检查字符串中的（仅整个）单词

Question

Checkio 培训。该任务称为流行词。任务是从给定字符串的（字符串）列表中搜索单词。例如：textt="When I was One I had just begun When I was Two I was nearly new"wwords=['i', 'was', 'three', 'near']我的代码是这样的：def popular_words(text: str, words: list) -> dict:    # your code here    occurence={}    text=text.lower()    for i in words:        occurence[i]=(text.count(i))    # incorrectly takes "nearly" as "near"    print(occurence)    return(occurence)popular_words(textt,wwords)几乎可以正常工作，返回{'i': 4, 'was': 3, 'three': 0, 'near': 1} 因此将“接近”视为“接近”的一部分。这显然是作者的意图。但是，除了"search for words that are not first (index 0) or last (last index) and for these that begin/end with whitespace"请问可以帮忙吗？请以这个相当幼稚的代码为基础。

Answer 1

你最好拆分你的句子，然后计算单词，而不是子字符串：

textt="When I was One I had just begun When I was Two I was nearly new"

wwords=['i', 'was', 'three', 'near']

text_words = textt.lower().split()

result = {w:text_words.count(w) for w in wwords}

print(result)

印刷：

{'three': 0, 'i': 4, 'near': 0, 'was': 3}

如果文本现在有标点符号，最好使用正则表达式根据非字母数字拆分字符串：

import re

textt="When I was One, I had just begun.I was Two when I was nearly new"

wwords=['i', 'was', 'three', 'near']

text_words = re.split("\W+",textt.lower())

result = {w:text_words.count(w) for w in wwords}

结果：

{'was': 3, 'near': 0, 'three': 0, 'i': 4}

（另一种替代方法是使用findall在字字符：text_words = re.findall(r"\w+",textt.lower())）

现在，如果您的“重要”单词列表很大，也许最好计算所有单词，然后使用经典过滤器进行过滤collections.Counter：

text_words = collections.Counter(re.split("\W+",textt.lower()))

result = {w:text_words.get(w) for w in wwords}

Answer 2

您的简单解决方案是：

from collections import Counter

textt="When I was One I had just begun When I was Two I was nearly new".lower()

wwords=['i', 'was', 'three', 'near']

txt = textt.split()

keys = Counter(txt)

for i in wwords:

    print(i + ' : ' + str(keys[i]))