push() a two-dimensional array

Question

I'm trying to push to a two-dimensional array without it messing up, currently My array is:

var myArray = [
[1,1,1,1,1],
[1,1,1,1,1],
[1,1,1,1,1]
]

And my code I'm trying is:

var r = 3; //start from rows 3
var c = 5; //start from col 5

var rows = 8;
var cols = 7;

for (var i = r; i < rows; i++)
{
    for (var j = c; j < cols; j++)
    {
        myArray[i][j].push(0);
    }
}

That should result in the following:

var myArray = [
[1,1,1,1,1,0,0],
[1,1,1,1,1,0,0],
[1,1,1,1,1,0,0],
[0,0,0,0,0,0,0],
[0,0,0,0,0,0,0],
[0,0,0,0,0,0,0],
]

But it doesn't and not sure whether this is the correct way to do it or not.

So the question is how would I accomplish this?

score 47 · Accepted Answer · edited May 06 '18 at 17:49

You have some errors in your code:

Use myArray[i].push( 0 ); to add a new column. Your code (myArray[i][j].push(0);) would work in a 3-dimensional array as it tries to add another element to an array at position [i][j].
You only expand (col-d)-many columns in all rows, even in those, which haven't been initialized yet and thus have no entries so far.

One correct, although kind of verbose version, would be the following:

var r = 3; //start from rows 3

var rows = 8;
var cols = 7;

// expand to have the correct amount or rows
for( var i=r; i<rows; i++ ) {
  myArray.push( [] );
}

// expand all rows to have the correct amount of cols
for (var i = 0; i < rows; i++)
{
    for (var j =  myArray[i].length; j < cols; j++)
    {
        myArray[i].push(0);
    }
}

score 5 · Answer 2 · answered Jul 05 '12 at 14:17

5

In your case you can do that without using push at all:

var myArray = [
    [1,1,1,1,1],
    [1,1,1,1,1],
    [1,1,1,1,1]
]

var newRows = 8;
var newCols = 7;

var item;

for (var i = 0; i < newRows; i++) {
    item = myArray[i] || (myArray[i] = []);

    for (var k = item.length; k < newCols; k++)
        item[k] = 0;    
}

answered Jul 05 '12 at 14:17

ZER0

24,846
5
51
54

1

Thanks very useful, is it possible to remove the added rows? Maybe I can use the same method above but instead splice so item[k].splice(k, 1) to remove the added rows/cols – Jo3la Jul 05 '12 at 15:40
1

Because the added row are appended, you can use `length` to remove them: `myArray.length = 3`; makes the last `5` to be discarded (assumes you have `8` rows). – ZER0 Jul 05 '12 at 16:23
1

Ah ha didn't think it was possible with the .length, got it working now... thanks. – Jo3la Jul 05 '12 at 19:01
1

`length` is a good method to clear an array without create a new instance, so if other objects points to the same array you're safe. – ZER0 Jul 06 '12 at 16:12

score 4 · Answer 3 · answered Jul 05 '12 at 14:09

You have to loop through all rows, and add the missing rows and columns. For the already existing rows, you loop from c to cols, for the new rows, first push an empty array to outer array, then loop from 0 to cols:

var r = 3; //start from rows 3
var c = 5; //start from col 5

var rows = 8;
var cols = 7;

for (var i = 0; i < rows; i++) {
  var start;
  if (i < r) {
    start =  c;
  } else {
    start = 0;
    myArray.push([]);
  }
  for (var j = start; j < cols; j++) {
        myArray[i].push(0);
    }
}

score 4 · Answer 4 · answered Jul 05 '12 at 14:16

Iterating over two dimensions means you'll need to check over two dimensions.

assuming you're starting with:

var myArray = [
    [1,1,1,1,1],
    [1,1,1,1,1],
    [1,1,1,1,1]
]; //don't forget your semi-colons

You want to expand this two-dimensional array to become:

var myArray = [
    [1,1,1,1,1,0,0],
    [1,1,1,1,1,0,0],
    [1,1,1,1,1,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
];

Which means you need to understand what the difference is.

Start with the outer array:

var myArray = [
    [...],
    [...],
    [...]
];

If you want to make this array longer, you need to check that it's the correct length, and add more inner arrays to make up the difference:

var i,
    rows,
    myArray;
rows = 8;
myArray = [...]; //see first example above
for (i = 0; i < rows; i += 1) {
    //check if the index exists in the outer array
    if (!(i in myArray)) {
        //if it doesn't exist, we need another array to fill
        myArray.push([]);
    }
}

The next step requires iterating over every column in every array, we'll build on the original code:

var i,
    j,
    row,
    rows,
    cols,
    myArray;
rows = 8;
cols = 7; //adding columns in this time
myArray = [...]; //see first example above
for (i = 0; i < rows; i += 1) {
    //check if the index exists in the outer array (row)
    if (!(i in myArray)) {
        //if it doesn't exist, we need another array to fill
        myArray[i] = [];
    }
    row = myArray[i];
    for (j = 0; j < cols; j += 1) {
        //check if the index exists in the inner array (column)
        if (!(i in row)) {
            //if it doesn't exist, we need to fill it with `0`
            row[j] = 0;
        }
    }
}

score 2 · Answer 5 · edited Jul 05 '12 at 15:13

2

var r = 3; //start from rows 3
var c = 5; //start from col 5

var rows = 8;

var cols = 7;


for (var i = 0; i < rows; i++)

