我有一个这样的数组：
var array = [
{"id":"A","type":"blue","rng":"50"},
{"id":"A","type":"blue","rng":"75"},
{"id":"A","type":"grey","rng":"76"},
{"id":"B","type":"blue","rng":"50"},
{"id":"B","type":"grey","rng":"85"},
{"id":"B","type":"grey","rng":"86"},
{"id":"C","type":"blue","rng":"50"},
{"id":"C","type":"grey","rng":"65"}
]
注意：对象按随机顺序排列。
我需要用较高的 "id":"*","type":"blue"过滤掉重复的"id":"*","type":"grey"和"rng" 。
所以最后的结果是：
var result = [
{"id":"A","type":"blue","rng":"50"},
{"id":"A","type":"grey","rng":"76"},
{"id":"B","type":"blue","rng":"50"},
{"id":"B","type":"grey","rng":"86"},
{"id":"C","type":"blue","rng":"50"},
{"id":"C","type":"grey","rng":"65"}
]
我热衷于使用下划线，但也欢迎任何其他解决方案。
答案 0 :(得分：2)
您可以使用哈希表和组合键作为索引的引用。
var data = [{ id: "A", type: "blue", rng: "50" }, { id: "A", type: "blue", rng: "75" }, { id: "A", type: "grey", rng: "76" }, { id: "B", type: "blue", rng: "50" }, { id: "B", type: "grey", rng: "85" }, { id: "B", type: "grey", rng: "86" }, { id: "C", type: "blue", rng: "50" }, { id: "C", type: "grey", rng: "65" }],
result = data.reduce(function (hash) {
return function (r, o) {
var key = ['id', 'type'].map(function (k) { return o[k]; }).join('|');
if (!(key in hash)) {
hash[key] = r.push(o) - 1;
} else if (r[hash[key]].rng < o.rng) {
r[hash[key]] = o;
}
return r;
};
}(Object.create(null)), []);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 1 :(得分：1)
您可以使用forEach()循环和一个对象作为thisArg参数来检查是否存在具有相同id|type的对象。
var array = [{"id":"A","type":"blue","rng":"50"},{"id":"A","type":"blue","rng":"75"},{"id":"A","type":"grey","rng":"76"},{"id":"B","type":"blue","rng":"50"},{"id":"B","type":"grey","rng":"85"},{"id":"B","type":"grey","rng":"86"},{"id":"C","type":"blue","rng":"50"},{"id":"C","type":"grey","rng":"65"}]
var result = []
array.forEach(function(e) {
// Create key with id and type
var key = e.id + '|' + e.type;
// Check if key exists in this object and if it doesn't create property with value of current value and push that value to result array
if(!this[key]) this[key] = e, result.push(this[key])
else {
// Otherwise check if rng of current element is < of rng of previous object with same key and if it is set rng to rng of current object
if(e.rng < this[key].rng) this[key].rng = e.rng
}
}, Object.create(null))
console.log(result)