How to find the max/min of a nested array in javascript?

Question

I want to find the maximum of a nested array, something like this:

a = [[1,2],[20,3]]
d3.max(d3.max(a)) // 20

but my array contains a text field that I want to discard:

a = [["yz",1,2],["xy",20,3]]
d3.max(a) // 20

Answer 1

If you have a nested array of numbers ( arrays = [[1, 2], [20, 3]] ), nest d3.max :

var max = d3.max(arrays, function(array) {
  return d3.max(array);
});

Or equivalently, use array.map :

var max = d3.max(arrays.map(function(array) {
  return d3.max(array);
}));

If you want to ignore string values, you can use array.filter to ignore strings:

var max = d3.max(arrays, function(array) {
  return d3.max(array.filter(function(value) {
    return typeof value === "number";
  }));
});

Alternatively, if you know the string is always in the first position, you could use array.slice which is a bit more efficient:

var max = d3.max(arrays, function(array) {
  return d3.max(array.slice(1));
});

Yet another option is to use an accessor function which returns NaN for values that are not numbers. This will cause d3.max to ignore those values. Conveniently, JavaScript's built-in Number function does exactly this, so you can say:

var max = d3.max(arrays, function(array) {
  return d3.max(array, Number);
});

Answer 2

Use this:

function arrmax(arrs) {
    var toplevel = [];

    var f = function(v) {
        return !isNaN(v);
    };

    for (var i = 0, l = arrs.length; i<l; i++) {
        toplevel.push(Math.max.apply(window, arrs[i].filter(f)));
    }
    return Math.max.apply(window, toplevel);
}

or better:

function arrmax(arrs) {
    if (!arrs || !arrs.length) return undefined;
    var max = Math.max.apply(window, arrs[0]), m,
        f = function(v){ return !isNaN(v); };
    for (var i = 1, l = arrs.length; i<l; i++) {
        if ((m = Math.max.apply(window, arrs[i].filter(f)))>max) max=m;
    }
    return max;
}

See MDN for Array.filter method details.

Answer 3

You can flatten an array and apply a function to each member

Array.prototype.flatten= function(fun){
    if(typeof fun!= 'function') fun= '';
    var A= [], L= this.length, itm;
    for(var i= 0; i<L; i++){
        itm= this[i];
        if(itm!= undefined){
            if(!itm.flatten){
                if(fun) itm= fun(itm);
                if(itm) A.push(itm);
            }
            else A= A.concat(itm.flatten(fun));
        }
    }
    return A;
}

var a= [["yz", 1, 2], ["xy", 20, 3]], max=-Infinity;

var max=Math.max.apply(a, a.flatten(Number));

Answer 4

If you now exactly what columns you want to test, you can use:

var columns = ["ColumnA", "ColumnB", "ColumnC"];

var max = selectedMax(columns,dataset);
var min = selectedMin(columns,dataset)

function selectedMax(columns, dataset) {
    var max;
    columns.forEach(function(element, index, array) {
        var tmpmax = d3.max(dataset, function(d) {
            return +d[element];
        });       
        max = (tmpmax > max || max === undefined) ? tmpmax : max;
    });
    return max;
}

function selectedMin(columns, dataset) {
    var min;
    columns.forEach(function(element, index, array) {
        var tmpmin = d3.min(dataset, function(d) {
            return +d[element];
        });
        min = (tmpmin < min || min === undefined) ? tmpmin : min;
    });  
return min;
}

Answer 5

It's a cruel hack, but looking at the source code for d3.max , your best bet might be to define a d3.max1 that discards the first element by copying that code, but replacing i=-1 with i=0 . The code at that link is excerpted here. Note that I'm not a regular d3.js user, but from what I know of the library, you're going to want make sure your version has an f.call case like this function does, so that it can respond to live updates correctly.

d3.max = function(array, f) {
  var i = -1,
      n = array.length,
      a,
      b;
  if (arguments.length === 1) {
    while (++i < n && ((a = array[i]) == null || a != a)) a = undefined;
    while (++i < n) if ((b = array[i]) != null && b > a) a = b;
  } else {
    while (++i < n && ((a = f.call(array, array[i], i)) == null || a != a)) a = undefined;
    while (++i < n) if ((b = f.call(array, array[i], i)) != null && b > a) a = b;
  }
  return a;
};

Then it would just be d3.max(d3.max1(a)) .

Answer 6

d3.array provides d3.merge which flattens an array of arrays.

Coupled with d3.max and javascript's Number as an accessor:

var max = d3.max(d3.merge(arrays), Number);

For example:

 var input = [["yz", 1, 2], ["xy", 20, 3]]; var max = d3.max(d3.merge(input), Number); console.log(max); 

 <script src="https://d3js.org/d3-array.v2.min.js"></script>