我有一个非常棘手的操作。所以就是这样。
我有一个名为＆＃39; data1＆＃39;
的对象数组[Object, Object, Object]
0:Object
id="00456145" //check this
name: "Rick"
upper:"0.67"
lower:"-0.34"
1:Object
id="00379321"
name:"Anjie"
upper:"0.46"
lower:"-0.56"
2:Object
id="00323113"
name:"dillan"
upper:"0.23"
lower:"-0.11"
我只对这些对象数组的id，upper和lower值感兴趣。
以下是名为＆＃39; data2＆＃39;
的第二个对象数组[Object, Object]
0:Object
id="0045614" //present here if we remove last element of '00456145'
cast="Rick"
Contact: "Yes"
upper:"0.11" //need to be updated to '0.67'
lower:"-0.11" //need to be updated to '-0.34'
1:Object
id="0032311" //present here if we remove last element of '00323113'
cast:"dillan"
Contact:"Maybe"
upper:"0.11"
lower:"-0.11"
所以，这就是我必须要做的。我将首先检查数据1＆＃39;。 id存在于＆＃39; data1＆＃39;检查。对于例如对象0具有id =＆＃34; 00456145＆＃34;
我删除了它中的最后一个数字。所以它变成id =＆＃34; 0045614＆＃34;。 然后我比较对象＆＃39; data2＆＃39;中是否存在此ID。
如果存在，则该对象0的上限和下限值出现在＆lt; data1＆＃39;被传递给对象＆quot; data2＆＃39; id存在的地方。在这种情况下，对象0的数据2＆＃39;有id =＆＃39; 0045614＆＃39;。
因此，上限和下限值将分别更新为0.67和-0.34。
与其他阵列类似。因此，最终输出应该类似于＆lt; data2＆＃39;
[Object, Object]
0:Object
id="0045614"
cast="Rick"
Contact: "Yes"
upper:"0.67" //updated
lower:"-0.34" //updated
1:Object
id="0032311"
cast:"dillan"
Contact:"Maybe"
upper:"0.23" //updated
lower:"-0.11" //updated
答案 0 :(得分：2)
在这种情况下我喜欢array#some，这样你就可以离开循环，所以没有不必要的迭代：
var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34" }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56" }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }];
var data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11" }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }];
data2 = data2.map(function(item) {
data1.some(function(a) {
if (item.id == a.id.slice(0, -1)) {
item.upper = a.upper;
item.lower = a.lower;
return true;
}
});
return item;
});
console.log(data2);
答案 1 :(得分：2)
我建议使用对象作为哈希表或地图。然后只需要两个循环，一个用于获取引用，一个用于分配。
大O： O（n + m）
虽然我实际上不知道哪一个（Map vs Object）更适合你，但你可能会得到自己的照片：
提案Object
var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34", }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56", }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }],
data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11", }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }],
hash = Object.create(null);
data1.forEach(function (a) {
hash[a.id.slice(0, -1)] = a;
});
data2.forEach(function (a) {
var o = hash[a.id];
o && Object.keys(o).forEach(function (k) {
if (k !== 'id' && a[k] !== o[k]) {
a[k] = o[k];
}
});
});
console.log(data2);
提案Map
var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34", }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56", }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }],
data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11", }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }],
map = new Map;
data1.forEach(function (a) {
map.set(a.id.slice(0, -1), a);
});
data2.forEach(function (a) {
var o = map.get(a.id);
o && Object.keys(o).forEach(function (k) {
if (k !== 'id' && a[k] !== o[k]) {
a[k] = o[k];
}
});
});
console.log(data2);
答案 2 :(得分：1)
你可以尝试这样的事情：
for (var i = 0; i < data1.length; i++) {
var item = data1[i];
var trim = item.id.substr(0, item.id.length - 1);
for (var j = 0; j < data2.length; j++) {
var item2 = data2[j];
if (item2.id === trim) {
item2.upper = item.upper;
item2.lower = item.lower;
}
}
}
如果您使用的是es6，则可以通过以下方式简化代码：
for (let item of data1) {
let trim = item.id.substr(0, item.id.length - 1);
for (let item2 of data2) {
if (item2.id === trim) {
item2.upper = item.upper;
item2.lower = item.lower;
}
}
}
答案 3 :(得分：1)
以下是使用data2的解决方案。循环遍历data1个对象，如果id与var data1 = [{
id: "00456145",
name: "Rick",
upper: "0.67",
lower: "-0.34"
}, {
id: "00379321",
name: "Anjie",
upper: "0.46",
lower: "-0.56"
}, {
id: "00323113",
name: "dillan",
upper: "0.23",
lower: "-0.11"
}];
var data2 = [{
id: "0045610", //present here if we remove last element of '00456145'
cast: "Rick",
Contact: "Yes",
upper: "0.11", //need to be updated to '0.67'
lower: "-0.11" //need to be updated to '-0.34'
}, {
id: "0032311", //present here if we remove last element of '00323113'
cast: "dillan",
Contact: "Maybe",
upper: "0.11",
lower: "-0.11"
}];
data2.forEach(function(item2){
data1.forEach(function(item1){
if (item1.id.substring(0,7) === item2.id){ // can be `item1.id.indexOf(item2.id) == 0
item2.upper = item1.upper;
item2.lower = item1.lower;
}
});
});
console.log(data2);
中的任何对象匹配，请更新匹配的对象。
{{1}}
答案 4 :(得分：1)
您可以使用类似
的内容 data1.forEach(function(obj) {
var search = obj.id.slice(0, -1);
data2.forEach(function(d) {
if (d.id === search) {
d.upper = obj.upper;
d.lower = obj.lower;
};
});
});
这样做。
答案 5 :(得分：1)
也许这就是你想要的。
var data1 = [{
id: "00456145",
name: "Rick",
upper: "0.67",
lower: "-0.34"
}, {
id: "00379321",
name: "Anjie",
upper: "0.46",
lower: "-0.56"
}, {
id: "00323113",
name: "dillan",
upper: "0.23",
lower: "-0.11"
}];
var data2 = [{
id:"0045614",
cast:"Rick",
Contact: "Yes",
upper:"0.11",
lower:"-0.11",
}, {
id:"0032311",
cast:"dillan",
Contact:"Maybe",
upper:"0.11",
lower:"-0.11",
}];
data2.forEach(function(data2Value) {
data1.forEach(function(data1Value) {
if(data1Value.id.substr(0, data1Value.id.length - 1) === data2Value.id) {
data2Value.upper = data1Value.upper;
data2Value.lower = data1Value.lower;
}
});
});
console.log(data2);