Question

给出的指示要求返回字符串中每个单词长度的代码。所以它会计算每个单词中的字母数量并将其打印在单词旁边我有这段代码：

def word_lengths(a):
    a = a.lower()
    c = list(a)
    a = ""
    for x in c:
        if x == "," or x == "." or x == "'" or x == "!" or x == "?":
            c[c.index(x)] = ""
    for x in c:
        a += x
    y = a.split()
    z = {}
    for x in y:
        z[x] = len(x)
    return z
print(word_lengths("I ate a bowl of cereal out of a dog bowl today."))

返回：

{'dog': 3, 'bowl': 4, 'a': 1, 'out': 3, 'of': 2, 'ate': 3, 'cereal': 6, 'i': 1, 'today': 5}

Answer 1

您可以将collections.defaultdict用于O（n）解决方案：

from collections import defaultdict
from string import punctuation

def word_lengths(x):
    table = str.maketrans(punctuation, ' ' * len(punctuation))
    # alternatively, table = str.maketrans({key: None for key in punctuation})
    x = x.translate(table).lower()
    d = defaultdict(list)
    for word in x.split():
        d[len(word)].append(word)
    return d

res = word_lengths("I ate a bowl of cereal out of a dog bowl today.")

# defaultdict(list,
#             {1: ['i', 'a', 'a'],
#              2: ['of', 'of'],
#              3: ['ate', 'out', 'dog'],
#              4: ['bowl', 'bowl'],
#              5: ['today'],
#              6: ['cereal']})

<强>解释

首先删除标点符号（根据@Patrick's solution）并将字符串设为小写。
初始化defaultdict个列表。
按空格拆分列表，迭代单词并将元素附加到字典列表值。

Answer 2

使用简单的迭代

<强>演示：

def word_lengths(s):
    d = {}
    for i in s.split():           #Split by space
        l = len(i)
        if l not in d:            #Create len as key
            d[l] = [i]
        else:
            d[l].append(i)  
    return d


print(word_lengths("I ate a bowl of cereal out of a dog bowl today."))

<强>输出：

{1: ['I', 'a', 'a'], 2: ['of', 'of'], 3: ['ate', 'out', 'dog'], 4: ['bowl', 'bowl'], 6: ['cereal', 'today.']}

Answer 3

这是一个使用str.translate

处理标点符号的版本

def word_lengths(s, remove='.,!?'):
    trans=str.maketrans('', '', remove)
    s = s.lower().translate(trans)
    d = defaultdict(list)
    for word in s.split():
        d[len(word)].append(word)
    return dict(d)  # Probably unnecessary and return d would work

word_lengths("I ate a bowl of cereal out of a dog bowl today.")

给我们

{1: ['i', 'a', 'a'],
 2: ['of', 'of'],
 3: ['ate', 'out', 'dog'],
 4: ['bowl', 'bowl'],
 5: ['today'],
 6: ['cereal']}

Answer 4

您可以使用defaultdict执行此操作，collections是标准库from collections import defaultdict import re def word_lengths(text): d = defaultdict(list) for word in re.findall(r'\w+', text.lower()): d[len(word)].append(word) return d模块中众多有用的数据结构之一。

re.findall

我们使用this.form.get("formControleName");仅匹配单词，没有空格和标点符号。如果要将连字符和撇号包含为单词字符，可以调整正则表达式。

Answer 5

可以简单地循环值以生成字典。

In [1]: c = defaultdict(list)

In [2]: for word in "I ate a bowl of cereal out of a dog bowl today.".split(' '):
...:     c[len(word)].append(word)
...:     

In [3]: c
Out[4]: 
defaultdict(list,
            {1: ['I', 'a', 'a'],
             2: ['of', 'of'],
             3: ['ate', 'out', 'dog'],
             4: ['bowl', 'bowl'],
             6: ['cereal', 'today.']})

将句子中的单词长度映射到单词列表

5 个答案: