如何从java上的txt文件中读取和获取方程中变量的系数

Question

如何从txt文件中读取方程并获得这些方程系数，例如3.2x-5.6y=10在txt文件中，我需要3.2、-5.6和10来制作图形gui程序。

我试过bufferedreader，但无法得到系数。

BufferedReader reader = null;
reader = new BufferedReader(new FileReader("input.txt"));
String line = reader.readLine();

Answer 1

这是解决方案。

import java.util.ArrayList;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {
    /*
     * Constructor just for test
     */
    Test()
    {
        Equation equation = getMembers("-3.2x - 5.6y + 34 = -10");
        System.out.print("Left members: ");
        for(EquationMember m : equation.leftMembers)
            System.out.print(m + " ");
        System.out.print("nRight members: ");
        for(EquationMember m : equation.rightMembers)
            System.out.print(m + " ");
    }
    /*
     * Main method
     */
    public static void main(String args[])
    {
        new Test();
    }
    /*
     * Transforms a strings representation of an equation into an Equation class
     */
    Equation getMembers(String equationString)
    {
        Equation equation = new Equation();
        String eq = equationString.replace(" ", "");
        char[] charArray = eq.toCharArray();
        boolean isLeft = true;
        String lastMember = "";
        for(int i = 0; i < charArray.length; i++)
        {   
            if(charArray[i] == '-' || charArray[i] == '+' || charArray[i] == '=')
            {
                if(lastMember.length() > 0)
                {
                    EquationMember m = new EquationMember();
                    m.constant = Float.parseFloat(match("([-+]?[\d\.]+)[A-Za-z]?", lastMember, 1));
                    m.variable = match("[\d\.]+([A-Za-z]?)", lastMember, 1);
                    if(isLeft)
                        equation.leftMembers.add(m);
                    else
                        equation.rightMembers.add(m);
                }
                lastMember = charArray[i] == '=' ? "" : String.valueOf(charArray[i]);
            }
            else
            {
                lastMember += charArray[i];
            }
            if(charArray[i] == '=')
            {
                isLeft = false;
            }
        }
        EquationMember m = new EquationMember();
        m.constant = Float.parseFloat(match("([-+]?[\d\.]+)[A-Za-z]?", lastMember, 1));
        m.variable = match("[\d\.]+([A-Za-z]?)", lastMember, 1);
        equation.rightMembers.add(m);
        return equation;
    }
    /*
     * Performs a regex match
     */
    String match(String regexp, String text, int group)
    {
        Pattern p1 = Pattern.compile(regexp);
        Matcher m1 = p1.matcher(text);
        if(m1.find())
            return m1.group(group);
        else
            return null;
    }
    /*
     * Class that represents a single member inside an equation
     */
    class EquationMember
    {
        float constant;
        String variable;
        public String toString()
        {
            return constant + "" + variable;
        }
    }
    /*
     * Class that represents an entire equation
     */
    class Equation
    {
        ArrayList<EquationMember> leftMembers;
        ArrayList<EquationMember> rightMembers;
        Equation()
        {
            leftMembers = new ArrayList<EquationMember>();
            rightMembers = new ArrayList<EquationMember>();
        }
    }
}

Answer 2

使用扫描仪对象。您可以使用它来使用RegExp或任何其他您想要的工具解析一行。

Answer 3

您提到了split，那么为什么不将字符串拆分为任何非数字的字符串呢？

final String s = "3.2x-5.6y=10";
String[] result = s.split("([^\d\.\+\-])+");
System.out.println(Arrays.toString(result));
// [3.2, -5.6, 10]

该正则表达式不适用于一般数字情况（考虑123456或无效输入+24-.2），但可以根据需要进行修复。

这并不能使变量与系数匹配，但您声明只需要值。