在Matlab中处理有理数而不失计算精度？

正如您所建议的那样,如果您绝对不希望在存储有理数时丢失准确性,则最佳解决方案可能是以整数组件的形式存储数字.

f = a + b/c您可以将代表减少为两个组件,而不是您的三个组件()f = n/d.因此,每个有理数将被定义(并存储)为双分量整数向量[n d].例如,示例中的数字f对应于n=5849991和d=16250.

为了简化处理以这种方式存储的有理数,您可以定义一个辅助函数,该函数从[n d]表示转换n/d为应用所需操作之前:

useInteger = @(x, nd, fun) fun(x,double(nd(1))/double(nd(2)));

然后

>> x = sqrt(pi);
>> nd = int64([5849991 16250]);
>> useInteger(x, nd, @plus)
ans =
  361.7719
>> useInteger(x, nd, @times)
ans =
  638.0824

如果要在计算中实现任意高精度,则应考虑将变精度arithmetic(vpa)与字符串参数一起使用.使用这种方法,您可以指定所需的位数:

>> vpa('sqrt(pi)*5849991/16250', 50)
ans =
638.08240465923757600307902117159072301901656248436

也许,创建一个Rational类,并定义所需的操作(plus,minus,times等).从这样的事情开始:

Rational.m

classdef Rational
    properties
        n;
        d;
    end
    methods
        function obj = Rational(n,d)
            GCD = gcd(n,d);
            obj.n = n./GCD;
            obj.d = d./GCD;
        end

        function d = dec(obj)
            d = double(obj.n)/double(obj.d);
        end

        % X .* Y
        function R = times(X,Y)
            chkxy(X,Y);
            if isnumeric(X),
                N = X .* Y.n; D = Y.d;
            elseif isnumeric(Y),
                N = X.n .* Y; D = X.d;
            else
                N = X.n .* Y.n; D = X.d .* Y.d;
            end
            R = Rational(N,D);
        end

        % X * Y
        function R = mtimes(X,Y)
            R = times(X,Y);
        end

        % X ./ Y
        function R = rdivide(X,Y)
            if isnumeric(Y),
                y = Rational(1,Y);
            else
                y = Rational(Y.d,Y.n);
            end
            R = times(X,y);
        end

        % X / Y
        function R = mrdivide(X,Y)
            R = rdivide(X,Y);
        end

        % X + Y
        function R = plus(X,Y)
            chkxy(X,Y);
            if isnumeric(X),
                N = X.*Y.d + Y.n; D = Y.d;
            elseif isnumeric(Y),
                N = Y.*X.d + X.n; D = X.d;
            else
                D = lcm(X.d,Y.d);
                N = sum([X.n Y.n].*(D./[X.d Y.d]));
            end
            R = Rational(N,D);
        end

        % X - Y
        function R = minus(X,Y)
            R = plus(X,-Y);
        end

        % -X
        function R = uminus(X)
            R = Rational(-X.n,X.d);
        end

        function chkxy(X,Y)
            if (~isa(X, 'Rational') && ~isnumeric(X)) || ...
                    (~isa(Y, 'Rational') && ~isnumeric(Y)),
                error('X and Y must be Rational or numeric.');
            end
        end
    end
end

例子

构造对象:

>> clear all % reset class definition
>> r1 = Rational(int64(1),int64(2))
r1 = 
  Rational with properties:

    n: 1
    d: 2
>> r2 = Rational(int64(3),int64(4))
r2 = 
  Rational with properties:

    n: 3
    d: 4

加减:

>> r1+r2
ans = 
  Rational with properties:

    n: 5
    d: 4
>> r1-r2
ans = 
  Rational with properties:

    n: -1
    d: 4

乘以除:

>> r1*r2
ans = 
  Rational with properties:

    n: 3
    d: 8
>> r1/r2
ans = 
  Rational with properties:

    n: 2
    d: 3

获取小数值:

>> r12 = r1/r2; % 2/3 ((1/2)/(3/4))
>> f = r12.dec
f =
    0.6667