我正在寻找一种有效的方法，从javascript数组中删除所有元素，如果它们存在于另一个数组中。
// If I have this array:
var myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
// and this one:
var toRemove = ['b', 'c', 'g'];
我想对myArray进行操作，使其处于这种状态:['a'， 'd'， 'e'， 'f']
与jQuery，我使用grep()和inArray()，这工作得很好:
myArray = $.grep(myArray, function(value) {
return $.inArray(value, toRemove) < 0;
});
有没有一个纯javascript的方法来做到这一点没有循环和剪接?
当前回答
//Using the new ES6 Syntax console.log(["a", "b", "c", "d", "e", "f", "g"].filter(el => !["b", "c", "g"].includes(el))); // OR // Main array let myArray = ["a", "b", "c", "d", "e", "f", "g"]; // Array to remove const toRemove = ["b", "c", "g"]; const diff = () => (myArray = myArray.filter((el) => !toRemove.includes(el))); console.log(diff()); // [ 'a', 'd', 'e', 'f' ] // OR const diff2 = () => { return myArray = myArray.filter((el) => !toRemove.includes(el)); }; console.log(diff2()); // [ 'a', 'd', 'e', 'f' ]
其他回答
如果您正在使用对象数组。然后，下面的代码将发挥神奇的作用，其中对象属性将作为删除重复项的标准。
在下面的示例中，比较每个项目的名称，已删除重复项。
试试这个例子。http://jsfiddle.net/deepak7641/zLj133rh/
var myArray = [ {name: 'deepak', place: 'bangalore'}, {name: 'chirag', place: 'bangalore'}, {name: 'alok', place: 'berhampur'}, {name: 'chandan', place: 'mumbai'} ]; var toRemove = [ {name: 'deepak', place: 'bangalore'}, {name: 'alok', place: 'berhampur'} ]; for( var i=myArray.length - 1; i>=0; i--){ for( var j=0; j<toRemove.length; j++){ if(myArray[i] && (myArray[i].name === toRemove[j].name)){ myArray.splice(i, 1); } } } alert(JSON.stringify(myArray));
过滤方法应该做到这一点:
const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
const toRemove = ['b', 'c', 'g'];
// ES5 syntax
const filteredArray = myArray.filter(function(x) {
return toRemove.indexOf(x) < 0;
});
如果toRemove数组很大，这种查找模式可能效率很低。创建一个映射，使查找是O(1)而不是O(n)，这样会更有性能。
const toRemoveMap = toRemove.reduce(
function(memo, item) {
memo[item] = memo[item] || true;
return memo;
},
{} // initialize an empty object
);
const filteredArray = myArray.filter(function (x) {
return toRemoveMap[x];
});
// or, if you want to use ES6-style arrow syntax:
const toRemoveMap = toRemove.reduce((memo, item) => ({
...memo,
[item]: true
}), {});
const filteredArray = myArray.filter(x => toRemoveMap[x]);
//Using the new ES6 Syntax console.log(["a", "b", "c", "d", "e", "f", "g"].filter(el => !["b", "c", "g"].includes(el))); // OR // Main array let myArray = ["a", "b", "c", "d", "e", "f", "g"]; // Array to remove const toRemove = ["b", "c", "g"]; const diff = () => (myArray = myArray.filter((el) => !toRemove.includes(el))); console.log(diff()); // [ 'a', 'd', 'e', 'f' ] // OR const diff2 = () => { return myArray = myArray.filter((el) => !toRemove.includes(el)); }; console.log(diff2()); // [ 'a', 'd', 'e', 'f' ]
最简单的方法如何:
var myArray = [a, b, c, d, e, f, g的); var toRemove = ['b'， 'c'， 'g']; var myArray = myArray.filter((item) => ! console.log (myArray)
这已经很晚了，但添加这个来解释@mojtaba roohi的回答。第一个代码块将不起作用，因为每个数组都有不同的对象，即df[0] != nfl[2]。这两个对象看起来相似，但完全不同，当我们使用像数字这样的基本类型时，情况就不是这样了。
let df = [{'name': 'C'}，{'name': 'D'}] 让nfl =[{'名字':' A '},{'名字':' B '},{'名字':' C '},{'名字':' D '}) Let res = nfl。filter(x => df.indexOf(x)<0) console.log (res)
下面是工作代码:
let df = [{'name': 'C'}，{'name': 'D'}] 让nfl =[{'名字':' A '},{'名字':' B '},{'名字':' C '},{'名字':' D '}); 让res = nfl.filter ((o1群)= > ! df.some ((o2) = > o1.name = = = o2.name)); console.log (res)
