Python 3.x 将文件转换为dict

预期产出：

my_file = "The Itsy Bitsy Spider went up the water spout.
Down came the rain & washed the spider out.
Out came the sun & dried up all the rain,
And the Itsy Bitsy Spider went up the spout again. "

我的代码：

{'the': ['itsy', 'water', 'rain', 'spider', 'sun', 'rain', 'itsy', 'spout'], 'itsy': ['bitsy', 'bitsy'], 'bitsy': ['spider', 'spider'], 'spider': ['went', 'out', 'went'], 'went': ['up', 'up'], 'up': ['the', 'all', 'the'], 'water': ['spout'], 'spout': ['down', 'again'], 'down': ['came'], 'came': ['the', 'the'], 'rain': ['washed', 'and'], 'washed': ['the'], 'out': ['out', 'came'], 'sun': ['dried'], 'dried': ['up'], 'all': ['the'], 'and': ['the'], 'again': []}

---

您可以通过以下几个概念重现预期结果：

给定的

import string

words_set = {}
    for line in my_file:
        lower_text = line.lower()
        for word in lower_text.split():
            word = word.strip(string.punctuation + string.digits)
            if word:
                if word in words_set:
                    words_set[word] = words_set[word] + 1
                else:
                    words_set[word] = 1

代码

import string
import itertools as it
import collections as ct


data = """\
The Itsy Bitsy Spider went up the water spout.
Down came the rain & washed the spider out.
Out came the sun & dried up all the rain,
And the Itsy Bitsy Spider went up the spout again.
"""

演示

def clean_string(s:str) -> str:
    """Return a list of lowered strings without punctuation."""
    table = str.maketrans("","", string.punctuation)
    return s.lower().translate(table).replace("  ", " ").replace("\n", " ")


def get_neighbors(words:list) -> dict:
    """Return a dict of right-hand, neighboring words."""
    dd = ct.defaultdict(list)
    for word, nxt in it.zip_longest(words, words[1:], fillvalue=""):
        dd[word].append(nxt)
    return dict(dd)

{'the': ['itsy', 'water', 'rain', 'spider', 'sun', 'rain', 'itsy', 'spout'],
 'itsy': ['bitsy', 'bitsy'],
 'bitsy': ['spider', 'spider'],
 'spider': ['went', 'out', 'went'],
 'went': ['up', 'up'],
 'up': ['the', 'all', 'the'],
 'water': ['spout'],
 'spout': ['down', 'again'],
 'down': ['came'],
 'came': ['the', 'the'],
 'rain': ['washed', 'and'],
 'washed': ['the'],
 'out': ['out', 'came'],
 'sun': ['dried'],
 'dried': ['up'],
 'all': ['the'],
 'and': ['the'],
 'again': ['']}

结果

words = clean_string(data).split()
get_neighbors(words)

ct.Counter(words)

---

详细信息

clean_字符串

• 您可以使用任意数量的方法来实现。这里我们使用翻译表来替换大部分标点符号。其他可通过

  str.replace（）

  直接删除

获取邻居

• A把清单记下来。如果缺少键，则生成新的列表值
• 我们通过迭代两个并列的单词列表，一个在另一个之前，来完成口述
• 这些列表是列表，用空字符串填充较短的列表

• dict（dd）

  确保返回一个简单的dict

---

如果您只想计算字数：

演示

def clean_string(s:str) -> str:
    """Return a list of lowered strings without punctuation."""
    table = str.maketrans("","", string.punctuation)
    return s.lower().translate(table).replace("  ", " ").replace("\n", " ")


def get_neighbors(words:list) -> dict:
    """Return a dict of right-hand, neighboring words."""
    dd = ct.defaultdict(list)
    for word, nxt in it.zip_longest(words, words[1:], fillvalue=""):
        dd[word].append(nxt)
    return dict(dd)

{'the': ['itsy', 'water', 'rain', 'spider', 'sun', 'rain', 'itsy', 'spout'],
 'itsy': ['bitsy', 'bitsy'],
 'bitsy': ['spider', 'spider'],
 'spider': ['went', 'out', 'went'],
 'went': ['up', 'up'],
 'up': ['the', 'all', 'the'],
 'water': ['spout'],
 'spout': ['down', 'again'],
 'down': ['came'],
 'came': ['the', 'the'],
 'rain': ['washed', 'and'],
 'washed': ['the'],
 'out': ['out', 'came'],
 'sun': ['dried'],
 'dried': ['up'],
 'all': ['the'],
 'and': ['the'],
 'again': ['']}

结果

words = clean_string(data).split()
get_neighbors(words)

ct.Counter(words)

---

我能用上面的代码数一数重复的单词。你的问题是什么？你的代码都没有格式化。你的输出有什么逻辑要求吗