我试图写一个Java例程来计算数学表达式从字符串值，如:
"5 + 3" "10-4 * 5" "(1 + 10) * 3"
我想避免很多如果-then-else语句。 我该怎么做呢?
当前回答
这实际上是对@Boann给出的答案的补充。它有一个轻微的错误，导致“-2 ^ 2”给出一个错误的结果-4.0。这里的问题是求幂的点。只需将取幂运算移到parseTerm()的块中，就可以了。看看下面，这是@Boann的回答略有修改。修改意见见评论。
public static double eval(final String str) {
return new Object() {
int pos = -1, ch;
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : -1;
}
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
double parse() {
nextChar();
double x = parseExpression();
if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
return x;
}
// Grammar:
// expression = term | expression `+` term | expression `-` term
// term = factor | term `*` factor | term `/` factor
// factor = `+` factor | `-` factor | `(` expression `)`
// | number | functionName factor | factor `^` factor
double parseExpression() {
double x = parseTerm();
for (;;) {
if (eat('+')) x += parseTerm(); // addition
else if (eat('-')) x -= parseTerm(); // subtraction
else return x;
}
}
double parseTerm() {
double x = parseFactor();
for (;;) {
if (eat('*')) x *= parseFactor(); // multiplication
else if (eat('/')) x /= parseFactor(); // division
else if (eat('^')) x = Math.pow(x, parseFactor()); //exponentiation -> Moved in to here. So the problem is fixed
else return x;
}
}
double parseFactor() {
if (eat('+')) return parseFactor(); // unary plus
if (eat('-')) return -parseFactor(); // unary minus
double x;
int startPos = this.pos;
if (eat('(')) { // parentheses
x = parseExpression();
eat(')');
} else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
while ((ch >= '0' && ch <= '9') || ch == '.') nextChar();
x = Double.parseDouble(str.substring(startPos, this.pos));
} else if (ch >= 'a' && ch <= 'z') { // functions
while (ch >= 'a' && ch <= 'z') nextChar();
String func = str.substring(startPos, this.pos);
x = parseFactor();
if (func.equals("sqrt")) x = Math.sqrt(x);
else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
else throw new RuntimeException("Unknown function: " + func);
} else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
//if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation -> This is causing a bit of problem
return x;
}
}.parse();
}
其他回答
对于JDK1.6，您可以使用内置的Javascript引擎。
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
public class Test {
public static void main(String[] args) throws ScriptException {
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
String foo = "40+2";
System.out.println(engine.eval(foo));
}
}
像RHINO或NASHORN这样的外部库可以用来运行javascript。javascript可以计算简单的公式，而不需要对字符串进行分割。如果代码写得好，也不会对性能造成影响。 下面是一个使用RHINO -的示例
public class RhinoApp {
private String simpleAdd = "(12+13+2-2)*2+(12+13+2-2)*2";
public void runJavaScript() {
Context jsCx = Context.enter();
Context.getCurrentContext().setOptimizationLevel(-1);
ScriptableObject scope = jsCx.initStandardObjects();
Object result = jsCx.evaluateString(scope, simpleAdd , "formula", 0, null);
Context.exit();
System.out.println(result);
}
看来应该由JEP来做这项工作
package ExpressionCalculator.expressioncalculator;
import java.text.DecimalFormat;
import java.util.Scanner;
public class ExpressionCalculator {
private static String addSpaces(String exp){
//Add space padding to operands.