{

 for (var j = 0; j < cols; j++)

 {
    if(j <= c && i <= r) {
      myArray[i][j] = 1;
    } else {
      myArray[i][j] = 0;
    }
}

}

edited Jul 05 '12 at 15:13

Sirko

72,589
19
149
183

answered Jul 05 '12 at 14:07

Daniel

601
1
4
13

2

(1) you have a syntax error: missing `{` before `else`. (2) This code wont work, if there are rows to be added: `ReferenceError: myArray is not defined`. – Sirko Jul 05 '12 at 14:15
1

yea add the {. There is a reference error, because I guess this a piece of inline code and the arrays already exist as they should, regarding the snippet above. – Daniel Jul 05 '12 at 14:38

score 0 · Answer 6 · answered Jul 05 '12 at 14:14

you are calling the push() on an array element (int), where push() should be called on the array, also handling/filling the array this way makes no sense you can do it like this

for (var i = 0; i < rows - 1; i++)
{
  for (var j = c; j < cols; j++)
  {
    myArray[i].push(0);
  }
}


for (var i = r; i < rows - 1; i++)
{

  for (var j = 0; j < cols; j++)
  {
      col.push(0);
  }
}

you can also combine the two loops using an if condition, if row < r, else if row >= r

score 0 · Answer 7 · answered Jan 17 '18 at 20:00

Create am array and put inside the first, in this case i get data from JSON response

$.getJSON('/Tool/GetAllActiviesStatus/',
   var dataFC = new Array();
   function (data) {
      for (var i = 0; i < data.Result.length; i++) {
          var serie = new Array(data.Result[i].FUNCAO, data.Result[i].QT, true, true);
          dataFC.push(serie);
       });

RWasson · Answer 8 · 2020-03-08T15:28:49.340

The solution below uses a double loop to add data to the bottom of a 2x2 array in the Case 3. The inner loop pushes selected elements' values into a new row array. The outerloop then pushes the new row array to the bottom of an existing array (see Newbie: Add values to two-dimensional array with for loops, Google Apps Script).

In this example, I created a function that extracts a section from an existing array. The extracted section can be a row (full or partial), a column (full or partial), or a 2x2 section of the existing array. A new blank array (newArr) is filled by pushing the relevant section from the existing array (arr) into the new array.

 function arraySection(arr, r1, c1, rLength, cLength) {      
   rowMax = arr.length;
    if(isNaN(rowMax)){rowMax = 1};
   colMax = arr[0].length;
   if(isNaN(colMax)){colMax = 1};
   var r2 = r1 + rLength - 1;
   var c2 = c1 + cLength - 1;

   if ((r1< 0 || r1 > r2 || r1 > rowMax || (r1 | 0) != r1) || (r2 < 0 || 
     r2 > rowMax || (r2 | 0) != r2)|| (c1< 0 || c1 > c2 || c1 > colMax || 
   (c1 | 0) != c1) ||(c2 < 0 || c2 > colMax || (c2 | 0) != c2)){
        throw new Error(
        'arraySection: invalid input')       
        return;
     };
   var newArr = [];

// Case 1: extracted section is a column array, 
//          all elements are in the same column
 if (c1 == c2){
   for (var i = r1; i <= r2; i++){ 
         // Logger.log("arr[i][c1] for i = " + i);
         // Logger.log(arr[i][c1]);
     newArr.push([arr[i][c1]]); 
     };
    };  

// Case 2: extracted section is a row array, 
//          all elements are in the same row
  if (r1 == r2 && c1 != c2){
    for (var j = c1; j <= c2; j++){ 
      newArr.push(arr[r1][j]); 
      };
   };  

// Case 3: extracted section is a 2x2 section 
    if (r1 != r2 && c1 != c2){     
     for (var i = r1; i <= r2; i++) {
          rowi = [];
       for (var j = c1; j <= c2; j++) {
          rowi.push(arr[i][j]);
        }
       newArr.push(rowi)
      };
    };
   return(newArr);
  };

score 0 · Answer 9 · answered Jun 07 '21 at 11:14

You can also try like this.

var r = 3; //start from rows 3
var c = 5; //start from col 5

var rows = 8;
var cols = 7;

for (var i = r; i < rows; i++)
{
    for (var j = c; j < cols; j++)
    {
        myArray.push([var[i],var[j])
    }
}

this will create a 2d array for you.

score 0 · Answer 10 · answered Jun 07 '22 at 17:40

0

<script>
    let test = new Array;
    for(let i = 1; i < 10; i++){
        test[i] = new Array;
        
        test[i]['type']     = 'test-type'+i;
        test[i]['content']  = 'test-content'+i;
    }
    console.log(test);
</script>

answered Jun 07 '22 at 17:40

Mohammad Koochaki

35
1
6

1

Your answer could be improved by adding more information on what the code does and how it helps the OP. – Tyler2P Jun 07 '22 at 19:28

push() a two-dimensional array

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