前端 · 深入理解 transform 函数的计算原理
在涉及到前端图形学的时候,几乎避免不了 transform 属性的应用。
而 transform 一共内置了五种不同大类的函数(矩阵变形、平移、缩放、旋转、倾斜,具体细节有九个),开发者经常容易被不同函数的组合变换,搞到晕头转向。
当面对需要精准定位的需求时,如果对 transform 的计算原理理解不透彻,就会导致代码冗长、复杂度增加,易读性也会迅速下降。
事实上,前端里的 transform 有很多种,比如 CSS 和 SVG 中的 transform 属性就有些许不同。不过万变不离其宗,它们底层的数学原理大体是一致的。
所以为了方便描述,本篇这里以 SVG transform 为主。
一来,可以免去 CSS 中大量关于单位不同的换算,排开很多跟原理无关的细节; 二来,作为矢量格式的 SVG 足够精简,用来描述数学计算方式,矢量化参数拥有天生的优势;
① transform: matrix(a, b, c, d, e, f)
说到图形学,那必然会涉及到矩阵运算。
matrix 函数可以说是最本源的存在,如果将前端页面想象成一块画布,matrix 就是这块画布的改造者。只需要设定不同的参数,就可以用 matrix 将图形随意变换。
同时,matrix 函数还是其他四类功能函数的核心,这四类分别是平移、缩放、旋转、倾斜,他们的实现方式都可以用 matrix 等价替换。
matrix 函数的参数是一个 3x3 的方阵矩阵,只不过这个矩阵中只有六个变量,所以函数声明里显式的参数列表长度为 6。
矩阵形式如下(假设为 M):
M=(acebdf001)M = \begin{pmatrix} a & c & e \\ b & d & f \\ 0 & 0 & 1 \\ \end{pmatrix}M=⎝⎛ab0cd0ef1⎠⎞
怎么用呢? 答案是:矩阵乘法
假设页面上有一个点 point_old 的坐标为 (oldX, oldY),转换后新的点 point_new 坐标为 (newX, newY)。
在运算过程中,点的矩阵描述方式如下:
pointold=(oldXoldY1)pointnew=(newXnewY1)point_{old} = \begin{pmatrix} oldX \\ oldY \\ 1 \end{pmatrix} \\ point_{new} = \begin{pmatrix} newX \\ newY \\ 1 \end{pmatrix}pointold=⎝⎛oldXoldY1⎠⎞pointnew=⎝⎛newXnewY1⎠⎞
计算方式为:
pointnew=M∗pointoldpoint_{new} = M * point_{old}pointnew=M∗pointold (newXnewY1)=(acebdf001)(oldXoldY1)=(a∗oldX+c∗oldY+eb∗oldX+d∗oldY+f1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} a & c & e \\ b & d & f \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} a*oldX+c*oldY+e \\ b*oldX+d*oldY+f \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛ab0cd0ef1⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛a∗oldX+c∗oldY+eb∗oldX+d∗oldY+f1⎠⎞
所以:
pointnew{newX=a∗oldX+c∗oldY+enewY=b∗oldX+d∗oldY+fpoint_{new} \begin{cases} newX = a*oldX + c*oldY + e \\ newY = b*oldX + d*oldY + f \\ \end{cases}pointnew{newX=a∗oldX+c∗oldY+enewY=b∗oldX+d∗oldY+f
在这六个参数中,e、f 主要负责偏移量,其余 a、b、c、d 则代表不同的放大倍数。
现在我们知道,可以通过对这六个参数的控制,实现不同的效果了。
比如默认状态下,matrix(1, 0, 0, 1, 0, 0) 代表了什么也不动,因为套用上述计算公式,
(newXnewY1)=(100010001)(oldXoldY1)=(1∗oldX+0∗oldY+00∗oldX+1∗oldY+01)=(oldXoldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1*oldX+0*oldY+0 \\ 0*oldX+1*oldY+0 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛100010001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛1∗oldX+0∗oldY+00∗oldX+1∗oldY+01⎠⎞=⎝⎛oldXoldY1⎠⎞
结果可以发现,点坐标没有任何变换。
到这里,transform 的核心函数 matrix() 是如何计算的,应该已经清楚了。
那么接下来看看剩下其他所有的函数是如何实现和 matrix 转换的。
