javascript 在多维数组javascript或coffeescript中获取最大值
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Get largest value in multi-dimensional array javascript or coffeescript
提问by Fresheyeball
I have an array that looks like the following:
我有一个如下所示的数组:
array = [[1, 5], [4, 7], [3, 8], [2, 3],
[12, 4], [6, 6], [4, 1], [3, 2],
[8, 14]]
What I need is the largest number from the first value of the sets, so in this case 12
.
Looking at some examples online, the best way I saw to accomplish this is :
我需要的是集合的第一个值中的最大数,所以在这种情况下12
. 在网上查看一些示例,我看到的实现此目的的最佳方法是:
Math.max.apply Math, array
Problem is, this only works with single dimensional arrays. How would I impliment this for my senario? (jquery allowed)
问题是,这只适用于一维数组。我将如何为我的 senario 暗示这一点?(jquery 允许)
The end solution:
最终解决方案:
It wasn't part of the question, but I needed both the min and max from the array, and that changes things a little.
这不是问题的一部分,但我需要数组中的最小值和最大值,这会稍微改变一些事情。
unless device.IE
justTheDates = magnitudeArray.map (i) -> i[0]
@earliest = Math.min.apply Math, justTheDates
@latest = Math.max.apply Math, justTheDates
else
@earliest = magnitudeArray[0][0]
@latest = magnitudeArray[0][0]
for magnitudeItem in magnitudeArray
@earliest = magnitudeItem[0] if magnitudeItem[0] < @earliest
@latest = magnitudeItem[0] if magnitudeItem[0] > @latest
回答by
You can use .reduce()
...
您可以使用.reduce()
...
array.reduce(function(max, arr) {
return Math.max(max, arr[0]);
}, -Infinity)
Here's a version that doesn't use Math.max
...
这是一个不使用的版本Math.max
......
array.reduce(function(max, arr) {
return max >= arr[0] ? max : arr[0];
}, -Infinity);
...and a jsPerf test.
...和jsPerf 测试。
回答by zerkms
http://jsfiddle.net/zerkms/HM7es/
http://jsfiddle.net/zerkms/HM7es/
var max = Math.max.apply(Math, arr.map(function(i) {
return i[0];
}));?
So at first you use array.map()
to convert the 2-dimensional array to a flat one, and after that use Math.max()
所以一开始你使用array.map()
将二维数组转换为平面数组,然后使用Math.max()
回答by epidemian
回答by Sarah-Jane
Array.prototype.maxX = function(){
return Math.max.apply(Math,this.map(function(o){return o[0];}));
};
回答by BitAtWork
I know this is an old post, but if you (or someone else) want the largest number in the whole array, try with:
我知道这是一篇旧帖子,但如果您(或其他人)想要整个数组中的最大数字,请尝试:
var array = [[1, 5], [4, 7], [3, 8], [2, 3],
[12, 4], [6, 6], [4, 1], [3, 2],
[8, 14]];
var max = array.reduce(function (max, arr) {
return max >= Math.max.apply(max, arr) ? max : Math.max.apply(max, arr);
}, -Infinity);
console.log(max);
In this examlpe, it will return the value 14.
在本例中,它将返回值 14。
回答by jimr
Using a comprehension in CoffeeScript:
在 CoffeeScript 中使用理解:
Math.max.apply Math, (x[0] for x in array)
回答by Hyman Stone
Also, look at _underscore.js. Here is a link to the function _max().
另外,看看_underscore.js。这是函数 _max() 的链接。
- It is simply more efficient to read, write and maintain.
- 它只是更有效地读取、写入和维护。
The best part about _underscore is that there are about another hundred helper functions similar to _max. Like sort.
_underscore 最好的部分是还有大约一百个类似于 _max 的辅助函数。喜欢排序。
Compare the syntax below:
比较以下语法:
var sortedObject = _.sortBy(object, function(val, key, object) {
return val;
});
They are easy to chain, and interpret! (Like Douglas Crockford might suggest)
它们易于链接和解释!(就像道格拉斯·克罗克福德(Douglas Crockford)可能建议的那样)
An excellent JSFIDDLE, was provided in this post by @Raynos.
@Raynos在这篇文章中提供了一个优秀的 JSFIDDLE。
If you are consistently conducting array operations with raw JavaScript, check out _underscore.js, it can greatly simplify your code.
如果您一直使用原始 JavaScript 进行数组操作,请查看 _underscore.js,它可以极大地简化您的代码。
Hope that helps, All the best! Nash
希望能帮到你,祝一切顺利!纳什
回答by dorgo
Sample Input: largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
样本输入:maximumOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
function largestOfFour(arr) {
var largest = 0;
var largestArr = [];
for(var i=0; i<arr.length; i++){
for(var j=0; j<arr[i].length; j++){
if(largest < arr[i][j]){
largest = arr[i][j];
}
largestArr[i] = largest;
}
largest = 0;
}
return largestArr;
}
You can populate largest numbers into new Array from two dim array.
您可以将最大的数字从两个暗数组填充到新数组中。