InDesign JS arrays quesion...

[deckarduk]

Hi everyone,

Just a quick question about arrays...

Is there a quick way to pull out all the even numbered elements separately and the same for the odd numbered ones?

Or, is it a question of looping through the array with an increment of 2 ?

Thanks in advance.

[Manan Joshi]

I suppose whatever you do would involve a loop mostly, unless you join the array values into a string and do some Grep magic. Is running a loop troublesome in your case?

-Manan

[deckarduk]

Hi Manan,

Thanks for the reply. Interesting idea regarding using GREP.

I haven't tried the loop method yet, just thought I'd check in advance that I'm not missing a nice arrays method.

The array I'm dealing with will only have a few hundred items in it, nothing too big.

Thanks again.

[Laubender]

Hi @deckarduk ,

let's see:

"Looping the array with increments of 2 would be one solution."

Exactly this.

Examples:

var a =
[
	0 ,
	1 ,
	2 ,
	3 ,
	4,
	5

];

var resultArrayStartWithZero = [];

for( var n=0; n<a.length; n=n+2 )
{
	resultArrayStartWithZero[resultArrayStartWithZero.length++] = n ;
};

// Result: 0,2,4

resultArrayStartWithOne = [];

for( var n=1; n<a.length; n=n+2 )
{
	resultArrayStartWithOne[resultArrayStartWithOne.length++] = n ;
};

// Result: 1,3,5

Regards,
Uwe Laubender
( Adobe Community Expert )

[deckarduk]

Hi Uwe,

Thanks for the post.

You've echoed the method I was going to use. A loop with an increment of 2 and changing the start point i.e. n=0 or n=1.

Thank you again for the help 🙂

[FRIdNGE]

Hi Uwe,

Weird approach! …

var a =
[
	"a",
	"b",
	"c",
	"d",
	"e",
	"f",
    "g"
];

var resultArrayStartWithZero = [];

for( var n=0; n<a.length; n=n+2 ) resultArrayStartWithZero.push(a[n]);

alert( "1/ " + resultArrayStartWithZero )

var resultArrayStartWithOne = [];

for( var n=1; n<a.length; n=n+2 ) resultArrayStartWithOne.push(a[n]);

alert( "2/ " + resultArrayStartWithOne )

(^/)  The Jedi

[FRIdNGE]

Variation:

var a =
[
	"a",
	"b",
	"c",
	"d",
	"e",
	"f",
    "g"
];

for ( var i = a.length-1; i >= 0; i=i-2 ) a.splice(i,1);

alert( "1/ " + a )

var a =
[
	"a",
	"b",
	"c",
	"d",
	"e",
	"f",
    "g"
];

for ( var i = a.length-2; i >= 0; i=i-2 ) a.splice(i,1);

alert( "2/ " + a )

(^/)

[deckarduk]

Nice!

Thanks for the posts and alternative solutions 🙂

[FRIdNGE]

Other variant!  =D

var X =
[
	"a",
	"b",
	"c",
	"d",
	"e",
	"f",
    "g"
];
var Y = [],  Z = [];
for ( var i = 0; i < X.length; i=i+2 ) Y.push(X.slice(i,i+1));
for ( var i = 1; i < X.length; i=i+2 ) Z.push(X.slice(i,i+1));
alert( "X = " + X + "\rY = " + Y + "\rZ = " + Z )

(^/)

[rob day]

Thanks for the posts and alternative solutions

There’s also the modulus operator:

var a = ["a", "b", 3, 4, "dog", "cat", 10, 20];
var even = [];
var odd = [];
for (var i = 0; i < a.length; i++){
    if (i % 2 == 1) {
        even.push(a[i])
    } else {
        odd.push(a[i])
    }
};   

$.writeln(even)
//returns b,4,cat,20
$.writeln(odd)
//returns a,3,dog,10

[Loic.Aigon]

Why on Earth is Uwe's solution weird? Would be curious.