Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
338 views
in Technique[技术] by (71.8m points)

javascript - Why is lodash.each faster than native forEach?

I was trying to find the fastest way of running a for loop with its own scope. The three methods I compared were:

var a = "t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t,t".split();

// lodash .each -> 1,294,971 ops/sec
lodash.each(a, function(item) { cb(item); });

// native .forEach -> 398,167 ops/sec
a.forEach(function(item) { cb(item); });

// native for -> 1,140,382 ops/sec
var lambda = function(item) { cb(item); };
for (var ix = 0, len = a.length; ix < len; ix++) {
  lambda(a[ix]);
}

This is on Chrome 29 on OS X. You can run the tests yourself here:

http://jsben.ch/BQhED

How is lodash's .each almost twice as fast as native .forEach? And moreover, how is it faster than the plain for? Sorcery? Black magic?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

_.each() is not fully compatible to [].forEach(). See the following example:

var a = ['a0'];
a[3] = 'a3';
_.each(a, console.log); // runs 4 times
a.forEach(console.log); // runs twice -- that's just how [].forEach() is specified

http://jsfiddle.net/BhrT3/

So lodash's implementation is missing an if (... in ...) check, which might explain the performance difference.


As noted in the comments above, the difference to native for is mainly caused by the additional function lookup in your test. Use this version to get more accurate results:

for (var ix = 0, len = a.length; ix < len; ix++) {
  cb(a[ix]);
}

http://jsperf.com/lo-dash-each-vs-native-foreach/15


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...