在没有嵌套循环的情况下迭代矩阵

Question

我想在不使用嵌套循环的情况下迭代大小为 n 的方阵主对角线上方的条目。

例如，如果矩阵 M 的大小为 n=3，则对角线上方有 choose(3,2)=3 个条目。其中choose是二项式系数。

for(i=1 to choose(n,2))
  row = getRow(i)
  col = getCol(i)
  M[row,col] = some value

我还没有想出一个基于索引 i 获取行和列的公式。

例如：

对于大小为 3 且索引从 1 开始的矩阵，

i=1 对应 row = 1 和 col = 2

i=2 对应 row = 1 和 col = 3

i=3 对应 row = 2 和 col = 3

Answer 1

您可以使用MATLAB的triu命令如下：

n=5; %user input

mat=magic(n);
nRows=size(mat,1);
nCols=size(mat,2);  

%extract the elements above the main diagonal  
u=triu(mat).*(~eye(n));
uT=u.'; %transpose to get the indices in the order you mentioned

%arrange it in a vector
u_ind=uT(uT~=0);

u_ind 将以所需格式包含上三角形上方的元素，即 u_ind(3) 将包含 row=1 和 col=4 处的元素。

要获取这些行列索引，可以按如下方式获取：

%You can easily get this if you make simple observations. For a matrix of size 5x5
%the number of total elements above main diagonal are: (4+3+2+1)
%i.e. sum of first n-1 elements -> hence the formula below
totalIndices=0.5*n*(n-1);

%number of elements per row you get
indicesPerRow=n-1:-1:1

%I observed that there is some relation between its index and its (row,col) subscript.
%If the index is 5 in a 5x5 matrix, then you can imagine that it is the first non-zero 
%element in the second row -> because first row has 4 elements. If the index was 8, 
%similarly, it would have been first non-zero element in the third row in the upper 
%triangular matrix we have formed. This is what I have translated into code below.

ind1=cumsum(indicesPerRow);
ind2=ind1;

%Enter the number whose (row,col) index you want to find.
myInd=9;
ind2(ind1<myInd)=[];
pos=find(ind1<myInd,1,'last');
ind2=ind2(1);
ind3=rem(ind2,myInd);
detRow=pos+1;
detCol=nCols-ind3;

fprintf('Index %d corresponds to (row,col)=(%d,%d)\n',myInd,detRow,detCol);

Answer 2

所以你想要一个公式！行 ！让我们来做一些简单的数学运算：

第 r 行的条目数是r*(r+1)/2。因此row 是满足row*(row+1)/2 <= i 的最大整数。所以row是方程正解的整数部分

row*(row+1)/2 = i

它改写为

row^2 + row - 2*i = 0

这是一个二次方程，因此您可以使用平方根来求解它。正解是(sqrt(1+8*i) - 1)/2。

所以你有：

row(i) = floor((sqrt(1+8*i) - 1)/2)
col(i) = i - row*(row+1)/2

Python 演示：

def rc(i):
    r = floor((-1 + sqrt(1+8*i))/2)
    return (r, i-r*(r+1)/2)

测试：

print [rc(i) for i in range(20)]
[(0, 0), (1, 0), (1, 1), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (3, 3), 
 (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4)]

正确的演示文稿

(0, 0), 
(1, 0), (1, 1), 
(2, 0), (2, 1), (2, 2), 
(3, 0), (3, 1), (3, 2), (3, 3), 
(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), 
...

注意：我的所有索引都从 0 开始。如果您想坚持通常的编号，则必须将 i、r 和 c 移动一位。

Answer 3

你可以试试：

n = 3;

for(int i = 1; i <= n; i++) 
{
    for(int j=i+1; j<=n; j++) 
    {
        row = i;
        col = j;
    }
}

结果：
1,2
1,3
2,3