我需要在矩阵中获取所有对角线的索引。矩阵可以是非正方形。
diag函数给出了值，但我需要coords。例如，[1 2 3; 4 5 6; 7 8 9]，我希望[1 1; 2 2; 3;3]加[2 6]和3，因为它们是矩阵的上对角线，下面也是[4 8] }和7。所以索引的完整列表是：[1 1; 2 2; 3 3], [1 2; 2 3], [1 3], [2 1], [3 2], [3 1]。
我也需要在另一个对角线方向......
它需要适用于任何形状的矩阵，即不仅仅是方形矩阵，包括矢量（即1xn和nx1） - 但我想我可以测试最后2个案例并单独处理。
编辑： 非方矩阵的一个例子是这个3x2矩阵：
1 2
3 4
5 6
在这种情况下，我需要的索引是：
[1 1; 2 2]
[2 1; 3 2]
[1 2]
[3 1]
然后对于以对方方式运行的对角线也一样......
我知道diag函数并且可以编写一些代码来迭代矩阵，但我不能看到如何轻松地将这些值转换为索引。
至于为什么我需要这个：这是我在MatLab做的在线课程。使用MatLab介绍编程。我是一位经验丰富的程序员，但我不熟悉MatLab，我想学习更多。
这个问题要求我们找到任何方向上任何n个数字的最大乘积，即行，列和对角线（双向）。事情是我们不返回产品，我们返回构成该产品的元素的索引。
我希望这能提供更多信息。
编辑2： 很抱歉没有回复你们，我一直忙于做其他事情。当我第一次发布这里时，我认为我错过了一些东西，这将是一些需要几行的东西。但它比那更棘手，所以你们发布的代码将花费我更多的时间来了解并理解等等。但我希望今天这样做并接受一个作为答案。感谢所有的回复。
答案 0 :(得分：0)
不要忘记Matlab理解线性索引到多维矩阵。对于OP的3*3（我们称之为a）矩阵的示例，表达式a(3)标识与a(3,1)相同的元素（Matlab使用列主要布局）。因此，对于k*k矩阵，表达式
[1, k+2, 2k+3, 3k+4, ...]
给出主对角线上元素的索引。有多种方法可以生成该向量。然后，ind2sub的明智应用将使2-d指数成为同一矩阵。
将这种方法概括为其他对角线和非方形矩阵应该不会太难。
答案 1 :(得分：0)
这是“西北到东南”对角线的解决方案。
A = [1,2,3,4,5;6,7,8,9,10;11,12,13,14,15];
rows = size (A,1); cols = size (A,2); padding = rows - 1;
padded = [ones(rows, padding), A, ones(rows, padding)];
% Create a Helper array containing column/row pairs and a product
Helper = zeros (cols + padding, rows * 2 + 1);
for i = 1 : cols + padding; % i moves along columns of the padded array
i0 = i - padding; % "actual" index w.r.t A, where leftmost index is 1-padding
c = 1; % a counter used for filling current "Helper" row appropriately
for j = 1 : rows; % j moves along rows
Helper(i, c) = i0; % store current "actual" column
Helper(i, c+1) = j; % store current row
c = c + 2; % move to next 2 slots in "Helper"
i0 = i0 + 1; % move to next "actual" column for processing
end
% Process product at appropriate indices
Helper(i, end) = prod (padded (sub2ind ( ...
size (padded), ...
Helper(i,2:2:end-1), ...
Helper(i,1:2:end-1)+padding ...
)));
end
% Isolate the correct indices for which the product is largest
[~, biggest] = max (Helper(:, end)); % column index of largest product in Helper
Pairs = reshape (Helper(biggest, 1:end-1), 2, []); % collect indices
Pairs = fliplr (Pairs.'); % express as rows/cols
% Remove pairs where column occurs in the "padded" zone
for i = 1 : size (Pairs, 1)
if Pairs(i, 1) < 1 || Pairs(i, 1) > cols; Pairs(i, :) = []; end
end
display(Pairs)
如果您了解这背后的逻辑，那么“NorthEast to SouthWest”对角线也应该是一块蛋糕。我会留下让你弄明白：）
答案 2 :(得分：0)
执行此操作的简单方法是构造一个与输入形状相同的数组，其中每个元素的值都是其索引。然后你可以得到每个对角线的值（索引）。
function diag_cell = alldiags(A)
[m,n] = size(A);
M = reshape(1:m*n, m, n); % construct array of indices
% there are (m-1) diagonals below the main (k parameter negative)
% and (n-1) diagonals above, plus 0 for the main diagonal
for d = 1-m:n-1
C1{d+m} = diag(M, d); % store in C1{1}..C1{m+n-1}
end
% to get the antidiagonals, rotate the index matrix `M` and repeat
% (remember to swap `m` and `n`):
M = rot90(M);
for d = 1-n:m-1
C2{d+n} = diag(M, d);
end
diag_cell = {C1{:}, C2{:}}; % concatenate the cell arrays and return
end
这是使用3x3样本矩阵的测试运行：
A =
1 2 3
4 5 6
7 8 9
>> alldiags(A)
ans =
{
[1,1] = 3
[1,2] =
2
6
[1,3] =
1
5
9
[1,4] =
4
8
[1,5] = 7
[1,6] = 1
[1,7] =
4
2
[1,8] =
7
5
3
[1,9] =
8
6
[1,10] = 9
}
如果你真的需要，你可以随时将所有索引转换为下标，但是如果你正在访问原始矩阵，那么使用索引可能更容易。这是一个将主对角线转换为下标的示例：
>> C{3}
ans =
1
5
9
>> [r,c] = ind2sub(size(A), C{3});
>> subs = [r c]
subs =
1 1
2 2
3 3
这是一个例子来说明为什么会这样。我们采用随机的6x3矩阵A：
A =
0.634825 0.560609 0.926716
0.049504 0.226152 0.748278
0.746754 0.493896 0.773868
0.993245 0.457522 0.412092
0.430003 0.695042 0.641917
0.935298 0.960166 0.022872
函数做的第一件事是获取输入矩阵A的大小，并创建一个相同大小的新矩阵，使每个元素的值等于其线性索引。线性索引从下标1的{{1}}开始，将列向下增加到[1,1]的{{1}}，然后转到m寻找下标[m,1]，在下标m+1的索引[1,2]处结束。
m*n此矩阵的输出[m,n]如下所示（重新格式化，因此更容易一次看到所有内容）：
>> [m,n] = size(A)
m = 6
n = 3
>> M = reshape(1:m*n, m, n)
M =
1 7 13
2 8 14
3 9 15
4 10 16
5 11 17
6 12 18
条目的前半部分是正常对角线（左上角到右下角），后半部分是反对角线（从右上角到左下角）。
举个例子，走对角线diag_cell。您可以查看矩阵diag_cell =
{
[1,1] = 6
[1,2] = 5 12
[1,3] = 4 11 18
[1,4] = 3 10 17
[1,5] = 2 9 16
[1,6] = 1 8 15
[1,7] = 7 14
[1,8] = 13
[1,9] = 1
[1,10] = 7 2
[1,11] = 13 8 3
[1,12] = 14 9 4
[1,13] = 15 10 5
[1,14] = 16 11 6
[1,15] = 17 12
[1,16] = 18
}
以查看这些索引引用的元素。
diag_cell{4} = [3; 10; 17]此对角线从M开始，到A =
0.634825 0.560609 0.926716
0.049504 0.226152 0.748278
*0.746754* 0.493896 0.773868
0.993245 *0.457522* 0.412092
0.430003 0.695042 *0.641917*
0.935298 0.960166 0.022872
结束。如果我们从[3,1]恢复值，我们会得到：
[5,3]答案 3 :(得分：0)
这个解决方案一直到你的最终任务 - 找到矩阵中任何向量的最大乘积的索引：
function out = maxVecProd(A)
% validate the input is a vector\matrix:
assert(ismatrix(A),'Input must be a vector or matrix')
[cmax,cind] = max(prod(A,1)); % find where the maximum prod in the columns
[rmax,rind] = max(prod(A,2)); % find where the maximum prod in the rows
if all(size(A))>1 % if its a matrix:
all_d = -(size(A,1)-1):(size(A,1)-1); % get all the valid diagonals
pd = zeros(2*numel(all_d),2); % the solution of all products
c = 1;
for d = all_d
pd(c,1) = prod(diag(A,d)); % get the diagonal product
pd(c,2) = prod(diag(fliplr(A),d)); % get the anti-diagonal product
c = c+1;
end
[dmax,dind] = max(pd(:)); % find where the maximum prod in the diagonals
% get the orientation of the maximum prod (column, row, diagonal)
[~,dirmax] = max([rmax,cmax,dmax]);
else
[~,dirmax] = max([rmax,cmax]);
end
switch dirmax
case 1
% create the indices for the maximum row:
out = [repelem(rind,size(A,2)).' (1:size(A,2)).'];
case 2
% create the indices for the maximum column:
out = [(1:size(A,1)).' repelem(cind,size(A,1)).'];
case 3
[r,dir] = ind2sub(size(pd),dind); % convert to [row,column]
d_ind = all_d(r); % find which value of d gave the maximum product
[R,C] = ndgrid(1:size(A,1),1:size(A,2)); % create a grid of all indices
if dir==1
% find the indices for a diagonal:
out = [diag(R,d_ind),diag(C,d_ind)];
else
% find the indices for an anti-diagonal:
out = [diag(fliplr(R),d_ind),diag(fliplr(C),d_ind)];
end
end
end
你得到maxVecProd(A)：
out =
1 2
2 2
3 2
这意味着最大的产品是（1,2），（2,2）和（3,2）中的元素。
答案 4 :(得分：0)
好的，所以我昨晚自己编了代码。它很长，我相信它可以以更好的方式完成。我不太了解MatLab，所以我的解决方案很天真。但它确实有效。
function indicies = maxproduct(A, n)
function diagIndicies = getAllReverseDiagonalIndicies()
% This function returns lists of indicies in the reverse diagonal direction (top-right to bottom-left)
if rows == 1
numCells = 0;
for j=1:length(A)
temp = [1, j];
numCells = numCells + 1;
diagIndicies{numCells} = temp;
end
return
end
% This loop adds all diagonals down and to the right, to the end of the matrix.
%fprintf('Diagonals down to main one are:\n');
numCells = 0;
for x=1:rows
rowNum = x;
colNum = 1;
temp = [];
while rowNum >= 1 && rowNum <= rows && colNum <= cols
temp = [temp; [rowNum colNum]];
rowNum = rowNum - 1;
colNum = colNum + 1;
end
numCells = numCells + 1;
temp = flipud(temp); % Need row major order for assignment
%disp(temp);
diagIndicies{numCells} = temp;
end
% Now go along bottom row
%fprintf('Diagonals along bottom are:\n');
for y=2:cols
rowNum = rows;
colNum = y;
temp = [];
while rowNum >= 1 && colNum <= cols
temp = [temp; [rowNum colNum]];
rowNum = rowNum - 1;
colNum = colNum + 1;
end
numCells = numCells + 1;
temp = flipud(temp); % Need row major order for assignment
%disp(temp);
diagIndicies{numCells} = temp;
end
end
function diagIndicies = getAllDiagonalIndicies()
% This function returns lists of indicies in the main diagonal direction (top-left to bottom-right)
if rows == 1
numCells = 0;
for j=1:length(A)
temp = [1, j];
numCells = numCells + 1;
diagIndicies{numCells} = temp;
end
return
end
% This loop adds all diagonals down and to the left, to the lhs of the matrix.
%fprintf('Diagonals down to main one are:\n');
numCells = 0;
for x=1:rows
rowNum = x;
colNum = cols;
temp = [];
while rowNum >= 1 && rowNum <= rows && colNum >= 1
temp = [temp; [rowNum colNum]];
rowNum = rowNum - 1;
colNum = colNum - 1;
end
numCells = numCells + 1;
temp = flipud(temp); % Need row major order for assignment
%disp(temp);
diagIndicies{numCells} = temp;
end
% Now go along bottom row...
%fprintf('Diagonals along bottom are:\n');
for y=cols-1:-1:1
rowNum = rows;
colNum = y;
temp = [];
while rowNum >= 1 && colNum >= 1
temp = [temp; [rowNum colNum]];
rowNum = rowNum - 1;
colNum = colNum - 1;
end
numCells = numCells + 1;
temp = flipud(temp); % Need row major order for assignment
%disp(temp);
diagIndicies{numCells} = temp;
end
end
% Main function starts here.
[rows, cols] = size(A);
theMaxProduct = -10000000;
indicies = [];
if isscalar(A)
if A == 1 && n == 1
indicies = [1,1];
return;
else
return;
end
end
% Find max product in each row of A
for i=1:rows
for j=1:cols-n+1
theProduct = 1;
for k=j:j+n-1
theProduct = theProduct * A(i, k);
end
if theProduct > theMaxProduct
theMaxProduct = theProduct;
indicies = [];
for k=j:j+n-1
indicies = [indicies; [i, k]];
end
end
end
end
fprintf('theMaxProduct after examining rows = %15.15f\n', theMaxProduct);
% Find max product in each column of A
for i=1:cols
for j=1:rows-n+1
theProduct = 1;
for k=j:j+n-1
theProduct = theProduct * A(k, i);
end
if theProduct > theMaxProduct
theMaxProduct = theProduct;
indicies = [];
for k=j:j+n-1
indicies = [indicies; [k, i]];
end
end
end
end
fprintf('theMaxProduct after examining cols = %15.15f\n', theMaxProduct);
% Find max product along reverse diagonals of A
diagIndicies = getAllReverseDiagonalIndicies();
%disp(diagIndicies);
for i=1: length(diagIndicies)
[numIndicies, ~] = size(diagIndicies{i});
for j=1:numIndicies-n+1
theProduct = 1;
for k=j:j+n-1
theProduct = theProduct * A(diagIndicies{i}(k, 1), diagIndicies{i}(k, 2));
end
if theProduct > theMaxProduct
theMaxProduct = theProduct;
indicies = [];
for k=j:j+n-1
indicies = [indicies; [diagIndicies{i}(k, 1), diagIndicies{i}(k, 2)]];
end
end
end
end
fprintf('theMaxProduct after examining reverse diag = %15.15f\n', theMaxProduct);
% Find max product along diagonals of A
diagIndicies = getAllDiagonalIndicies();
%disp(diagIndicies);
for i=1: length(diagIndicies)
[numIndicies, ~] = size(diagIndicies{i});
for j=1:numIndicies-n+1
theProduct = 1;
for k=j:j+n-1
theProduct = theProduct * A(diagIndicies{i}(k, 1), diagIndicies{i}(k, 2));
end
if theProduct > theMaxProduct
theMaxProduct = theProduct;
indicies = [];
for k=j:j+n-1
indicies = [indicies; [diagIndicies{i}(k, 1), diagIndicies{i}(k, 2)]];
end
end
end
end
fprintf('theMaxProduct after examining main diag = %15.15f\n', theMaxProduct);
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