我有问题添加一个数组的所有元素以及平均它们。我将如何做到这一点，并实现它与我目前的代码?元素的定义如下所示。
<script type="text/javascript">
//<![CDATA[
var i;
var elmt = new Array();
elmt[0] = "0";
elmt[1] = "1";
elmt[2] = "2";
elmt[3] = "3";
elmt[4] = "4";
elmt[5] = "7";
elmt[6] = "8";
elmt[7] = "9";
elmt[8] = "10";
elmt[9] = "11";
// Problem here
for (i = 9; i < 10; i++){
document.write("The sum of all the elements is: " + /* Problem here */ + " The average of all the elements is: " + /* Problem here */ + "<br/>");
}
//]]>
</script>
当前回答
在常绿浏览器上，你可以使用箭头函数 Avg = [1,2,3].reduce((a,b) => (a+b);
运行10万次，for循环方法和reduce方法之间的时间差可以忽略不计。
s=Date.now();for(i=0;i<100000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length }; console.log("100k reduce took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<100000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl }; console.log("100k for loop took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<1000000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl }; console.log("1M for loop took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<1000000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length }; console.log("1M reduce took " + (Date.now()-s) + "ms."); /* * RESULT on Chrome 51 * 100k reduce took 26ms. * 100k for loop took 35ms. * 10M for loop took 126ms. * 10M reduce took 209ms. */
其他回答
将for循环计数器设置为0....你得到了元素9，然后你就完成了。其他答案都是基础数学。使用一个变量来存储你的和(需要将字符串转换为整数)，然后除以你的数组长度。
这是你的一句台词:
var average = arr.reduce((sum,item,index,arr)=>index !== arr.length-1?sum+item:sum+item/arr.length,0)
你也可以使用lodash， _.sum(数组)和_.mean(数组)在数学部分(也有其他方便的东西)。
_.sum([4, 2, 8, 6]);
// => 20
_.mean([4, 2, 8, 6]);
// => 5
在常绿浏览器上，你可以使用箭头函数 Avg = [1,2,3].reduce((a,b) => (a+b);
运行10万次，for循环方法和reduce方法之间的时间差可以忽略不计。
s=Date.now();for(i=0;i<100000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length }; console.log("100k reduce took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<100000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl }; console.log("100k for loop took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<1000000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl }; console.log("1M for loop took " + (Date.now()-s) + "ms."); s=Date.now();for(i=0;i<1000000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length }; console.log("1M reduce took " + (Date.now()-s) + "ms."); /* * RESULT on Chrome 51 * 100k reduce took 26ms. * 100k for loop took 35ms. * 10M for loop took 126ms. * 10M reduce took 209ms. */
一种狡猾的方法可以做到这一点，尽管它确实需要使用(非常讨厌的)eval()。
var sum = eval(elmt.join('+')), avg = sum / elmt.length;
document.write("The sum of all the elements is: " + sum + " The average of all the elements is: " + avg + "<br/>");
我只是想把这个作为“盒子外”的选项之一。你永远不知道，狡猾可能会给你加分(或扣分)。
