turn the numbers in the console / print

Question

I get data in console in this form

0:(8) [70, 61, 74, 79, 76, 63, 51, 51]
1:(4) [64, 84, 84, 66]
2:(2) [79, 69]
3:(3) [77, 100, 98]

I would like to print my data in console like this:

0:(4) [70, 64, 79,  77]
1:(4) [61, 84, 69, 100]
2:(4) [74, 84,  0,   0]
3:(4) [79, 66, 98,   0]
4:(4) [76,  0,  0,   0]
5:(4) [63,  0,  0,   0]
6:(4) [51,  0,  0,   0]
7:(4) [51,  0,  0,   0]

With this two function I get max of number in row and I fill empty column with 0, but I would like to show data like in example that I write.

function fillWithZeros(arr, length) {
for (var i = 0; i < length; i++) {
    if (arr[i] == null) {
        arr[i] = 0;
    }
}

return arr;
};
function findMaxEvents(item){
var maxNumEvents = 0;
for (var i = 0; i < item.length; i++){
    if (item[i] > maxNumEvents){
        maxNumEvents = item[i];
    }
}

return maxNumEvents;
}

Question, Why is `98` on row `3` and not on `2`? – Rajesh Oct 13 '17 at 09:44 — Rajesh, Oct 13 '17 at 09:44

JEEVAN GEORGE ANTONY · Accepted Answer · 2017-10-13T11:25:05.710

1

I hope you are actually looking for Transposing the existing array. Thanks to Transposing a 2D-array in JavaScript..

The obtained answer is a bit different from yours. Hope this helps to try and find the right way to reach your goal.

var array = [[70, 61, 74, 79, 76, 63, 51, 51], 
[64, 84, 84, 66], 
[79, 69], 
[77, 100, 98]];

var arrayWithMaxlen = array.map(function(a){return a.length;}).indexOf(Math.max.apply(Math, array.map(function(a){return a.length;})));
console.log("arrayWithMaxlen : " +arrayWithMaxlen); //gets index of array with max length

var newArray = array[arrayWithMaxlen].map(function(col, i) { 
  return array.map(function(row) { 
    if(row[i]!=undefined)return row[i];
    else return 0;
  })
});

console.log(newArray);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Oct 13 '17 at 11:25

answered Oct 13 '17 at 09:51

JEEVAN GEORGE ANTONY

168
5
21

1

Please be aware with this solution that if the first array isn't the longest, then the result wont contain alle the original values. Thus this solution is very dependent on the order of the arrays for it to return the correct result. – Nikolaj Dam Larsen Oct 13 '17 at 11:12
1

Thanks @Nikolaj Dam Larsen. Didn't notice that. Hope the Updated code resolves that issue. – JEEVAN GEORGE ANTONY Oct 13 '17 at 11:20
1

Yes, perfect. :-) It can be done a bit cleaner, but that's merely details. – Nikolaj Dam Larsen Oct 13 '17 at 11:21

score 1 · Answer 2 · answered Oct 13 '17 at 09:51

The operation you're describing is basically a simple matrix transpose. There are many ways to do that. Here's one alternative:

var array = [
  [70, 61, 74, 79, 76, 63, 51, 51],
  [64, 84, 84, 66],
  [79, 69],
  [77, 100, 98]
];

function transpose(arr){
  var largest = arr.reduce(function(max, i){ return max.length > i.length ? max : i; }, []);
  return largest.map(function(col, i){
    return arr.map(function(row){
      return row[i] || 0;
    });
  });
}

console.log(transpose(array));

Nina Scholz · Answer 3 · 2017-10-13T10:20:44.810

0

Just transpose it.

BTW, the accepted answer does rely on the first inner array length. If it is shorter then the others all greater indices are gone.

var array = [[70, 61, 74, 79, 76, 63, 51, 51], [64, 84, 84, 66], [79, 69], [77, 100, 98]],
    result = array.reduce(function (r, a, i) {
        a.forEach(function (b, j) {
            r[j] = r[j] || Array.apply(null, { length: array.length }).map(function () { return 0; });
            r[j][i] = b;
        });
        return r;
    }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Oct 13 '17 at 10:20

answered Oct 13 '17 at 09:23

Nina Scholz

376,160
25
347
392

1

Though I agree with the implementation, there is either a mistake in expected output or some weird reason, but `3:(4) [79, 66, 98, 0]` makes your output incorrect and might attract unwanted comments. – Rajesh Oct 13 '17 at 09:46

turn the numbers in the console / print

3 Answers3