我正在尝试使用这个resample_hermite函数,我在各种堆栈溢出答案中找到了这个函数,它似乎来自这个项目。不幸的是,当我尝试过多地向下采样时,它会破坏图像,如这里的jsfiddle所示。这是原始的fiddle,直到我将其向下采样1/2。
以下是代码(也在jsfiddle中):
var img = document.getElementById("mm");
var W = 425; //works if 850 X 638
var H = 319; //works if 850 X 638
var canvas = document.getElementById("cc");
var ctx = canvas.getContext("2d");
canvas.width = W;
canvas.height = H;
var img = new Image();
img.crossOrigin = "Anonymous"; //cors support
img.onload = function(){
ctx.drawImage(img, 0, 0); //draw image
//now we can resize
resample_hermite(canvas, W, H, W/2, H/2);
}
img.src = 'http://i.imgur.com/8VsK7gS.png';
function resample_hermite(canvas, W, H, W2, H2){
var time1 = Date.now();
var img = canvas.getContext("2d").getImageData(0, 0, W, H);
var img2 = canvas.getContext("2d").getImageData(0, 0, W2, H2);
var data = img.data;
var data2 = img2.data;
var ratio_w = W / W2;
var ratio_h = H / H2;
var ratio_w_half = Math.ceil(ratio_w/2);
var ratio_h_half = Math.ceil(ratio_h/2);
for(var j = 0; j < H2; j++){
for(var i = 0; i < W2; i++){
var x2 = (i + j*W2) * 4;
var weight = 0;
var weights = 0;
var weights_alpha = 0;
var gx_r = gx_g = gx_b = gx_a = 0;
var center_y = (j + 0.5) * ratio_h;
for(var yy = Math.floor(j * ratio_h); yy < (j + 1) * ratio_h; yy++){
var dy = Math.abs(center_y - (yy + 0.5)) / ratio_h_half;
var center_x = (i + 0.5) * ratio_w;
var w0 = dy*dy //pre-calc part of w
for(var xx = Math.floor(i * ratio_w); xx < (i + 1) * ratio_w; xx++){
var dx = Math.abs(center_x - (xx + 0.5)) / ratio_w_half;
var w = Math.sqrt(w0 + dx*dx);
if(w >= -1 && w <= 1){
//hermite filter
weight = 2 * w*w*w - 3*w*w + 1;
if(weight > 0){
dx = 4*(xx + yy*W);
//alpha
gx_a += weight * data[dx + 3];
weights_alpha += weight;
//colors
if(data[dx + 3] < 255)
weight = weight * data[dx + 3] / 250;
gx_r += weight * data[dx];
gx_g += weight * data[dx + 1];
gx_b += weight * data[dx + 2];
weights += weight;
}
}
}
}
data2[x2] = gx_r / weights;
data2[x2 + 1] = gx_g / weights;
data2[x2 + 2] = gx_b / weights;
data2[x2 + 3] = gx_a / weights_alpha;
}
}
console.log("hermite = "+(Math.round(Date.now() - time1)/1000)+" s");
canvas.getContext("2d").clearRect(0, 0, Math.max(W, W2), Math.max(H, H2));
canvas.getContext("2d").putImageData(img2, 0, 0);
}
项目所有者发布了一个对该项目的更新,修复了它。您只需要四舍五入宽度和高度输入。