Getting the min and max value in JavaScript, but f

问题:

I know this gets asked again and again but hear me out - this question is slightly different.

I can get a max or min from a 1D array like this:

var w_max = Math.max.apply(Math, myArray); 
var w_min = Math.min.apply(Math, myArray); 

But my array is of the type:

[[[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]]]
//max = 219.9, min = 5.12

and I need to get min and max of the second value, with the first value being an index of sorts. I've tried

myArray[][], myArray[0][0], myArray[[1]] 

and more. Console logging shows I'm getting NaN or -Infinity!!

回答1:

You can map the array to the second values of the elements:

var arr = [[[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]]];
var values = arr[0].map(function(elt) { return elt[1]; });
var max = Math.max.apply(null, values);
var min = Math.min.apply(null, values);

回答2:

I really admired the sleek readability of @TedHopp 's solution -- so much so that I decided to adapt it into a function. The practicality of the function is still yet to be seen, but I liked the idea of being able to identify the min / max of a specified index within a 2D array.

function getMinMaxOf2DIndex (arr, idx) {
    return {
        min: Math.min.apply(null, arr.map(function (e) { return e[idx]})),
        max: Math.max.apply(null, arr.map(function (e) { return e[idx]}))
    }
} 

getMinMaxOf2DIndex() takes two parameters, arr -- an array, and idx the index of the values to compare.

Usage Examples:

// using the OP's array as an example
var array = [[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]];

getMinMaxOf2DIndex(array, 0); // {min: 1, max: 6}
getMinMaxOf2DIndex(array, 1); // {min: 5.12, max: 219.9}

// and so on...
var array = [[1, 9, 6 , 3], [2, 4, 7, 2], [6, 5, 9, 4]];
getMinMaxOf2DIndex(array, 2); // {min: 1, max: 6}
getMinMaxOf2DIndex(array, 3); // {min: 6, max: 9}

回答3:

In ES6 you can use spread operator:

var max = Math.max(...arrayOfValues);

Working example:

var arr = [[[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]]];

var v = arr[0].map(e => e[1]);

var max = Math.max(...v);
var min = Math.min(...v);

console.log('max: ' + max);
console.log('min: ' + min);

回答4:

If you are saying it is 2D array then it should have one less []

var myArray = [[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]];
var w_max=+Infinity;
var w_min=-Infinity;

for(var i=0;i< myArray.length;i++){
    elementArray = myArray[i];
    if(w_max < Math.max.apply(Math, elementArray)){
        w_max = Math.max.apply(Math, elementArray);
    } 
    if (w_min > Math.min.apply(Math, elementArray)){
        w_min = Math.min.apply(Math, elementArray);
        }
}

回答5:

That array you have in the post is interesting, but this works for it.

var array = [[[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]]];
function getMax(a){
    var max = a[0][0][1];
    for(var i = 0; i< a[0].length; i++){
        if(a[0][i][1] > max)
            max = a[0][i][1];
    }
    return max;
}

回答6:

var list = [
  [1, 112.0],
  [2, 5.12],
  [3, 113.1],
  [4, 33.6],
  [5, 85.9],
  [6, 219.9]
];

function getMinMax(list) {

  var min = +Infinity;
  var max = -Infinity;

  for (var i = 0; i < list.length; i++) {
    var touple = list[i];

    var val = touple[1];
    console.log(touple);
    console.log(val);

    min = Math.min(min, val);
    max = Math.max(max, val);
  }

  return {
    min: min,
    max: max
  };

}

console.log(getMinMax(list));

回答7:

You can loop through the first array (the only one here) with a forEach:

var A= [
    [[1, 112.0], [2, 5.12], [3, 113.1], [4, 33.6], [5, 85.9], [6, 219.9]]
];
var mm= [Infinity, -Infinity];

A[0].forEach(function(itm){
    var n= itm[1];
    if(n<mm[0]) mm[0]= n;
    if(n>mm[1]) mm[1]= n;   
});

mm

/* returned value: (Array) [5.12, 219.9] */

回答8:

for big 2D arrays use this avoid function.prototype.apply as in above almost all answers i can see it. because it can handle only limited amount of array length .

function MaxMin2dray(arr, idx){
    var max = Number.MIN_VALUE;
    var min = Number.MAX_VALUE;
     arr.forEach(function(e) {
     if (max < e[idx]) {
      max = e[idx];
    }
    if (min > e[idx]) {
      min = e[idx];
    }
  });
  return {max: max, min: min};
}

回答9:

The first answer is not correct!!!

let arr = [[1, 112.0],[2,5.12],[3,113.1],[4,33.6],[5,85.9],[6,219.9]];
let values = Array.concat.apply([],arr);
let max = Math.max(...values);
let min = Math.min(...values);
console.log(arr)
console.log(min)//1
console.log(max)//219.9
console.log(values)//[ 1, 112, 2, 5.12, 3, 113.1, 4, 33.6, 5, 85.9, 6, 219.9 ]