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es6 How to use set: 1. Define from var to let through "let a=[1,2,3,4,5];"; 2. Through "let set=new Set( );" syntax is set.
The operating environment of this article: Windows 7 system, ECMAScript version 6, Dell G3 computer.
How to use es6 set?
The usage of set in es6
is as follows:
//定义从var变为let 不可重复定义 let a=[1,2,3,4,5]; // Set 它类似于数组,但是成员的值都是唯一的,没有重复的值。 let set=new Set();
Set instance methods are divided into two categories: Operation methods (for operating data) and traversal methods (for traversing members).
The following will introduce four operation methods.
add(value): Add a value and return the Set structure itself.
delete(value): Delete a value and return a Boolean value indicating whether the deletion was successful.
has(value): Returns a Boolean value indicating whether the value is a member of Set.
clear(): Clear all members, no return value.
The Array.from method can convert the Set structure into an array.
Instances of the Set structure have four traversal methods that can be used to traverse members.
keys(): Returns the traverser of key names
values(): Returns the traverser of key values
entries(): Returns the key-value pair Traverser
forEach(): Use the callback function to traverse each member
It should be noted that the traversal order of Set is the insertion order.
Since the Set structure has no key names, only key values (or the key name and key value are the same value), the behaviors of the keys method and the values method are exactly the same.
Instances of the Set structure can be traversed by default, and its default traverser generation function is its values method.
This means that the values method can be omitted and the for...of loop is used to traverse the Set directly.
The spread operator (...) uses for...of loop internally, so it can also be used for Set structure.
Example
//1 数组去重 let a=[1,2,3,4,5,6,7,1,2,3]; let b=new Set([...a]); b=[...b]; // 2 求交集 并集 差集 let a=new Set[1,2,3]; let b=new Set[3,4,5]; //交集 let c=new Set([...a,...b]); //并集 let d=new Set([...a].filter(x=>b.has[x])); //交集 let d=new Set([...a].filter(x=>!b.has[x]));
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