<h1>default</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9"></rect> </svg> 复制代码
② transform: translate(x)
translate 为平移函数,当只有一个参数时,表示图形水平移动了多少的距离。 即:
newX=x+oldXnewX = x + oldXnewX=x+oldX
那么很简单的,构造矩阵 matrix(1, 0, 0, 1, x, 0) 即可实现 translate(x) 的效果:
(newXnewY1)=(10x010001)(oldXoldY1)=(1∗oldX+0∗oldY+x0∗oldX+1∗oldY+01)=(x+oldXoldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1*oldX+0*oldY+x \\ 0*oldX+1*oldY+0 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} x + oldX\\ oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛100010x01⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛1∗oldX+0∗oldY+x0∗oldX+1∗oldY+01⎠⎞=⎝⎛x+oldXoldY1⎠⎞
<div> <h1>transform: translate(x)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="translate(100)"></rect> </svg> <h1>transform: matrix(1, 0, 0, 1, x, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(1,0,0,1,100,0)"></rect> </svg> </div> 复制代码
③ transform: translate(x, y)
这里可以看做单一参数的 translate 函数的重载函数,第二个参数 y 值,代表在笛卡尔坐标系下的二维平面中,y 轴方向的平移运动。
即:
{newX=x+oldXnewY=y+oldY\begin{cases} newX = x + oldX \\ newY = y + oldY \end{cases}{newX=x+oldXnewY=y+oldY
同理,可构造矩阵 matrix(1, 0, 0, 1, x, y) 实现 translate(x, y) 的效果:
(newXnewY1)=(10x01y001)(oldXoldY1)=(1∗oldX+0∗oldY+x0∗oldX+1∗oldY+y1)=(x+oldXy+oldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & x \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1*oldX+0*oldY+x \\ 0*oldX+1*oldY+y \\ 1 \\ \end{pmatrix} = \begin{pmatrix} x + oldX \\ y + oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛100010xy1⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛1∗oldX+0∗oldY+x0∗oldX+1∗oldY+y1⎠⎞=⎝⎛x+oldXy+oldY1⎠⎞
<div> <h1>transform: translate(x, y)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="translate(100,50)"></rect> </svg> <h1>transform: matrix(1, 0, 0, 1, x, y)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(1,0,0,1,100,50)"></rect> </svg> </div> 复制代码
④ transform: scale(s)
scale 为缩放函数,当只有一个参数时,表示图形在水平和纵向两个轴上,实现等比例的放大缩小。
即:
{newX=s∗oldXnewY=s∗oldY\begin{cases} newX = s*oldX \\ newY = s*oldY \end{cases}{newX=s∗oldXnewY=s∗oldY
由于这里是成比例放大,所以可得变换矩阵 matrix(s, 0, 0, s, 0, 0):
(newXnewY1)=(s000s0001)(oldXoldY1)=(s∗oldX+0∗oldY+00∗oldX+s∗oldY+01)=(s∗oldXs∗oldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix}= \begin{pmatrix} s & 0 & 0 \\ 0 & s & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix}= \begin{pmatrix} s*oldX+0*oldY+0 \\ 0*oldX+s*oldY+0 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} s*oldX \\ s*oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛s000s0001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛s∗oldX+0∗oldY+00∗oldX+s∗oldY+01⎠⎞=⎝⎛s∗oldXs∗oldY1⎠⎞
