Creating a matrix from spaced out lines of another matrix

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Patrick O'Mahony 2020-3-31

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评论： Patrick O'Mahony 2020-4-1

I have a large 3 column matrix of xyz elevation data in 1 meter intervals. The data corresponds to a 100 meter by 100 meter grid, and lists the data row by row, meaning that the y term changes by 1 every 100 values. The data begins in the top right corner of the 'grid', as shown:

x y z

0 100 34

1 100 35

2 100 23

...

0 99 54

1 99 45

2 99 37

...

etc.

I would like to write a function that creates a smaller matrix corresponding to a 3 meter by 3 meter portion of the grid, but because of the large matrix's construction, the nine lines of data are grouped in three sets of three, each spaced around 100 lines from the next. Any advice and assistance would be appreciated.

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the cyclist 2020-3-31

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Is your matrix stored in a numeric data type (such as a double), or in a cell array, table, or other data type?

So, you want to enter the center coordinates (x,y) -- suppose they are 43, 52 -- and then get output

?

Patrick O'Mahony 2020-3-31

The data is stored as a double.

The eventual goal is to plot the 3x3 grid on a mesh, but what you're suggesting as input and output would be really helpful, yes.

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Answer 1

the cyclist 2020-4-1

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编辑：the cyclist 2020-4-1

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Here is a function definition. Put it in a file named smallFromBig.m.

function [xyzSmall] = smallFromBig(x,y,xyzBig)
    
    isNearX = ismember(xyzBig(:,1),[x-1 x x+1]);
    isNearY = ismember(xyzBig(:,2),[y-1 y y+1]);
    
    xyzSmall = xyzBig(isNearX & isNearY,:);
    
end

Here is an example of how I called it.

% Make up some input data
m = 100;
n = 100;
xyzBig = [repelem(1:m,n)' mod(0:m*n-1,n)' rand(m*n,1)];
 
% Find the small grid around x==47, y==52
[xyzSmall] = smallFromBig(47,52,xyzBig)

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the cyclist 2020-4-1

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I'm not sure why your version is working, but I would do it like this:

xd = 2;
yd = 3;
isNearX = ismember(xyzBig(:,1),[x-xd:x+xd]);
isNearY = ismember(xyzBig(:,2),[y-yd:y+yd]);
xyzSmall = xyzBig(isNearX & isNearY,:);

which would give a 5x7 grid, for example.

Patrick O'Mahony 2020-4-1

That makes sense. It works great, thanks.

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Answer 2

darova 2020-4-1

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Try this

dx = max(x)-min(x);
ix = round((x-min(x))/max(x)*2) + 1;     % result 1 2 3
ind = cell(3,1);
for i = 1:length(ind)
    ind{i} = x(ix==i);                   % write data
end