以下是使用循环的解决方案。它还增加了定义步长的可能性。

% Input.
x = round(rand(20, 1) * 100)

% Window size.
ws = 5;

% Step size.
ss = 1;

% Number of elements.
nx = numel(x);

% Number of result elements.
ny = numel(1:ss:(nx - ws + 1));

% Initialize output.
y = zeros(ny, 1);

% Calculate output.
jj = 1;
for ii = 1:ss:(nx - ws + 1)
  ww = x(ii:(ii + ws - 1));
  [~, yLoop(jj)] = min(ww);
  yLoop(jj) = yLoop(jj) + ii - 1;
  jj = jj + 1;
end

% Output.
y

存储在y中的最小值的索引相对于原始输入x。
ws = 5和ss = 1的输出：

x =
   88
   74
   96
   31
    6
   67
   98
   92
   69
   49
   12
   28
   43
   87
   68
   49
   20
   98
   83
   62

y =
    5
    5
    5
    5
    5
   10
   11
   11
   11
   11
   11
   12
   17
   17
   17
   17

编辑

我添加了一个使用索引的版本，这对于大输入来说要快得多。

% Input.
x = round(rand(20000, 1) * 100);

% Window size.
ws = 5;

% Step size.
ss = 1;

% Number of elements.
nx = numel(x);

% --- Solution using loop --- %

% Number of result elements.
ny = numel(1:ss:(nx - ws + 1));

% Initialize output (loop).
yLoop = zeros(ny, 1);

tic
% Calculate output.
jj = 1;
for ii = 1:ss:(nx - ws + 1)
  ww = x(ii:(ii + ws - 1));
  [~, yLoop(jj)] = min(ww);
  yLoop(jj) = yLoop(jj) + ii - 1;
  jj = jj + 1;
end
tLoop = toc;

% Output (loop).
yLoop;

% --- Solution using indexing --- %

tic
% Calculate indices.
ind = 1:ss:(nx - ws + 1);
ind = repmat(ind, ws, 1) + ([0:(ws - 1)].' * ones(1, numel(ind)));

% Calculate output (indexing).
[~, yIndexing] = min(x(ind));
yIndexing = (yIndexing + (0:ss:(ss * (size(ind, 2) - 1)))).';
tIndexing = toc;

% Compare loop and indexing version.
fprintf("yLoop == yIndexing: %d\n", sum(yLoop == yIndexing) == ny);
fprintf("yLoop time: %f s\n", tLoop);
fprintf("yIndeing time: %f s\n", tIndexing);

对于n = 20000、ws = 5和ss = 1，我们得到以下时间：

yLoop time: 0.672510 s
yIndeing time: 0.004466 s

让我们

data = [6 7 7 3 6 1 3 7 2 1 0 8]; % example data
N = 4; % block size

对于不相交的块，只需重塑一个矩阵，使每个块成为一列，然后计算每列的Arg min：

[~, indices] = min(reshape(data,N,[]), [], 1);

赠送

indices =
     4     2     3

对于滑动块，创建索引矩阵，然后沿每列计算Arg min：

[~, indices] = min(data(bsxfun(@plus, 1:numel(data)-N+1, (0:N-1).')), [], 1);

赠送

indices =
     4     3     4     3     2     1     4     4     3