I'm currently facing a problem where I have to shift data in a multidimensional JS array in different directions, but I think the first solution I whipped up is not as efficient as it could be (some fancy math I don't know of maybe?).
Let me explain the problem a bit better with some examples of my data.
This is the array in question:
[1][0][1][0][1][0]
[0][1][1][0][0][0]
[0][0][1][0][1][1]
[0][0][0][0][1][1]
[0][0][1][1][0][0]
[1][0][0][1][1][1]
If I shifted it to the left, it should look like this:
[1][1][1][0][0][0]
[1][1][0][0][0][0]
[1][1][1][0][0][0]
[1][1][0][0][0][0]
[1][1][0][0][0][0]
[1][1][1][1][0][0]
Or down
[0][0][0][0][0][0]
[0][0][0][0][0][0]
[0][0][1][0][1][0]
[0][0][1][0][1][1]
[1][0][1][1][1][1]
[1][1][1][1][1][1]
To put it simple - shift 1's as far as possible.
The code I wrote is fairly simple
for (w = 0; w < arraySize; w++) {
for (h = 0; h < arraySize; h++) {
tiles = $('.arrayContainer .row').eq(w).children();
if (tiles.eq(h).html() != "" && tiles.eq(h - 1).html() == "" && (h-1) >= 0) {
tiles.eq(h - 1).html(tiles.eq(h).html());
tiles.eq(h).html("");
}
}
}
This will shift values left-to-right by one step. arraySize
is the variable that determines the size of the array, tiles
gets each row of the array output.
If I wasn't here, posting and wouldn't think this could be done better I'd create four of these structures and just live with it, but I'm pretty sure this can be done better.
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\$\begingroup\$ Are you always dealing with just 0s and 1s? \$\endgroup\$samanime– samanime2014年03月25日 22:55:01 +00:00Commented Mar 25, 2014 at 22:55
1 Answer 1
From a once over:
You should consider MVC, your data should live in an object you created, you should update your model, and then from your model update the UI
If you dont want to do this, then you should at least find all cells and assign each cell to a 2d JS array
var tiles = []; for (w = 0; w < arraySize; w++) { tiles[w] = []; for (h = 0; h < arraySize; h++) { tiles[w][h] = $('.arrayContainer .row').eq(w).children().eq(h); } }
Otherwise you are executing
$('.arrayContainer .row')
far too many times, and that operation takes a lot of time, then you can do your logic like this:for (w = 0; w < arraySize; w++) { for (h = 0; h < arraySize; h++) { tile = tiles[w][h] if (tile.html() != '' && tiles[w][h-1].html() == '' && (h-1) >= 0) { tiles[w][h-1].html(tile.html()); tile.html(''); } } }
You could consider assigning
tiles[w][h-1]
to a variable as well to make it more readableConsider
c
andr
as variables (column
androw
if you are feeling verbose ). Those are better names thanh
eight andw
idth.I hope you declared
tiles
withvar
somewhere else, otherwise you have global namespace pollution.As for not having to repeat the same 2 loops for every direction, you could think about passing a columnDelta and RowDelta ( for your 2 loops, columnDelta would be -1, rowDelta would be 0 ) I am not sure it is worth it, since you only have 4 cardinal directions, up to you.
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1\$\begingroup\$ Thanks for all the great suggestions, they truly make sense. What about the for loops, though. Is there any way to avoid creating this structure for every direction I want the data to be shifted? \$\endgroup\$user2978301– user29783012014年03月25日 20:39:23 +00:00Commented Mar 25, 2014 at 20:39
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\$\begingroup\$ Updated question. \$\endgroup\$konijn– konijn2014年03月25日 21:01:52 +00:00Commented Mar 25, 2014 at 21:01
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\$\begingroup\$ Thanks for the great answer! I'll stick to these suggestions. \$\endgroup\$user2978301– user29783012014年03月27日 09:26:37 +00:00Commented Mar 27, 2014 at 9:26