获取给定索引前后的矢量索引(窗口+/- 1) [英] Get vector indices before-and-after (window +/- 1) given indices

问题描述

给定idx一个索引向量，以获得idx +/-1的排序向量，最好的Matlab/Octave惯用法是什么?

What's the best Matlab/Octave idiom, given idx a vector of indices, to get the sorted vector of idx +/-1 ?

我有一个n x 7数据矩阵，第3列是一个整数标签，并且我有兴趣查看其上不连续的区域. 因此，我得到了相应的索引:

I have an n x 7 data matrix, column 3 is an integer label, and I'm interested in viewing the neighborhood of discontinuities on it. Hence I get the corresponding indices:

idx = find(diff(data(:,3)) > 0)

5297
6275
6832
...
20187

然后，如果我想在我的列上(例如(mx2)矩阵[idx-1; idx+1]上)查看该邻域+/- 1，则需要形成按顺序连接或重新排序的idx-1, idx+1向量. 我发现了一些笨拙的方法，正确的方法是什么? (我尝试了所有有关重新排列的八度章节矩阵)

Then if I want to view that neighborhood +/- 1 on my column (e.g. on the (mx2) matrix [idx-1; idx+1]), I need to form the vector of idx-1, idx+1 either concatenated in-order, or resorted. I found some clunky ways of doing this, what's the proper way? (I tried all of the octave chapter on Rearranging Matrices)

% WAY 1: this works, but is ugly - a needless O(n) sort
sort([idx-1; idx+1])

% horzcat,vertcat,vec only stack it vertically
horzcat([idx-1; idx+1])
horzcat([idx-1; idx+1]')

% WAY 2?
%One of vec([idx-1; idx+1]) or vec([idx-1; idx+1]') should work? but doesn't, they always stack columnwise
horzcat([idx-1; idx+1]')

ans =
Columns 1 through ...
5297    6275    6832 ...  20187    5299    6277    6834 ... 20189

% TRY 3...
reshape([idx-1; idx+1], [36,1]) doesn't work either

您希望只有两种方法可以解开2xm矩阵，但是...

You would expect there are only two ways to unstack a 2xm matrix, but ...

推荐答案

您可以使用隐式单例扩展(R2016b或更新的MATLAB，Octave本身)进行扩展

You can do this with implicit singleton expansion (R2016b or newer MATLAB, native to Octave)

idx = [2, 6, 9]; % some vector of integers
% Use reshape with [] to tell MATLAB "however many rows it takes"
neighbours = reshape( idx + [-1;1], [], 1 );

>> neighbours = [1; 3; 6; 8; 8; 10];

如果您不知道idx是行还是列，则可以通过使用

If you don't know whether idx is a row or column, you can be more robust by using

neighbours = reshape( idx(:)' + [-1,1], [], 1)

如果您不想使用隐式扩展(并再次处理行或列idx)，则可以使用像这样的重塑

If you don't want to use implicit expansion (and again coping with either row or column idx), you can use reshape like so

neighbours = reshape( [idx(:)-1, idx(:)+1]', [], 1 )

---

注意:您可能还希望将整个内容包装在对unique的调用中.在我的示例中，您两次获得了索引8，我不确定这是否符合您的情况.

---

Note: you may also want to wrap the whole thing in a call to unique. In my example, you get the index 8 twice, I'm not sure if this is desirable or not in your situation.

但是，unique会执行排序(除非您使用'stable'标志，但这会使它变慢)，因此，如果要删除重复项，也可以使用原始方法:

However, unique performs a sort (unless you use the 'stable' flag but that can make it even slower), so you might as well use your original approach if you want to remove duplicates:

% Remove duplicates and sort the result using unique 
neighbours = unique( [idx-1, idx+1] );

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