Javascript get object sequence out of array

Question

I have different kind of javascript objects, they all have a property 'row'.

var row1 = {
    row: 1
};

var row2 = {
    row: 2
};

var row3 = {
    row: 3
};

var row4 = {
    row: 4
};

...

I have an array defined as follow:

var objArray = [];

In this array it's possible to have multiple 'rows'. The sequence is always te same starting from the lower row to a higher row.

Now I want to get the objects that are linked next to each other (like 4 in a row). In my case it's also possible to have 3 in a row, 5 in a row and so on..

Example:

objArray.push(row0);
objArray.push(row1);
objArray.push(row2);
objArray.push(row3);
objArray.push(row5);
objArray.push(row6);
objArray.push(row7);
objArray.push(row9);
objArray.push(row10);
objArray.push(row12);

In this case I need 2 lists, 1 containing row0 to 3 and one containing 5 to 7.

I've tried a bit in this JSFiddle: Here we can see the console output If you need more clarification, please ask.

Thanks in advance!

In your example, surely you should expect 4 lists out: 0-3, 5-7, 9-10 and 12? no? — Jamiec
– Jamiec, Commented Nov 28, 2013 at 11:45
Let me get this straight - do you want to do array slice of n elements, starting from the object that has the row property equal to certain value? Or do you want to do array chunk of n elements (then why the row property should matter at all)? — eithed
– eithed, Commented Nov 28, 2013 at 11:46
@Jamiec: I don't need to have objects 9, 10 and 12 because the sequence is incorrect. — Dieterg
– Dieterg, Commented Nov 28, 2013 at 11:49
@Eithedog: there are more properties than just 'row'. Each object has a different position in the row. I need to find the objects that are positioned next to each other. — Dieterg
– Dieterg, Commented Nov 28, 2013 at 11:50
They're all "next to each other" row1 has property row of 1, row2 has property row 2 etc.. What is your actual problem that you're trying to solve with this, as this as it stands make no sense? — Reinstate Monica Cellio
– Reinstate Monica Cellio, Commented Nov 28, 2013 at 11:51

woutervs · Accepted Answer · 2013-11-28 11:57:08Z

1

I made a fiddle that does just this for you:

    for(var i = 0; i < objArray.length; i++) {
    if(currentnum !== -1)
    {
         var result = objArray[i].row - currentnum;
        currentnum = objArray[i].row;
        if(result === 1)
        {
         currentarray.push(objArray[i]);   
        } else {
            arrayofarrays.push(currentarray);
            currentarray = [];
            currentarray.push(objArray[i]);
        }
    }  else {
        currentnum = objArray[i].row;
        currentarray.push(objArray[i]);
    }
}
arrayofarrays.push(currentarray);

http://jsfiddle.net/B76a8/6/

answered Nov 28, 2013 at 11:57

woutervs

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Gabriele Petrioli · Accepted Answer · 2013-11-28 11:58:41Z

Since you have already figured how to keep the counter and reset it for each new group you can do

var counter = 1,
    lastIndex = 0;

//see chrome dev tools, I need to return objects beginning at the 3rd place untill the 7th place in the array
//how do I get these objects? 
for (var i = 0; i < objArray.length; i++) {

    if ((i < objArray.length - 1 && objArray[i].row + 1 == objArray[i + 1].row) ||  
       (i == objArray.length - 1 && objArray[i - 1].row == objArray[i].row - 1)) {
        counter++;
    } else {
        // here we output the grouped items
        console.log(objArray.slice(lastIndex, counter+lastIndex));
        lastIndex = counter+lastIndex;
        counter = 1;
    }
}

Demo at http://jsfiddle.net/B76a8/7/

output

[Object { row=0}, Object { row=1}]
[Object { row=3}, Object { row=4}, Object { row=5}, Object { row=6}, Object { row=7}, Object { row=8}]
[Object { row=10}]

eithed · Accepted Answer · 2013-11-28 12:07:04Z

0

First, let's sort the array:

objArray.sort(function(a,b){ return a.row - b.row });

Then, for the given n, this should return you next and previous elements:

function getElement(array, n)
{
 var ret = [];
 for (var i=0; i<array.length; i++)
   if (array[i].row == n)
   {
    if (i > 0) ret.push(array[i-1]);
    if (i < array.length-1) ret.push(array[i+1]);
   }

 return ret;
}

As getting all the other options with the same color is a different thing let's do it through:

function getByColor(array, color)
{
 var ret = [];
 for (var i=0; i<array.length; i++)
  if (array[i].color == color)
   ret.push(array[i]);

 return ret;
}

Then you can merge both of the arrays by using concat

answered Nov 28, 2013 at 12:07

eithed

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