并且迷路了.

I am trying to solve a difficult problem and am getting lost.

这是我应该做的:

INPUT: file
OUTPUT: dictionary

Return a dictionary whose keys are all the words in the file (broken by
whitespace). The value for each word is a dictionary containing each word
that can follow the key and a count for the number of times it follows it.

You should lowercase everything.
Use strip and string.punctuation to strip the punctuation from the words.

Example:
>>> #example.txt is a file containing: "The cat chased the dog."
>>> with open('../data/example.txt') as f:
...     word_counts(f)
{'the': {'dog': 1, 'cat': 1}, 'chased': {'the': 1}, 'cat': {'chased': 1}}

到目前为止，我想尽一切办法来至少拉出正确的单词:

Here's all I have so far, in trying to at least pull out the correct words:

def word_counts(f):
    i = 0
    orgwordlist = f.split()
    for word in orgwordlist:
        if i<len(orgwordlist)-1:
            print orgwordlist[i]
            print orgwordlist[i+1]

with open('../data/example.txt') as f:
    word_counts(f)

我想我需要以某种方式使用.count方法并最终将一些字典压缩在一起，但是我不确定如何为每个第一个单词计算第二个单词.

I'm thinking I need to somehow use the .count method and eventually zip some dictionaries together, but I'm not sure how to count the second words for each first word.

我知道我距离解决问题还差得很远，但是想一次迈出一步.感谢您提供任何帮助，甚至只是指向正确方向的提示.

I know I'm nowhere near solving the problem, but trying to take it one step at a time. Any help is appreciated, even just tips pointing in the right direction.

推荐答案

我们可以通过两次通过解决此问题:

We can solve this in two passes:

1. 在第一遍中，我们构造一个Counter，并使用zip(..)计算两个连续单词的元组；和
2. 然后我们将Counter放入词典字典中.

1. in a first pass, we construct a Counter and count the tuples of two consecutive words using zip(..); and
2. then we turn that Counter in a dictionary of dictionaries.

这将导致以下代码:

from collections import Counter, defaultdict

def word_counts(f):
    st = f.read().lower().split()
    ctr = Counter(zip(st,st[1:]))
    dc = defaultdict(dict)
    for (k1,k2),v in ctr.items():
        dc[k1][k2] = v
    return dict(dc)

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