Léo*_* 준영 3 matlab computation
我想在计算中使用这个有理数,而不会丢失Matlab中图片的准确性:
f = 359.0 + 16241/16250.0
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我认为存储,例如f = uint64(359.0 + 16241/16250.0)失去准确性,在Matlab中被视为360.
我认为处理这个问题的最好方法是永远不要存储价值,而是存储它的因素
% f = a + b/c
a = 359
b = 16241
c = 16250
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然后通过变量a,b和c进行计算,并将结果作为图片给出.
这是保持准确性的好方法吗?
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)));
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然后
>> x = sqrt(pi);
>> nd = int64([5849991 16250]);
>> useInteger(x, nd, @plus)
ans =
361.7719
>> useInteger(x, nd, @times)
ans =
638.0824
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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
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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
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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
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加减:
>> r1+r2
ans =
Rational with properties:
n: 5
d: 4
>> r1-r2
ans =
Rational with properties:
n: -1
d: 4
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乘以除:
>> r1*r2
ans =
Rational with properties:
n: 3
d: 8
>> r1/r2
ans =
Rational with properties:
n: 2
d: 3
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获取小数值:
>> r12 = r1/r2; % 2/3 ((1/2)/(3/4))
>> f = r12.dec
f =
0.6667
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