Javascript: Problem adding array into 2D array

Question

I am trying to shuffle an array and add the shuffled-set every time into 2D array using a loop. the problem is that the resulted 2D array contains the same shuffled-set for all rows, while the output of each shuffled-set is different.

So; what do I miss here?

var b=[1,2,3,4,5,6,7,8,9];
var matrix=[];

for(var i=0; i <5; i ++){
    shuffle(b);
    matrix[i]=b;
    //matrix.push(b);
    console.log(b);
}
console.log(matrix);


function shuffle(array) {
    var ctr = array.length, temp, index;
    while (ctr > 0) {
        index = Math.floor(Math.random() * ctr);
        ctr--;
        temp = array[ctr];
        array[ctr] = array[index];
        array[index] = temp;
    }
    return array;
}

the output:

Array(9) [ 7, 3, 4, 9, 2, 1, 6, 8, 5 ]
Array(9) [ 5, 6, 3, 9, 8, 4, 1, 7, 2 ]
Array(9) [ 9, 5, 2, 4, 3, 6, 8, 7, 1 ]
Array(9) [ 3, 5, 2, 1, 4, 9, 6, 8, 7 ]
Array(9) [ 5, 9, 4, 6, 7, 3, 2, 8, 1 ]

(5) […]
0: Array(9) [ 5, 9, 4, … ]
1: Array(9) [ 5, 9, 4, … ]
2: Array(9) [ 5, 9, 4, … ]
3: Array(9) [ 5, 9, 4, … ]
4: Array(9) [ 5, 9, 4, … ]
length: 5

trincot · Accepted Answer · 2019-10-10 19:31:23Z

1

You only have one 1D array object, i.e. b. And you keep shuffling and adding that same object reference to matrix:

 matrix[i]=b;

So even after you have added such a row, you will keep mutating that added row when you call shuffle again.

You should take a copy of the array so you have different arrays that get added to matrix:

 matrix[i]=b.slice();

There are several ways to get a copy: b.slice() can also be Array.from(b), or [...b].

Alternatively, you could put the initalisation of b as new array, inside the loop:

var matrix = [];
for (var i = 0; i < 5; i++){
    var b = [1,2,3,4,5,6,7,8,9];
    shuffle(b);
    matrix[i] = b;
}

As you have designed shuffle to return the array also, you can reduce the above code to:

var matrix = [];
for(var i = 0; i < 5; i++){
    matrix[i] = shuffle([1,2,3,4,5,6,7,8,9]);
}

edited Oct 10, 2019 at 19:31

answered Oct 10, 2019 at 19:25

trincot

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Comments

Bryan Dellinger · Accepted Answer · 2019-10-10 19:31:15Z

1

you could take a shallow clone of your shuffle.

 for(var i=0; i <5; i ++){
    const clone = {...shuffle(b)};
    matrix[i]=clone;
    //matrix.push(b);
    console.log(clone);
}
console.log(matrix);

answered Oct 10, 2019 at 19:31

Bryan Dellinger

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1 Comment

trincot Over a year ago

Note that {...shuffle(b)} is not an array instance.

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