python - 将句子中的单词长度映射到单词列表

Question

给出的说明要求返回字符串中每个单词长度的代码。所以就像它会计算每个单词中的字母数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']})

解释

• 首先删除标点符号(根据 @Patrick's solution )并使您的字符串小写。
• 初始化列表的 defaultdict。
• 按空格拆分您的列表，迭代单词并将元素附加到您的字典列表值。