Question

我正在用JavaScript编写一段代码，该代码应该替换JSON obj中多个字符串中所有出现的char。

并非所有字符串都包含特定字符，我们正在讨论很多字符串。所以我的问题是：在讨论效率时，是否最好进行替换或搜索字符串中的char，并且只有在找到make替换时？

换句话说：

var obj = ["str","str2","tr3","str","tr2","str3","str","s22tr2","str3","st","rtr2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3"];

选项1：

for(var i=0;i<obj.length;i++){
  if(obj[i].indexOf("s")!=-1){
      document.write(obj[i].replace(/s/gi,"*"));
  }
}

选项2：

for(var i=0;i<obj.length;i++){
    document.write(obj[i].replace(/s/gi,"*"));
}

思想？

感谢。

Answer 1

它取决于obj中元素的数量和每个元素的大小，在大多数情况下直接替换更快。到目前为止，您提供的样本最快的是“加入和替换”。检查这个非常懒惰的例子：

var obj = ["str","str2","tr3","str","tr2","str3","str","s22tr2","str3","st","rtr2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3","str","str2","str3"];
var b = obj.join(',');
b +=',' + b; //  * 2
b +=',' + b; //  * 4
b +=',' + b; //  * 8
b +=',' + b; //  * 16
b +=',' + b; //  * 32
b +=',' + b; //  * 64
b +=',' + b; //  * 128
obj = b.split(',');
var t1 = new Date().getTime();

for(var i=0;i<obj.length;i++)
    if(obj[i].indexOf("s")!=-1 || obj[i].indexOf("S")!=-1)
        document.write(obj[i].replace(/s/gi,"*"));
    else
        document.write(obj[i]);
document.write('<br>');

var t2 = new Date().getTime();

for(var i=0;i<obj.length;i++)
    document.write(obj[i].replace(/s/gi,"*"));
document.write('<br>');

var t3 = new Date().getTime();

document.write(obj.join('|').replace(/s/gi,"*").split('|').join('')); // see note
document.write('<br>');

var t4 = new Date().getTime();

alert((t2-t1) + ' vs ' + (t3-t2) + ' vs ' + (t4-t3));

注意：'|'表示未包含在对象元素中的char（或甚至是标记），有助于避免错误。

选择并选择。

更新：

在第一次测试中添加了资本S.

有趣的案例：/s/gi.test vs indexOf

搜索与搜索＆amp;更换

1 个答案: