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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
鑒於我有一個數組,如下所示:
var array = [["001"," 003"],["002"," 001"],["001"," 002"],["002"," 001"], ["", "001"]]
條件是數組必須相互匹配2個或更多個匹配項,然后它將結果存儲到新數組中。
我正在使用以下javascript方法,該方法將針對數組中的每個數組進行比較。
我正在嘗試做的是使javascript方法不循環數組中的每個數組來查找匹配項,但這將基於上述條件(2個匹配項或更多)
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; }, []) )
上面的數組的結果將為空,根據上述條件(至少2個匹配項),我希望它是:
["001", "002", " 001"]
您的回答將不勝感激!
謝謝
有關O(N)
解決方案(當前的.every
里面.reduce
為O(N^2)
可以使用reduce
創建的字符串,其值是字符串的出現次數索引的對象。 然后,對對象的keys
進行.filter
,以獲得至少具有兩個數的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);
請注意,由於輸入中包含兩個" 001"
字符串,因此它們也包含在輸出中。
對於ES5版本,我通過Babel運行了以上內容,並得到:
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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