Python 检查字符串中的（仅整）字

关于Checkio的培训。这项任务被称为流行语。任务是从给定字符串中的（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']

我的代码如下：

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)

它几乎可以正常工作，返回

{'i': 4, 'was': 3, 'three': 0, 'near': 1} 

因此，将“近”视为“近”的一部分。这显然是作者的意图。一、 然而，除了这个，我找不到其他办法

"search for words that are not first (index 0) or last (last index) and for these that begin/end with whitespace"

我可以请你帮忙吗？请以这段相当幼稚的代码为基础。

你最好把句子分开，然后数一数单词，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)

印刷品：

{'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}

（另一种选择是对单词字符使用

findall

：

text\u words=re.findall（r“\w+”，textt.lower（））

）

现在，如果您的“重要”单词列表很大，那么最好先统计所有单词，然后使用经典的

集合进行过滤。Counter

：

text_words = collections.Counter(re.split("\W+",textt.lower()))
result = {w:text_words.get(w) for w in wwords}

---

您的简单解决方案如下：

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]))

---

嘿，你想让你的代码做什么？要计算单词出现的次数吗？任务是返回一个字典，其中单词作为键，出现的次数作为值。就像现在一样。只忽略单词中的单词。值得注意的是，collections模块有一个内置的计数器类：建议您阅读感谢！我应该自己来的。很好用（当教程给了我一个包含撇号的单词，这个单词不能正确计算，但是我会在早上解决）。无论如何谢谢你！