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
高性能和不可变的解决方案
Javascript
const excludeFromArr = (arr, exclude) => {
const excludeMap = exclude.reduce((all, item) => ({ ...all, [item]: true }), {});
return arr.filter((item) => !excludeMap?.[item]);
};
打字稿:
const excludeFromArr = (arr: string[], exclude: string[]): string[] => {
const excludeMap = exclude.reduce<Record<string, boolean>>((all, item) => ({ ...all, [item]: true }), {});
return arr.filter((item) => !excludeMap?.[item]);
};
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));
如果你正在使用Typescript并且想要匹配单个属性值，这应该基于上面Craciun Ciprian的答案。
您还可以通过允许非对象匹配和/或多属性值匹配使其更通用。
/**
*
* @param arr1 The initial array
* @param arr2 The array to remove
* @param propertyName the key of the object to match on
*/
function differenceByPropVal<T>(arr1: T[], arr2: T[], propertyName: string): T[] {
return arr1.filter(
(a: T): boolean =>
!arr2.find((b: T): boolean => b[propertyName] === a[propertyName])
);
}
//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' ]
你可以使用_。by和lodash的区别
const myArray = [
{name: 'deepak', place: 'bangalore'},
{name: 'chirag', place: 'bangalore'},
{name: 'alok', place: 'berhampur'},
{name: 'chandan', place: 'mumbai'}
];
const toRemove = [
{name: 'deepak', place: 'bangalore'},
{name: 'alok', place: 'berhampur'}
];
const sorted = _.differenceBy(myArray, toRemove, 'name');
示例代码:CodePen
高性能和不可变的解决方案
Javascript
const excludeFromArr = (arr, exclude) => {
const excludeMap = exclude.reduce((all, item) => ({ ...all, [item]: true }), {});
return arr.filter((item) => !excludeMap?.[item]);
};
打字稿:
const excludeFromArr = (arr: string[], exclude: string[]): string[] => {
const excludeMap = exclude.reduce<Record<string, boolean>>((all, item) => ({ ...all, [item]: true }), {});
return arr.filter((item) => !excludeMap?.[item]);
};