<div> <h1>transform: scale(s)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="scale(2)"></rect> </svg> <h1>transform: matrix(s, 0, 0, s, 0, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(2,0,0,2,0,0)"> </rect> </svg> </div> 复制代码
⑤ transform: scale(sx, sy)
这里同样的,也是拥有两个参数的重载函数,由此可以分开控制不同轴向的缩放倍率。
即:
{newX=sx∗oldXnewY=sy∗oldY\begin{cases} newX = sx*oldX \\ newY = sy*oldY \end{cases}{newX=sx∗oldXnewY=sy∗oldY
同理可得变换矩阵 matrix(sx, 0, 0, sy, 0, 0):
(newXnewY1)=(sx000sy0001)(oldXoldY1)=(sx∗oldX+0∗oldY+00∗oldX+sy∗oldY+01)=(sx∗oldXsy∗oldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix}= \begin{pmatrix} sx & 0 & 0 \\ 0 & sy & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} = \begin{pmatrix} sx*oldX+0*oldY+0 \\ 0*oldX+sy*oldY+0 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} sx*oldX \\ sy*oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛sx000sy0001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛sx∗oldX+0∗oldY+00∗oldX+sy∗oldY+01⎠⎞=⎝⎛sx∗oldXsy∗oldY1⎠⎞
<div> <h1>transform: scale(sx, sy)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="scale(0.5,2)"></rect> </svg> <h1>transform: matrix(sx, 0, 0, sy, 0, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(0.5,0,0,2,0,0)"></rect> </svg> </div> 复制代码
⑥ transform: rotate(a)
rotate 为旋转函数,当参数个数为 1 时,表示以当前元素坐标系原点为旋转点,旋转角度为 a 度。
需要提前注意的是,这里的单位为 deg,角度制。
而在接下来换算成 matrix 的过程中,需要用到三角函数。
所以在数值上,需要将角度制,转换成弧度制:
a′=π180∗aa'=\frac{\pi}{180}*aa′=180π∗a
此外,由于在二维平面旋转运动下,任意点到旋转圆心的距离不变。所以为了方便计算,我们在这里使用极坐标系,推导笛卡尔坐标系下物体运动的方式。
根据极坐标系,我们用有序数对 *(ρ, θ) * 表示任意点 P 的坐标,ρ 代表极径,θ 代表极角(弧度制)。
记为 P(ρ, θ)
那么,任意点旋转 a 角度(*a' * 弧度)后的坐标即为:P(ρ, θ + a')
利用坐标系间的映射关系:
{X=ρ∗cos(θ)Y=ρ∗sin(θ)\begin{cases} X = \rho*cos(\theta) \\ Y = \rho*sin(\theta) \\ \end{cases}{X=ρ∗cos(θ)Y=ρ∗sin(θ)
可得:
newP=oldP(ρ,θ+a′)newP = oldP(\rho,\theta + a')newP=oldP(ρ,θ+a′) {newX=ρ∗cos(θ+a′)newY=ρ∗sin(θ+a′)\begin{cases} newX = \rho*cos(\theta+a') \\ newY = \rho*sin(\theta+a') \\ \end{cases}{newX=ρ∗cos(θ+a′)newY=ρ∗sin(θ+a′)
进一步展开可得:
newX=ρ∗cos(θ+a′)=ρ∗cos(θ)∗cos(a′)−ρ∗sin(θ)∗sin(a′)=oldX∗cos(a′)−oldY∗sin(a′)=cos(a′)∗oldX+(−1)∗sin(a′)∗oldY\begin{aligned} newX &= \rho*cos(\theta+a') \\ &= \rho*cos(\theta)*cos(a')-\rho*sin(\theta)*sin(a') \\ &= oldX*cos(a')-oldY*sin(a') \\ &= cos(a')*oldX + (-1)*sin(a')*oldY \\ \end{aligned}newX=ρ∗cos(θ+a′)=ρ∗cos(θ)∗cos(a′)−ρ∗sin(θ)∗sin(a′)=oldX∗cos(a′)−oldY∗sin(a′)=cos(a′)∗oldX+(−1)∗sin(a′)∗oldY newY=ρ∗sin(θ+a′)=ρ∗sin(θ)∗cos(a′)+ρ∗cos(θ)∗sin(a′)=oldY∗cos(a′)+oldX∗sin(a′)=sin(a′)∗oldX+cos(a′)∗oldY\begin{aligned} newY & = \rho*sin(\theta+a') \\ & = \rho*sin(\theta)*cos(a')+\rho*cos(\theta)*sin(a') \\ & = oldY * cos(a') + oldX * sin(a') \\ & = sin(a') * oldX + cos(a') * oldY \end{aligned}newY=ρ∗sin(θ+a′)=ρ∗sin(θ)∗cos(a′)+ρ∗cos(θ)∗sin(a′)=oldY∗cos(a′)+oldX∗sin(a′)=sin(a′)∗oldX+cos(a′)∗oldY
根据上式,可以推出变换矩阵为 matrix(cos(a'), sin(a'), -sin(a'), cos(a'), 0, 0)
(newXnewY1)=(cos(a′)−sin(a′)0sin(a′)cos(a′)0001)(oldXoldY1)=(cos(a′)∗oldX−sin(a′)∗oldY+0sin(a′)∗oldX+cos(a′)∗oldY+01)=(ρ∗cos(a′)∗cos(θ)−ρ∗sin(a′)∗sin(θ)ρ∗sin(a′)∗cos(θ)+ρ∗cos(a′)∗sin(θ)1)=(ρ∗cos(θ+a′)ρ∗sin(θ+a′)1)\begin{aligned} \begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} & = \begin{pmatrix} cos(a') & -sin(a') & 0 \\ sin(a') & cos(a') & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} cos(a')*oldX-sin(a')*oldY+0 \\ sin(a')*oldX+cos(a')*oldY+0 \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} \rho*cos(a')*cos(\theta)-\rho*sin(a')*sin(\theta) \\ \rho*sin(a')*cos(\theta)+\rho*cos(a')*sin(\theta) \\ 1 \\\end{pmatrix} \\ & = \begin{pmatrix} \rho*cos(\theta + a') \\ \rho*sin(\theta + a') \\ 1 \\ \end{pmatrix} \end{aligned}⎝⎛newXnewY1⎠⎞=⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛cos(a′)∗oldX−sin(a′)∗oldY+0sin(a′)∗oldX+cos(a′)∗oldY+01⎠⎞=⎝⎛ρ∗cos(a′)∗cos(θ)−ρ∗sin(a′)∗sin(θ)ρ∗sin(a′)∗cos(θ)+ρ∗cos(a′)∗sin(θ)1⎠⎞=⎝⎛ρ∗cos(θ+a′)ρ∗sin(θ+a′)1⎠⎞
<div> <h1>transform: rotate(a)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="rotate(30)"></rect> </svg> <h1>transform: matrix(cos(a'), sin(a'), -sin(a'), cos(a'), 0, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(0.866025,0.5,-0.5,0.866025,0,0)"></rect> </svg> </div> 复制代码
⑦ transform: rotate(a, x, y)
当 rotate 函数被指定旋转点后,情况稍微复杂了一点。
由于函数本质上控制的是画布本身,也可以理解为坐标系本身。
所以,如果想要坐标系上的某一个图形围绕具体一个点旋转,则需要以下三个步骤:
第一、将旋转点从坐标系原点,移动至指定点; 第二、该指定点默认为坐标系原点,开始旋转; 第三、为了保持旋转时其他图案的不变,将坐标系原点从指定点移动回初始点位。
所以,通常指定点的旋转,会采用 <translate(x, y)><rotate(a)><translate(-x, -y)> 的方式。
translate 中的参数 x、y 即为 rotate(a, x, y) 中的指定点坐标。
那么这种情况,应当如何用 matrix 描述呢?
