我需要在 Canvas 上绘制图表。但是我如何使用代数方程作为输入，并基于方程，使用 javascript 绘制曲线？
例如:
x2+5y=250
方程绘制了一个包含正值和负值的图形。
<!DOCTYPE html>
<html>
<head>
<title>Interactive Line Graph</title>
<script src="http://ajax.aspnetcdn.com/ajax/jQuery/jquery-1.6.1.min.js"></script>
<script>
var graph;
var xPadding = 30;
var yPadding = 30;
var data = { values:[
{ X: "1", Y: 15 },
{ X: "2", Y: 35 },
{ X: "3", Y: 60 },
{ X: "4", Y: 14 },
{ X: "5", Y: 20 },
{ X: "6", Y: 95 },
]};
// Returns the max Y value in our data list
function getMaxY() {
var max = 0;
for(var i = 0; i < data.values.length; i ++) {
if(data.values[i].Y > max) {
max = data.values[i].Y;
}
}
max += 10 - max % 10;
return max;
}
// Return the x pixel for a graph point
function getXPixel(val) {
return ((graph.width() - xPadding) / data.values.length) * val + (xPadding * 1.5);
}
// Return the y pixel for a graph point
function getYPixel(val) {
return graph.height() - (((graph.height() - yPadding) / getMaxY()) * val) - yPadding;
}
$(document).ready(function() {
graph = $('#graph');
var c = graph[0].getContext('2d');
c.lineWidth = 2;
c.strokeStyle = '#333';
c.font = 'italic 8pt sans-serif';
c.textAlign = "center";
// Draw the axises
c.beginPath();
c.moveTo(xPadding, 0);
c.lineTo(xPadding, graph.height() - yPadding);
c.lineTo(graph.width(), graph.height() - yPadding);
c.stroke();
// Draw the X value texts
for(var i = 0; i < data.values.length; i ++) {
c.fillText(data.values[i].X, getXPixel(i), graph.height() - yPadding + 20);
}
// Draw the Y value texts
c.textAlign = "right"
c.textBaseline = "middle";
for(var i = 0; i < getMaxY(); i += 10) {
c.fillText(i, xPadding - 10, getYPixel(i));
}
c.strokeStyle = '#f00';
// Draw the line graph
c.beginPath();
c.moveTo(getXPixel(0), getYPixel(data.values[0].Y));
for(var i = 1; i < data.values.length; i ++) {
c.lineTo(getXPixel(i), getYPixel(data.values[i].Y));
}
c.stroke();
// Draw the dots
c.fillStyle = '#333';
for(var i = 0; i < data.values.length; i ++) {
c.beginPath();
c.arc(getXPixel(i), getYPixel(data.values[i].Y), 4, 0, Math.PI * 2, true);
c.fill();
}
});
</script>
</head>
<body>
<canvas id="graph" width="200" height="150">
</canvas>
</body>
</html>最佳答案
解析线性方程。
或者问题可能是方程式的解析。
这个答案展示了如何解析一个简单的线性方程。
用户输入 x2+5y=230，您需要为 f(x) 求解和绘制 y，这就是函数函数(x){ 返回(3 * x -230)/-5;
将假设方程始终采用与 x 和 y 以及一些标量和常数相同的形式 scalar * x + const + scalar * y = const
定义规则
规则
- 只有 x 和 y 会被视为变量。
- 项是标量和变量
2x或常量+1。 - 将忽略所有其他字符，包括
*、/、% - 数字可以有小数位。有效数字
1+10.2-210e5 - 标量必须与变量相邻
3y2变为6y3y-2保持原样。
解析
要解析方程式，我们必须将其分解为明确且易于操作的单位。在这种情况下，我称之为术语的单元将具有 3 个属性。
- 标量一个数字
- variable 变量 x,y 的名称或常量为 null
- side 项在等式的左边或右边
方程式示例
2x + 2 + 3y = 4x - 1y
首先解析创建 条款
// shorthand not code
{2,x,true; // true is for left
{2,null,true; // null is a constant
{3,y,true;
{4,x,false;
{-1,y,false;
一旦所有的项都被解析，方程式就通过对 x、y 和常数的所有项求和并将所有项向左移动并翻转右侧任何值的符号来求解。
sumX = 2 + -4; //as 4x is on the right it becomes negative
sumY = 3 + 1;
const = 2;
做等式
-2x + 4y + 2 = 0
然后将 y 向右移出并用其标量除以左边。
-2x + 2 = 4y
(-2x + 2)/-4 = y
结果是一个我们可以从 javascript 调用的函数，它会计算 x 的值并获取 y 的值。
function(x){ return (-2 * x + 2) / 4; }
解析器
以下函数解析并返回 x 的输入方程式的函数。然后该函数用于在下面的演示中绘制点。
function parseEquation(input){
// Important that white spaces are removed first
input = input.replace(/\s+/g,""); // remove whitespaces
input = input.replace(/([\-\+])([xy])/g,"$11$2"); // convert -x -y or +x +y to -1x -1y or +1x +1y
// just to make the logic below a little simpler
var newTerm = () => {term = { val : null, scalar : 1, left : left, };} // create a new term
var pushTerm = () => {terms.push(term); term = null;} // push term and null current
// regExp [xy=] gets "x","y", or "="" or [\-\+]??[0-9\.]+ gets +- number with decimal
var reg =/[xy=]|[\-\+]??[0-9\.eE]+/g; // regExp to split the input string into parts
var parts = input.match(reg); // get all the parts of the equation
var terms = []; // an array of all terms parsed
var term = null; // Numbers as constants and variables with scalars are terms
var left = true; // which side of equation a term is
parts.forEach( p=> {
if (p === "x" || p === "y") {
if (term !== null && term.val !== null) { // is the variable defined
pushTerm(); // yes so push to the stack and null
}
if (term === null) { newTerm(); } // do we need a new term?
