Javascript 计算数组中的项目数,忽略重复项
我需要一个JavaScript函数,它能够计算数组中唯一项的数量(即忽略重复项)。比如说,Javascript 计算数组中的项目数,忽略重复项,javascript,arrays,Javascript,Arrays,我需要一个JavaScript函数,它能够计算数组中唯一项的数量(即忽略重复项)。比如说, var r = [car, car, bike, bike, truck] 结果应该是3而不是5。尝试以下方法: var r = ['car', 'car', 'bike', 'bike', 'truck']; var counts = [], count = 0, i = 0; for(i=0; i<r.length; i++) { if(counts[r[i]] == undefine
var r = [car, car, bike, bike, truck]
结果应该是3而不是5。尝试以下方法:
var r = ['car', 'car', 'bike', 'bike', 'truck'];
var counts = [], count = 0, i = 0;
for(i=0; i<r.length; i++) {
if(counts[r[i]] == undefined) {
counts[r[i]] = 1;
count++;
}
}
console.log(count);
var r=['car','car','bike','bike','truck'];
变量计数=[],计数=0,i=0;
对于(i=0;i使用
如果希望它适用于所有阵列:
Array.prototype.countUniques = function() {
return this.filter(function(elem, index, self) {
return index == self.indexOf(elem);
}).length;
};
用法:arr.countUniques()//3
Array.prototype.count_unique = function() {
var arr2 = [];
for (i = this.length; i--;) {
if (arr2.indexOf( this[i] ) == -1) arr2.push( this[i] );
}
return arr2.length;
}
使用
Array.prototype.count_unique = function() {
var arr2 = [];
for (i = this.length; i--;) {
if (arr2.indexOf( this[i] ) == -1) arr2.push( this[i] );
}
return arr2.length;
}
var r = ['car', 'car', 'bike', 'bike', 'truck'],
unique = r.count_unique();