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 retextt="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 Countertextt="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]))