我在代码中的HTML5 canvas元素上绘制了很多1px的线。绘图代码大致如下所示,在这种情况下,transform
变量是使用d3-zoom设置的。 instructions.f32
是一个Float32Array,其中包含我用于绘制线条的坐标。
context.setTransform(
transform.k,
0,
0,
transform.k,
transform.x,
transform.y
);
context.lineWidth = 1 / transform.k;
context.beginPath();
for (let i = from; i < to; ++i) {
let v1 = instructions.f32[i * 4 + 1];
let v2 = instructions.f32[i * 4 + 2];
// execute some moveTo/lineTo commands using v1 and v2 as coordinates
}
context.stroke();
此代码的一个问题是1px的线很模糊,因为我绘制的是像素边界。我尝试修改代码以将线对齐到最近的像素,如下所示:
let v1 = (Math.round(instructions.f32[i * 4 + 1] * transform.k) + 0.5) / transform.k;
let v2 = (Math.round(instructions.f32[i * 4 + 2] * transform.k) + 0.5) / transform.k;
但这仍会导致线条模糊,如下图所示(放大图像的屏幕截图):
如果我没有任何转换集,据我了解,我只需要将坐标舍入到最接近的像素并添加0.5即可获得清晰的线条。但是我不确定在整个画布转换后如何实现此目的,并且我没有绘制最终的坐标系。到目前为止,我的纠正尝试都失败了,看来我在这里遗漏了一些东西,或者在途中犯了其他错误。
使用setTransform转换整个画布时,如何在画布中绘制清晰的1px线?我该如何精确地四舍五入坐标才能将生成的线捕捉到像素?
答案 0 :(得分:0)
由于似乎您的变换没有倾斜或旋转属性,所以最简单的方法可能是不变换上下文,而是缩放并转换所有坐标。
当前,您将lineWidth设置为1 / zoom,考虑到Compiters在Math精度方面的出色表现,您将很难用它绘制一个完美的1px笔画,只有几个zoom值可以,如果您愿意要将缩放限制在这些值,您会得到断断续续的缩放。
相反,始终将lineWidth保持为1px,缩放并转换所有坐标,然后再将它们四舍五入到最近的像素边界。
context.setTransform(1,0,0,1,0,0);
context.lineWidth = 1;
context.beginPath();
for (let i = from; i < to; ++i) {
let v1 = instructions.f32[i * 4 + 1];
let v2 = instructions.f32[i * 4 + 2];
// scale and translate
v1 = (v1 + transform.x) * transform.k;
v2 = (v2 + transform.y) * transfrom.k;
// round
const r1 = Math.round(v1);
const r2 = Math.round(v2);
// to nearest px boundary
v1 = r1 + (0.5 * Math.sign(r1 - v1) || 0.5);
v2 = r2 + (0.5 * Math.sign(r2 - v2) || 0.5);
// lineTo...
}
const pts = [60, 60, 60, 110, 100,110, 100, 90, 220, 90];
const zoom = d3.behavior.zoom()
.scaleExtent([1, 10])
.on("zoom", zoomed);
const transform = {k: 1, x: 0, y: 0};
const context = canvas.getContext('2d');
d3.select('canvas')
.call(zoom);
draw();
function draw() {
context.setTransform(1,0,0,1,0,0);
context.clearRect(0,0,canvas.width, canvas.height);
context.lineWidth = 1;
context.beginPath();
for (let i = 0; i < pts.length; i+=2) {
let v1 = pts[i];
let v2 = pts[i + 1];
// scale and translate
v1 = (v1 + transform.x) * transform.k;
v2 = (v2 + transform.y) * transform.k;
// round
const r1 = Math.round(v1);
const r2 = Math.round(v2);
// to nearest px boundary
v1 = r1 + (0.5 * Math.sign(r1 - v1) || 0.5);
v2 = r2 + (0.5 * Math.sign(r2 - v2) || 0.5);
context.lineTo(v1, v2);
}
context.stroke();
}
function zoomed() {
const evt = d3.event;
transform.k = evt.scale;
transform.x = evt.translate[0];
transform.y = evt.translate[1];
draw();
}
canvas {border: 1px solid}
zoom with mousewheel and pan by dragging<br>
<canvas id="canvas"></canvas>
<script src="//d3js.org/d3.v3.min.js"></script>
但是您可能更喜欢精度较低,但锯齿不平整和简单的地板:
v1 = (v1 + transform.x) * transform.k;
v2 = (v2 + transform.y) * transform.k;
// floor
v1 = Math.floor(v1) + 0.5;
v2 = Math.floor(v2) + 0.5;
// lineTo
const pts = [60, 60, 60, 110, 100,110, 100, 90, 220, 90];
const zoom = d3.behavior.zoom()
.scaleExtent([1, 10])
.on("zoom", zoomed);
const transform = {k: 1, x: 0, y: 0};
const context = canvas.getContext('2d');
d3.select('canvas')
.call(zoom);
draw();
function draw() {
context.setTransform(1,0,0,1,0,0);
context.clearRect(0,0,canvas.width, canvas.height);
context.lineWidth = 1;
context.beginPath();
for (let i = 0; i < pts.length; i+=2) {
let v1 = pts[i];
let v2 = pts[i + 1];
// scale and translate
v1 = (v1 + transform.x) * transform.k;
v2 = (v2 + transform.y) * transform.k;
// floor
v1 = Math.floor(v1) + 0.5;
v2 = Math.floor(v2) + 0.5;
context.lineTo(v1, v2);
}
context.stroke();
}
function zoomed() {
const evt = d3.event;
transform.k = evt.scale;
transform.x = evt.translate[0];
transform.y = evt.translate[1];
draw();
}
canvas {border: 1px solid}
zoom with mousewheel and pan by dragging<br>
<canvas id="canvas"></canvas>
<script src="//d3js.org/d3.v3.min.js"></script>