[js] 第197天 写一个方法找出一段话里面出现频率最多的词

[haizhilin2013]

第197天 写一个方法找出一段话里面出现频率最多的词

[vkboo]

• 写一个比较简单粗暴易懂的方法

var str = 'ok,i am the danger ha ha,great,ok!'
var reg = /\s|,|!|\./;
var arrs = str.split(reg);
var strMaps = [];
var results = [];

arrs.forEach(e => {
    var keys = strMaps.map(e => Object.keys(e)[0]);
    if (!~keys.indexOf(e))
        strMaps.push({
            [e]: calcCount(arrs, e),
        })
});


var results = strMaps.sort((a, b) => {
    var aValue = Object.values(a)[0];
    var bValue = Object.values(b)[0];
    return bValue - aValue;
})

console.log(results) // 按字符重复量降序排序



/** 计算item在数组中出现的次数 */
function calcCount(ary, item) {
    var count = 0;
    var _ary = [...ary];
    while (_ary.includes(item)) {
        count++;
        let index = _ary.indexOf(item);
        _ary.splice(index, 1);
    }
    return count;
}

[wheatup]

一行代码实现方法：

const texts = 'I have a pen, I have an apple, Uh! apple pen. Pen pineapple apple pen.'
const mostFrequentWord = Object.entries(texts.toLowerCase()
	.split(/[\s\,\.\!\?;]/)
	.filter(word => word)
	.reduce((sum, word) => ({ ...sum, [word]: (sum[word] || 0) + 1 }), {}))
	.sort((a, b) => b[1] - a[1])[0];													

console.log(mostFrequentWord);		// ["pen", 4]

[Seasonley]

不排序，只保存最大的，降低时间复杂度

const fn = (texts) => (',' + texts).split(/\W+/).reduce((acc, i) => (
    i = i.toLowerCase(),
    acc[i] = (acc[i] || 0) + 1,
    acc.$.n < acc[i] ? (acc.$.w = i, acc.$.n = acc[i], acc) : acc
), { $: { w: '', n: -1 } }).$

fn('I have a pen, I have an apple, Uh! apple pen. Pen pineapple apple pen.')
//{ w: 'pen', n: 4 }

[xiaoqiangz]

const fn = (texts) => { let result = texts.split(/\W+/).reduce((acc, cur) => { return acc.has(cur) ? acc.set(cur, acc.get(cur) + 1) : acc.set(cur, 1) }, new Map()) return Array.from(result).sort((a, b) => b[1]-a[1])[0] } var s = fn('I have a pen, I have an apple, Uh! apple pen. Pen pineapple apple pen.'.toLocaleLowerCase()) console.log(s)