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[英]Evaluating an math expression in Postfix (RPN) Java. No stack, only strings
[英]Evaluating a math expression with variables. (java 8)
我想对这个问题的答案提供额外帮助, 评估以字符串形式给出的数学表达式 。 用户@Boann用一个非常有趣的算法回答了这个问题,他也指出可以改变它来接受变量。 我已经设法改变它并让它工作, 但不知道他如何分离编译和评估 。 这是我的代码:
import java.util.HashMap;
import java.util.Map;
public class EvaluateExpressionWithVariabels {
@FunctionalInterface
interface Expression {
double eval();
}
public static void main(String[] args){
Map<String,Double> variables = new HashMap<>();
for (double x = 100; x <= +120; x++) {
variables.put("x", x);
System.out.println(x + " => " + eval("x+(sqrt(x))",variables).eval());
}
}
public static Expression eval(final String str,Map<String,Double> variables) {
return new Object() {
int pos = -1, ch;
//if check pos+1 is smaller than string length ch is char at new pos
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : -1;
}
//skips 'spaces' and if current char is what was searched, if true move to next char return true
//else return false
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
Expression parse() {
nextChar();
Expression 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
Expression parseExpression() {
Expression x = parseTerm();
for (;;) {
if (eat('+')) { // addition
Expression a = x, b = parseTerm();
x = (() -> a.eval() + b.eval());
} else if (eat('-')) { // subtraction
Expression a = x, b = parseTerm();
x = (() -> a.eval() - b.eval());
} else {
return x;
}
}
}
Expression parseTerm() {
Expression x = parseFactor();
for (;;) {
if (eat('*')){
Expression a = x, b = parseFactor(); // multiplication
x = (() -> a.eval() * b.eval());
}
else if(eat('/')){
Expression a = x, b = parseFactor(); // division
x = (() -> a.eval() / b.eval());
}
else return x;
}
}
Expression parseFactor() {
if (eat('+')) return parseFactor(); // unary plus
if (eat('-')){
Expression b = parseFactor(); // unary minus
return (() -> -1 * b.eval());
}
Expression 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();
}
double xx = Double.parseDouble(str.substring(startPos, this.pos));
x = () -> xx;
} else if (ch >= 'a' && ch <= 'z') { // functions
while (ch >= 'a' && ch <= 'z') nextChar();
String func = str.substring(startPos, this.pos);
if ( variables.containsKey(func)){
x = () -> variables.get(func);
}else{
double xx = parseFactor().eval();
if (func.equals("sqrt")) x = () -> Math.sqrt(xx);
else if (func.equals("sin")) x = () -> Math.sin(Math.toRadians(xx));
else if (func.equals("cos")) x = () -> Math.cos(Math.toRadians(xx));
else if (func.equals("tan")) x = () -> Math.tan(Math.toRadians(xx));
else throw new RuntimeException("Unknown function: " + func);
}
} else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
if (eat('^')){
x = () -> {
double d = parseFactor().eval();
return Math.pow(d,d); // exponentiation
};
}
return x;
}
}.parse();
}
}
如果你看看他的答案他的主要
public static void main(String[] args) {
Map<String,Double> variables = new HashMap<>();
Expression exp = parse("x^2 - x + 2", variables);
for (double x = -20; x <= +20; x++) {
variables.put("x", x);
System.out.println(x + " => " + exp.eval());
}
}
他在这一行调用函数parse
Expression exp = parse("x^2 - x + 2", variables);
编译表达式一次并使用for来使用唯一的x值多次计算它。 parse
函数引用了什么。
ps:我对用户问题发表了评论,没有回复。
我认为持有地图参考的Expression
的实际代码没有在答案中公布。
为了使该示例代码起作用,表达式必须具有某种内存。 实际上代码操纵地图,并且当表达式保存对它的引用时,对表达式的eval
-method的每次后续调用都将采用调整后的值。
因此,为了使代码工作,您首先需要一个表达式实现,它包含一个map引用。 但要注意这种表达可能带来的任何副作用。
因此,对于该特定示例,代码必须类似于以下内容(没有时间完全查看它是否可行,但是您将得到这个想法:重要的是,表达式包含一个参考,该参考由外):
static Expression parse(String expression, Map<String, String> variables) {
return new PseudoCompiledExpression(expression, variables);
}
static class PseudoCompiledExpression implements Expression {
Map<String, String> variables;
Expression wrappedExpression;
PseudoCompiledExpression(String expression, Map<String, String> variables) {
this.variables = variables;
wrappedExpression = eval(expression, variables);
}
public double eval() {
// the state of the used map is altered from outside...