//https://regex101.com/r/sJ9gM7/73
exp = exp.replaceAll("(?<=[0-9()])[\\/]", " / ");
exp = exp.replaceAll("(?<=[0-9()])[\\^]", " ^ ");
exp = exp.replaceAll("(?<=[0-9()])[\\*]", " * ");
exp = exp.replaceAll("(?<=[0-9()])[+]", " + ");
exp = exp.replaceAll("(?<=[0-9()])[-]", " - ");
//Keep replacing double spaces with single spaces until your string is properly formatted
/*while(exp.indexOf(" ") != -1){
exp = exp.replace(" ", " ");
}*/
exp = exp.replaceAll(" {2,}", " ");
return exp;
}
public static Double evaluate(String expr){
DecimalFormat df = new DecimalFormat("#.####");
//Format the expression properly before performing operations
String expression = addSpaces(expr);
try {
//We will evaluate using rule BDMAS, i.e. brackets, division, power, multiplication, addition and
//subtraction will be processed in following order
int indexClose = expression.indexOf(")");
int indexOpen = -1;
if (indexClose != -1) {
String substring = expression.substring(0, indexClose);
indexOpen = substring.lastIndexOf("(");
substring = substring.substring(indexOpen + 1).trim();
if(indexOpen != -1 && indexClose != -1) {
Double result = evaluate(substring);
expression = expression.substring(0, indexOpen).trim() + " " + result + " " + expression.substring(indexClose + 1).trim();
return evaluate(expression.trim());
}
}
String operation = "";
if(expression.indexOf(" / ") != -1){
operation = "/";
}else if(expression.indexOf(" ^ ") != -1){
operation = "^";
} else if(expression.indexOf(" * ") != -1){
operation = "*";
} else if(expression.indexOf(" + ") != -1){
operation = "+";
} else if(expression.indexOf(" - ") != -1){ //Avoid negative numbers
operation = "-";
} else{
return Double.parseDouble(expression);
}
int index = expression.indexOf(operation);
if(index != -1){
indexOpen = expression.lastIndexOf(" ", index - 2);
indexOpen = (indexOpen == -1)?0:indexOpen;
indexClose = expression.indexOf(" ", index + 2);
indexClose = (indexClose == -1)?expression.length():indexClose;
if(indexOpen != -1 && indexClose != -1) {
Double lhs = Double.parseDouble(expression.substring(indexOpen, index));
Double rhs = Double.parseDouble(expression.substring(index + 2, indexClose));
Double result = null;
switch (operation){
case "/":
//Prevent divide by 0 exception.
if(rhs == 0){
return null;
}
result = lhs / rhs;
break;
case "^":
result = Math.pow(lhs, rhs);
break;
case "*":
result = lhs * rhs;
break;
case "-":
result = lhs - rhs;
break;
case "+":
result = lhs + rhs;
break;
default:
break;
}
if(indexClose == expression.length()){
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose);
}else{
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose + 1);
}
return Double.valueOf(df.format(evaluate(expression.trim())));
}
}
}catch(Exception exp){
exp.printStackTrace();
}
return 0.0;
}
public static void main(String args[]){
Scanner scanner = new Scanner(System.in);
System.out.print("Enter an Mathematical Expression to Evaluate: ");
String input = scanner.nextLine();
System.out.println(evaluate(input));
}
}
在我的大学项目中，我正在寻找一个既支持基本公式又支持更复杂方程(特别是迭代运算符)的解析器/求值器。我发现了一个非常好的JAVA和。net开源库，叫做mXparser。我将给出几个例子，让大家对语法有一些感觉，如需进一步指导，请访问项目网站(特别是教程部分)。
https://mathparser.org/
https://mathparser.org/mxparser-tutorial/
https://mathparser.org/api/
举几个例子
一个简单的开始
Expression e = new Expression("( 2 + 3/4 + sin(pi) )/2");
double v = e.calculate()
2 -用户定义的参数和常量
Argument x = new Argument("x = 10");
Constant a = new Constant("a = pi^2");
Expression e = new Expression("cos(a*x)", x, a);
double v = e.calculate()
3 -用户定义的函数
Function f = new Function("f(x, y, z) = sin(x) + cos(y*z)");
Expression e = new Expression("f(3,2,5)", f);
double v = e.calculate()
4 -迭代
Expression e = new Expression("sum( i, 1, 100, sin(i) )");
double v = e.calculate()
最近发现的-如果你想尝试语法(并查看高级用例)，你可以下载由mXparser支持的标量计算器应用程序。
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