term.val = p;
} else if( p === "=") { // is it the equals sign
if (!left) { throw new SyntaxError("Unxpected `=` in equation."); }
if (term === null) { throw new SyntaxError("No left hand side of equation."); }// make sure that there is a left side
terms.push(term); // push the last left side term onto the stack
term = null;
left = false; // everything on the right from here on in
} else { // all that is left are numbers (we hope)
if (isNaN(p)){ throw new SyntaxError("Unknown value '"+p+"' in equation"); }//check that there is a number
if (term !== null && (p[0] === "+" || p[0] === "-")) { // check if number is a new term
pushTerm(); // yes so push to the stack and null
}
if (term === null) { newTerm(); } // do we need a new term?
term.scalar *= Number(p); // set the scalar to the new value
}
});
if (term !== null) { // there may or may not be a term left to push to the stack
pushTerm();
}
// now simplify the equation getting the scalar for left and right sides . x on left y on right
var scalarX = 0;
var scalarY = 0
var valC = 0; // any constants
terms.forEach(t => {
t.scalar *= !t.left ? -1 : 1; // everything on right is negative
if (t.val === "y") {
scalarY += -t.scalar; // reverse sign
} else if (t.val === "x") {
scalarX += t.scalar;
} else {
valC += t.scalar;
}
})
// now build the code string for the equation to solve for x and return y
var code = "return (" + scalarX + " * x + (" + valC + ")) / "+scalarY +";\n";
var equation = new Function("x",code); // create the function
return equation;
}
下面的用法例子都是同一个方程
var equation = parseEquation("x2+5y+x=230");
var y = equation(10); // get y for x = 10;
equation = parseEquation("x2+x=230-5y");
equation = parseEquation("x2+x-30=200-2y-3y");
equation = parseEquation("200- 2y-3y = x2+x-30");
equation = parseEquation("200-2y- 3y - x2-x+30=0");
equation = parseEquation("100.0 + 100-2y- 3y - x2-x+30=0");
equation = parseEquation("1e2 + 10E1-2y- 3y - x2-x+30=0");
演示
我已将它添加到 markE 已经给出的答案中的代码中。 (希望你不介意 markE)
function plot(equation) {
var graph;
var xPadding = 30;
var yPadding = 30;
var data = {
values : [{
X : "1",
Y : 15
}, {
X : "2",
Y : 35
}, {
X : "3",
Y : 60
}, {
X : "4",
Y : 14
}, {
X : "5",
Y : 20
}, {
X : "6",
Y : -30
},
]
};
// Returns the max Y value in our data list
function getMaxY() {
var max = 0;
for (var i = 0; i < data.values.length; i++) {
if (data.values[i].Y > max) {
max = data.values[i].Y;
}
}
max += 10 - max % 10;
return max;
}
var scaleA = 1.4;
// Return the x pixel for a graph point
function getXPixel(val) {
return ((graph.width() / scaleA - xPadding) / data.values.length) * val + (xPadding * 1.5);
}
// Return the y pixel for a graph point
function getYPixel(val) {
return graph.height() / scaleA - (((graph.height() / scaleA - yPadding) / getMaxY()) * val) - yPadding;
}
graph = $('#graph');
var c = graph[0].getContext('2d');
c.clearRect(0,0,graph[0].width,graph[0].height);
c.lineWidth = 2;
c.strokeStyle = '#333';
c.font = 'italic 8pt sans-serif';
c.textAlign = "center";
// Draw the axises
c.beginPath();
c.moveTo(xPadding, 0);
c.lineTo(xPadding, graph.height() / scaleA - yPadding);
c.lineTo(graph.width(), graph.height() / scaleA - yPadding);
c.stroke();
// Draw the X value texts
for (var i = 0; i < data.values.length; i++) {
c.fillText(data.values[i].X, getXPixel(i), graph.height() / scaleA - yPadding + 20);
}
// Draw the Y value texts
c.textAlign = "right"
c.textBaseline = "middle";
for (var i = 0; i < getMaxY(); i += 10) {
c.fillText(i, xPadding - 10, getYPixel(i));
}
c.strokeStyle = '#f00';
// Draw the line graph
c.beginPath();
c.moveTo(getXPixel(0), getYPixel(equation(0)));
for (var i = 1; i < data.values.length; i++) {
c.lineTo(getXPixel(i), getYPixel(equation(i)));
}
c.stroke();
// Draw the dots
c.fillStyle = '#333';
for (var i = 0; i < data.values.length; i++) {
c.beginPath();
c.arc(getXPixel(i), getYPixel(equation(i)), 4, 0, Math.PI * 2, true);
c.fill();
}
}
var codeText = "";
function parseEquation(input){
// Important that white spaces are removed first
input = input.replace(/\s+/g,""); // remove whitespaces
input = input.replace(/([\-\+])([xy])/g,"$11$2"); // convert -x -y or +x +y to -1x -1y or +1x +1y
// just to make the logic below a little simpler
var newTerm = () => {term = { val : null, scalar : 1, left : left, };} // create a new term
var pushTerm = () => {terms.push(term); term = null;} // push term and null current
// regExp [xy=] gets "x","y", or "="" or [\-\+]??[0-9\.]+ gets +- number with decimal
var reg =/[xy=]|[\-\+]??[0-9\.eE]+/g; // regExp to split the input string into parts
var parts = input.match(reg); // get all the parts of the equation
var terms = []; // an array of all terms parsed
var term = null; // Numbers as constants and variables with scalars are terms
var left = true; // which side of equation a term is
parts.forEach(p=>{
if (p === "x" || p === "y") {
if (term !== null && term.val !== null) { // is the variable defined
pushTerm(); // yes so push to the stack and null
}
if (term === null) { newTerm(); } // do we need a new term?
term.val = p;
} else if( p === "="){ // is it the equals sign
if (!left) { throw new SyntaxError("Unxpected `=` in equation."); }
if (term === null) { throw new SyntaxError("No left hand side of equation."); }// make sure that there is a left side
terms.push(term); // push the last left side term onto the stack
term = null;
left = false; // everything on the right from here on in
} else { // all that is left are numbers (we hope)
if (isNaN(p)){ throw new SyntaxError("Unknown value '"+p+"' in equation"); }//check that there is a number
if (term !== null && (p[0] === "+" || p[0] === "-")){ // check if number is a new term
pushTerm(); // yes so push to the stack and null
}
if(term === null){ newTerm(); } // do we need a new term?
term.scalar *= Number(p); // set the scalar to the new value
}
});
if(term !== null){// there may or may not be a term left to push to the stack
pushTerm();
}
// now simplify the equation getting the scalar for left and right sides . x on left y on right
var scalarX = 0;
var scalarY = 0
var valC = 0; // any constants
terms.forEach(t => {
t.scalar *= !t.left ? -1 : 1; // everything on right is negative
if (t.val === "y") {
scalarY += -t.scalar; // reverse sign
} else if (t.val === "x") {
scalarX += t.scalar;
} else {
valC += t.scalar;
}
})
// now build the code string for the equation to solve for x and return y
var code = "return (" + scalarX + " * x + (" + valC + ")) / "+scalarY +";\n";
codeText = code;
var equation = new Function("x",code); // create the function
return equation;
}
function parseAndPlot(){
var input = eqInput.value;
try{
var equation = parseEquation(input);
plot(equation);
error.textContent ="Plot of "+input+ " as 'function(x){ "+codeText+"}'";
}catch(e){
error.textContent = "Error parsing equation. " + e.message;
}
}
var button = document.getElementById("plot");
var eqInput = document.getElementById("equation-text");
var error = document.getElementById("status");
button.addEventListener("click",parseAndPlot);
parseAndPlot();<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<canvas id="graph" width="200" height="150"></canvas> <br>
Enter a linear equation : <input id="equation-text" value="x2 + 5y = 250" type="text"></input><input id="plot" value="plot" type=button></input><div id="status"></div>关于javascript - 如何使用js绘制基于方程的图形，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39052890/