return wrappedException.eval();
}
对困惑感到抱歉。 我所提到的“ parse
”函数只是现有的eval
函数,但由于它返回了一个Expression
对象而重命名。
所以你有:
public static Expression parse(String str, Map<String,Double> variables) { ... }
并通过以下方式调用它:
Map<String,Double> variables = new HashMap<>();
Expression exp = parse("x+(sqrt(x))", variables);
for (double x = 100; x <= +120; x++) {
variables.put("x", x);
System.out.println(x + " => " + exp.eval());
}
另一件事:在解析时需要知道名称是指变量还是函数,以便知道它是否需要参数,但是在解析期间你不能在变量映射上调用containsKey
,因为在调用exp.eval()
之前,变量可能不会出现在映射中! 一种解决方案是将函数放在映射中,因此您可以在其上调用containsKey
:
} else if (ch >= 'a' && ch <= 'z') { // functions and variables
while (ch >= 'a' && ch <= 'z') nextChar();
String name = str.substring(startPos, this.pos);
if (functions.containsKey(name)) {
DoubleUnaryOperator func = functions.get(name);
Expression arg = parseFactor();
x = () -> func.applyAsDouble(arg.eval());
} else {
x = () -> variables.get(name);
}
} else {
然后在类级别的某处,初始化functions
映射:
private static final Map<String,DoubleUnaryOperator> functions = new HashMap<>();
static {
functions.put("sqrt", x -> Math.sqrt(x));
functions.put("sin", x -> Math.sin(Math.toRadians(x)));
functions.put("cos", x -> Math.cos(Math.toRadians(x)));
functions.put("tan", x -> Math.tan(Math.toRadians(x)));
}
(将functions
map定义为解析器中的局部变量也是可以的,但是在每次解析时会增加一些开销。)
我发现上面的语法不能用于取幂。 这个工作:
public class BcInterpreter {
static final String BC_SPLITTER = "[\\^\\(\\/\\*\\-\\+\\)]";
static Map<String,Double> variables = new HashMap<>();
private static final Map<String,DoubleUnaryOperator> functions = new HashMap<>();
static {
functions.put("sqrt", x -> Math.sqrt(x));
functions.put("sin", x -> Math.sin(Math.toRadians(x)));
functions.put("cos", x -> Math.cos(Math.toRadians(x)));
functions.put("tan", x -> Math.tan(Math.toRadians(x)));
functions.put("round", x -> Math.round(x));
functions.put("abs", x -> Math.abs(x));
functions.put("ceil", x -> Math.ceil(x));
functions.put("floor", x -> Math.floor(x));
functions.put("log", x -> Math.log(x));
functions.put("exp", x -> Math.exp(x));
// TODO: add more unary functions here.
}
/**
* Parse the expression into a lambda, and evaluate with the variables set from fields
* in the current row. The expression only needs to be evaluated one time.
* @param recordMap
* @param fd
* @param script
* @return
*/
static String materialize(Map<String, String> recordMap, FieldDesc fd, String script){
// parse the expression one time and save the lambda in the field's metadata
if (fd.get("exp") == null) {
fd.put("exp", parse(script, variables));
}
// set the variables to be used with the expression, once per row
String[] tokens = script.split(BC_SPLITTER);
for(String key : tokens) {
if (key != null) {
String val = recordMap.get(key.trim());
if (val != null)
variables.put(key.trim(), Double.parseDouble(val));
}
}
// evaluate the expression with current row's variables
return String.valueOf(((Expression)(fd.get("exp"))).eval());
}
@FunctionalInterface
interface Expression {
double eval();
}
static Map<String,Double> getVariables(){
return variables;
}
public static Expression parse(final String str,Map<String,Double> variables) {
return new Object() {
int pos = -1, ch;
//if check pos+1 is smaller than string length ch is char at new pos
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : -1;
}
//skips 'spaces' and if current char is what was searched, if true move to next char return true
//else return false
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
Expression parse() {
nextChar();
Expression 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 = base | base '^' base
// base = '-' base | '+' base | number | identifier | function factor | '(' expression ')'
Expression parseExpression() {
Expression x = parseTerm();
for (;;) {
if (eat('+')) { // addition
Expression a = x, b = parseTerm();
x = (() -> a.eval() + b.eval());
} else if (eat('-')) { // subtraction
Expression a = x, b = parseTerm();
x = (() -> a.eval() - b.eval());
} else {
return x;
}
}
}
Expression parseTerm() {
Expression x = parseFactor();
for (;;) {
if (eat('*')){
Expression a = x, b = parseFactor(); // multiplication
x = (() -> a.eval() * b.eval());
} else if(eat('/')){
Expression a = x, b = parseFactor(); // division
x = (() -> a.eval() / b.eval());
} else {
return x;
}
}
}
Expression parseFactor(){
Expression x = parseBase();
for(;;){
if (eat('^')){
Expression a = x, b = parseBase();
x = (()->Math.pow(a.eval(),b.eval()));
}else{
return x;
}
}
}
Expression parseBase(){
int startPos = this.pos;
Expression x;
if (eat('-')){
Expression b = parseBase();
x = (()-> (-1)*b.eval());
return x;
}else if (eat('+')){
x = parseBase();
return x;
}
if (eat('(')) { // parentheses
x = parseExpression();
eat(')');
return x;
} else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
while ((ch >= '0' && ch <= '9') || ch == '.'){
nextChar();
}
double xx = Double.parseDouble(str.substring(startPos, this.pos));
x = () -> xx;
return x;
} else if (ch >= 'a' && ch <= 'z' || ch >= 'A' && ch <= 'Z') { // functions and variables
while (ch >= 'a' && ch <= 'z' || ch >= 'A' && ch <= 'Z') nextChar();
String name = str.substring(startPos, this.pos);
if (functions.containsKey(name)) {
DoubleUnaryOperator func = functions.get(name);
Expression arg = parseFactor();
x = () -> func.applyAsDouble(arg.eval());
} else {
x = () -> variables.get(name);
}
return x;
}else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
}
}.parse();
}
}
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