我们假设上述三个变换矩阵分别为:
{translate(x,y)=T1=(10x01y001)rotate(a)=R=(cos(a′)−sin(a′)0sin(a′)cos(a′)0001)translate(−x,−y)=T2=(10−x01−y001)\begin{cases} translate(x,y)= T_1 = \begin{pmatrix} 1 & 0 & x \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{pmatrix} \\ rotate(a) = R = \begin{pmatrix} cos(a') & -sin(a') & 0 \\ sin(a') & cos(a') & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \\ translate(-x,-y)=T_2= \begin{pmatrix} 1 & 0 & -x \\ 0 & 1 & -y \\ 0 & 0 & 1 \\ \end{pmatrix} \end{cases}⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧translate(x,y)=T1=⎝⎛100010xy1⎠⎞rotate(a)=R=⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0001⎠⎞translate(−x,−y)=T2=⎝⎛100010−x−y1⎠⎞
则,根据函数执行方式可得矩阵计算方式为:
(newXnewY1)=T1∗R∗T2∗(oldXoldY1)\begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} = T_1 * R * T_2 * \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=T1∗R∗T2∗⎝⎛oldXoldY1⎠⎞
即:
M=T1∗R∗T2=(10x01y001)(cos(a′)−sin(a′)0sin(a′)cos(a′)0001)(10−x01−y001)=(cos(a′)−sin(a′)xsin(a′)cos(a′)y001)(10−x01−y001)=(cos(a′)−sin(a′)−x∗cos(a′)+y∗sin(a′)+xsin(a′)cos(a′)−x∗sin(a′)−y∗cos(a′)+y001)\begin{aligned} M & = T_1 * R * T_2 \\ & = \begin{pmatrix} 1 & 0 & x \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} cos(a') & -sin(a') & 0 \\ sin(a') & cos(a') & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & -x \\ 0 & 1 & -y \\ 0 & 0 & 1 \\ \end{pmatrix} \\ &= \begin{pmatrix} cos(a') & -sin(a') & x \\ sin(a') & cos(a') & y \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & -x \\ 0 & 1 & -y \\ 0 & 0 & 1 \\ \end{pmatrix} \\ &= \begin{pmatrix} cos(a') & -sin(a') & -x*cos(a')+y*sin(a')+x \\ sin(a') & cos(a') & -x*sin(a')-y*cos(a')+y \\ 0 & 0 & 1 \end{pmatrix} \end{aligned}M=T1∗R∗T2=⎝⎛100010xy1⎠⎞⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0001⎠⎞⎝⎛100010−x−y1⎠⎞=⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0xy1⎠⎞⎝⎛100010−x−y1⎠⎞=⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0−x∗cos(a′)+y∗sin(a′)+x−x∗sin(a′)−y∗cos(a′)+y1⎠⎞
也就是说,变换矩阵为 : matrix(cos(a'), sin(a'), -sin(a'), cos(a'), -xcos(a')+ysin(a')+x, -xsin(a')-ycos(a')+y)
(newXnewY1)=(cos(a′)−sin(a′)−x∗cos(a′)+y∗sin(a′)+xsin(a′)cos(a′)−x∗sin(a′)−y∗cos(a′)+y001)(oldXoldY1)\begin{pmatrix} newX \\ newY \\ 1 \end{pmatrix} = \begin{pmatrix} cos(a') & -sin(a') & -x*cos(a')+y*sin(a')+x \\ sin(a') & cos(a') & -x*sin(a')-y*cos(a')+y \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix}⎝⎛newXnewY1⎠⎞=⎝⎛cos(a′)sin(a′)0−sin(a′)cos(a′)0−x∗cos(a′)+y∗sin(a′)+x−x∗sin(a′)−y∗cos(a′)+y1⎠⎞⎝⎛oldXoldY1⎠⎞
<div> <h1>transform: rotate(a, x, y)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="rotate(30,0,100)"></rect> </svg> <h1>transform: matrix(cos(a'), sin(a'), -sin(a'), cos(a'), -x*cos(a')+y*sin(a')+x, -x*sin(a')-y*cos(a')+y)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(0.866025, 0.5, -0.5, 0.866025, 50.0, 13.39746)"></rect> </svg> </div> 复制代码
⑧ transform: skewX(a)
skewX 表示的是 x 轴方向上的倾斜,同样这里将使用三角函数,也同样的,存在弧度制下的:
a′=π180∗aa'=\frac{\pi}{180}*aa′=180π∗a
由于倾斜只发生在 x 轴方向,由此可得:
{newX=Δx+oldX=tan(a′)∗oldY+oldXnewY=oldY\begin{cases} newX = \Delta x + oldX = tan(a')*oldY + oldX\\ newY = oldY \end{cases}{newX=Δx+oldX=tan(a′)∗oldY+oldXnewY=oldY
故,变换函数为 matrix(1, 0, tan(a'), 1, 0, 0)
(newXnewY1)=(1tan(a′)0010001)(oldXoldY1)=(1∗oldX+tan(a′)∗oldY+00∗oldX+1∗oldY+01)=(tan(a′)∗oldY+oldXoldY1)\begin{aligned} \begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} & = \begin{pmatrix} 1 & tan(a') & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} 1*oldX+tan(a')*oldY+0 \\ 0*oldX+1*oldY+0 \\ 1 \\ \end{pmatrix}\\ &= \begin{pmatrix} tan(a')*oldY + oldX\\ oldY \\ 1 \\ \end{pmatrix} \end{aligned}⎝⎛newXnewY1⎠⎞=⎝⎛100tan(a′)10001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛1∗oldX+tan(a′)∗oldY+00∗oldX+1∗oldY+01⎠⎞=⎝⎛tan(a′)∗oldY+oldXoldY1⎠⎞
<div> <h1>transform: skewX(a)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="skewX(30)"></rect> </svg> <h1>transform: matrix(1, 0, tan(a'), 1, 0, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(1,0,0.577350,1,0,0)"></rect> </svg> </div> 复制代码
⑨ transform: skewY(a)
skewY 表示的是 y 轴方向的倾斜,原理同上:
{newX=oldXnewY=Δy+oldY=tan(a′)∗oldX+oldY\begin{cases} newX = oldX \\ newY = \Delta y + oldY = tan(a')*oldX + oldY \\ \end{cases}{newX=oldXnewY=Δy+oldY=tan(a′)∗oldX+oldY
可得变换函数 matrix(1, tan(a'), 0, 1, 0, 0)
(newXnewY1)=(100tan(a′)10001)(oldXoldY1)=(1∗oldX+0∗oldY+0tan(a′)∗oldX+1∗oldY+01)=(oldXtan(a′)∗oldX+oldY1)\begin{aligned} \begin{pmatrix} newX \\ newY \\ 1 \\ \end{pmatrix} & = \begin{pmatrix} 1 & 0 & 0 \\ tan(a') & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} oldX \\ oldY \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} 1*oldX + 0*oldY + 0 \\ tan(a')*oldX+1*oldY + 0 \\ 1 \\ \end{pmatrix}\\ &= \begin{pmatrix} oldX\\ tan(a')*oldX+oldY \\ 1 \\ \end{pmatrix} \end{aligned}⎝⎛newXnewY1⎠⎞=⎝⎛1tan(a′)0010001⎠⎞⎝⎛oldXoldY1⎠⎞=⎝⎛1∗oldX+0∗oldY+0tan(a′)∗oldX+1∗oldY+01⎠⎞=⎝⎛oldXtan(a′)∗oldX+oldY1⎠⎞
<div> <h1>transform: skewY(a)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="skewY(30)"></rect> </svg> <h1>transform: matrix(1, tan(a'), 0, 1, 0, 0)</h1> <svg x="0px" y="0px" width="600px" height="300px"> <line label="axisX" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="600" y2="0" /> <line label="axisY" fill="none" stroke="black" stroke-width="10" x1="0" y1="0" x2="0" y2="300" /> <rect x="0" y="0" width="200" height="100" fill="red" opacity="0.9" transform="matrix(1,0.577350,0,1,0,0)"></rect> </svg> </div> 复制代码
综上,就是 transform 全部函数的计算方式了。
或者也可以认为是它的矩阵运算描述。
当然,代码实现的时候可能会为了减少不必要的矩阵运算,从而做了最优化处理。但是理解它的运算原理,清楚底层的计算逻辑,却是十分有益的。
作者:心宿二_Sco
链接:https://juejin.cn/post/7031047012011212814