Question

以下代码生成随机点（x，y），然后针对每个点将画布（一个正方形）分成四个。对于迭代中的下一个点，它将搜索该点所在的正方形，并将其分成四个较小的正方形-达到一定的正方形大小。

问题在于，在Chrome中运行速度非常快，而在Ps中运行速度非常慢（在Chrome中1k点需要2秒，在Ps中需要30分钟！在1k点中Ps需要大约10秒。

是否有更好的重写方法？顺便说一句，PS不支持ES5

var squares = [];
var canvaswidth = app.activeDocument.width.as("px");
var canvasheight = app.activeDocument.height.as("px");
squares.push([{
    x: 0,
    y: 0
}, {
    x: canvaswidth,
    y: 0
}, {
    x: canvaswidth,
    y: canvasheight
}, {
    x: 0,
    y: canvasheight
}])
vertices = [];
for (i = 0; i < 8000; i++) {
    vertices.push({
        x: Math.floor(Math.random() * canvaswidth),
        y: Math.floor(Math.random() * canvasheight)
    })
}
var t0 = new Date().getTime();
var minsquaresize = 24;
for (v = 0; v < vertices.length; v++) {
    if (v > 0 && Math.abs(vertices[v].x - vertices[v - 1].x) > minsquaresize && Math.abs(vertices[v].y - vertices[v - 1].y) > minsquaresize) {
        r = 2;

           for (s = 0; s < squares.length; s++) {

            var squares_s = squares[s];
            if (squares_s != undefined && vertices[v].x >= squares_s[0].x && vertices[v].x <= squares_s[2].x && vertices[v].y >= squares_s[0].y && vertices[v].y <= squares_s[2].y && squares_s[1].x - squares_s[0].x > minsquaresize && squares_s[3].y - squares_s[0].y > minsquaresize) {
                var s1p1 = {
                    x: Math.round(squares_s[0].x),
                    y: Math.round(squares_s[0].y)
                };
                var s1p2 = {
                    x: Math.round((squares_s[0].x + squares_s[1].x) / 2),
                    y: Math.round((squares_s[0].y + squares_s[1].y) / 2)
                };
                var s1p3 = {
                    x: Math.round(((squares_s[1].x - squares_s[0].x) / r) + squares_s[0].x),
                    y: Math.round(((squares_s[3].y - squares_s[0].y) / r) + squares_s[0].y)
                }
                var s1p4 = {
                    x: (squares_s[0].x + squares_s[3].x) / 2,
                    y: Math.round((squares_s[0].y + squares_s[3].y) / 2)
                }
                var s2p2 = {
                    x: squares_s[1].x,
                    y: squares_s[1].y
                }
                var s2p3 = {
                    x: Math.round((squares_s[1].x + squares_s[2].x) / 2),
                    y: Math.round((squares_s[1].y + squares_s[2].y) / 2)
                }
                var s3p3 = {
                    x: squares_s[2].x,
                    y: squares_s[2].y
                }
                var s3p4 = {
                    x: Math.round((squares_s[2].x + squares_s[3].x) / 2),
                    y: Math.round(Math.round((squares_s[2].y + squares_s[3].y) / 2))
                }
                var s4p4 = {
                    x: squares_s[3].x,
                    y: squares_s[3].y
                }
                //alert(s4p4.y)
                delete squares[s];
                squares.push([s1p1, s1p2, s1p3, s1p4])
                squares.push([s1p2, s2p2, s2p3, s1p3])
                squares.push([s1p3, s2p3, s3p3, s3p4])
                squares.push([s1p4, s1p3, s3p4, s4p4])
                break;
            }
        }
    }
}
var t1 = new Date().getTime() - t0;
alert("time: "+t1)

Answer 1

通过使方块反向循环，可以显着提高性能。所以通常是：

for(vertices length, v++){
 for(squares length, s++){
   if vertex is within square then delete square from square array, split square into 4 equal squares and add them to array
 }
}

顶点是从路径中收集的，因此顶点4可能会接近顶点3，因此可能在从顶点3创建的最后一个正方形的区域中-在正方形数组的末尾。所以：

for(var s = squares.length; s--;){...}

这工作得快得多（也许10倍）。奇怪的是，对于随机放置的顶点来说，速度也更快。

在Photoshop中，Js比在Chrome中慢。我可以更快地进行两个for循环吗？

1 个答案: