Question

我有一个带有两个整数输入的函数，如下所示：

function f = func(n, m)
    a = 2;
    b = 1;
    f = sqrt((n/a)^2 + (m/b)^2);
end

m和n是整数且大于或等于零。 f的前几个值和它们发生的输入如下所示：

n ----- m ----- f

0 ----- 0 ----- 0

1 ----- 0 ----- 0.5

2 ----- 0 ----- 1

0 ----- 1 ----- 1

1 ----- 1 ----- 1.118

等等。我想得到f的前N个值和它们各自的n和m。在matlab中有一种简单的方法吗？

Answer 1

meshgrid和arrayfun可用于为输入范围生成输出数组

代码

nValues = 0:2 
mValues = 0:3

[ii,jj] = meshgrid(mValues,nValues)
output = arrayfun(@func,ii,jj)

可以修改两个值向量以获取所需值的范围

输出

output =

         0    0.5000    1.0000    1.5000
    1.0000    1.1180    1.4142    1.8028
    2.0000    2.0616    2.2361    2.5000

要给出问题中矩阵的结果，可以使用以下内容（感谢@Divakar）

[jj(:),ii(:),arrayfun(@func,jj(:),ii(:))]

ans =

         0         0         0
    1.0000         0    0.5000
    2.0000         0    1.0000
         0    1.0000    1.0000
    1.0000    1.0000    1.1180
    2.0000    1.0000    1.4142
         0    2.0000    2.0000
    1.0000    2.0000    2.0616
    2.0000    2.0000    2.2361
         0    3.0000    3.0000
    1.0000    3.0000    3.0414
    2.0000    3.0000    3.1623

Answer 2

<强>代码

%// Parameters
N = 5
a = 2;
b = 1;

%// Extents of n and m would be from 0 to N-1 to account for all possible
%// minimum values of f results resulting from their use 
len1 = N-1

%// Create n and m for maximum possible combinations scenario, but save
%// them as n1 and m1 for now, as the final ones would be chopped versions
%// of them.
[n1,m1] = ndgrid(0:len1,0:len1)

%// Get corresponding f values, but store as f1, for the same chopping reason
f1 = sqrt((n1(:)./a).^2 + (m1(:)./b).^2);

%// Sort f1 so that the smallest N values from it could be choosen and also
%// get the selected row indices based on the sorting as row1
[f1,row1] = sort(f1)

%// Choose n and m based on the sorted indices and also chop off at N.
%// Use these n and m values to finally get f
n = n1(row1(1:N))
m = m1(row1(1:N))
f = f1(1:N)

<强>输出

如果N = 5，你会得到 -

如果N = 9，你会得到 -

Answer 3

像这样的东西（可能效率很低）？

N = 100 % stop 
i = 0
n = 0
m = 0
nout = [n]
mout = [m]
fout = [f(n,m)]
while i ~= N
    a = f(n+1,m)
    b = f(n,m+1)
    if (a > b)
        m = m + 1
        nout = [nout n]
        mout = [mout m]
        fout = [fout b]
    else 
        n = n + 1
        nout = [nout n]
        mout = [mout m]
        fout = [fout a]
    end if
    i = i + 1
end while

具有两个输入的函数的前N个值

3 个答案:

代码

输出