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[英]JavaScript: Find Matches AND Differences in two Arrays of Objects
[英]Find matches from an array of arrays based on matches condition in javascript
鉴于我有一个数组,如下所示:
var array = [["001"," 003"],["002"," 001"],["001"," 002"],["002"," 001"], ["", "001"]]
条件是数组必须相互匹配2个或更多个匹配项,然后它将结果存储到新数组中。
我正在使用以下javascript方法,该方法将针对数组中的每个数组进行比较。
我正在尝试做的是使javascript方法不循环数组中的每个数组来查找匹配项,但这将基于上述条件(2个匹配项或更多)
JavaScript方法:
var array = [["001"," 003"],["002"," 001"],["001"," 002"],["002"," 001"], ["", "001"]] console.log( array.shift().reduce(function(res, v) { if (res.indexOf(v) === -1 && array.every(function(a) { return a.indexOf(v) !== -1; })) res.push(v); return res; }, []) )
上面的数组的结果将为空,根据上述条件(至少2个匹配项),我希望它是:
["001", "002", " 001"]
您的回答将不胜感激!
谢谢
有关O(N)
解决方案(当前的.every
里面.reduce
为O(N^2)
可以使用reduce
创建的字符串,其值是字符串的出现次数索引的对象。 然后,对对象的keys
进行.filter
,以获得至少具有两个数的keys
:
const input = [["001"," 003"],["002"," 001"],["001"," 002"],["002"," 001"], ["", "001"]]; const obj = input.reduce((a, arr) => { arr.forEach((str) => { a[str] = (a[str] || 0) + 1; }); return a; }, {}); const output = Object.keys(obj) .filter(key => obj[key] >= 2) console.log(output);
请注意,由于输入中包含两个" 001"
字符串,因此它们也包含在输出中。
对于ES5版本,我通过Babel运行了以上内容,并得到:
var input = [["001", " 003"], ["002", " 001"], ["001", " 002"], ["002", " 001"], ["", "001"]]; var obj = input.reduce(function (a, arr) { arr.forEach(function (str) { a[str] = (a[str] || 0) + 1; }); return a; }, {}); var output = Object.keys(obj).filter(function (key) { return obj[key] >= 2; }); console.log(output);
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