回显页面上出现的前 5 个单词

Question

是否有功能或插件可以获取页面上出现次数最多的前 5 个单词？ 我正在构建一个包含动态内容的网站，我想检查大多数出现的相同单词。

例如，“西班牙”这个词在页面上出现了大约 12 次，而“荷兰”这个词出现了大约 8 次。 我想要一个功能来自动检查这个并回应这个。

Answer 1

我不确定它会有多大用处。 你最终会得到毫无意义的词。

 let text = document.querySelector('#input').innerText.split(/\\s/); let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => { if (cv.length > 0) acc[cv] = acc[cv] + 1 || 1; return acc; }, {}); text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5); console.log(text);

 <div id="input"> <p> Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words. </p> <p> For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this. </p> </div>

添加过滤

我们可以在计算单词之前过滤数组，因此您可以向filters数组添加您喜欢的任何内容以删除这些单词。 我可能应该从一开始就想到这一点......

 const filters = ['a', 'to', 'the', 'for', 'i']; let text = document.querySelector('#input').innerText.split(/\\s/); let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => { if (cv.length > 0) acc[cv] = acc[cv] + 1 || 1; return acc; }, {}); text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5); console.log(text);

 <div id="input"> <p> Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words. </p> <p> For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this. </p> </div>