How to find a value in a matrix

3 visualizaciones (últimos 30 días)
Honey
Honey el 27 de Nov. de 2021
Comentada: Honey el 28 de Nov. de 2021
Hi
I have a matrix of latitude and another matrix of longtitude for the location of an area pixel to pixel. I have a location value (longtitue and latitude) that I want to find the position of the pixels which this value will be located in. How can I find this?
Look at below to the exemplary matrix of Latitude.
There is another matrix like this for longtitudeThe point that I am looking for is Latitude=35.6886 and Longtitude= 53.6113
  2 comentarios
David Hill
David Hill el 27 de Nov. de 2021
Can't you just use the find() command?
Honey
Honey el 27 de Nov. de 2021
I tried this but it doesnt work.

Iniciar sesión para comentar.

Respuesta aceptada

DGM
DGM el 27 de Nov. de 2021
Consider the simple example:
% two orthogonal grids
[x y] = meshgrid(linspace(0,1,10))
x = 10×10
0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000
y = 10×10
0 0 0 0 0 0 0 0 0 0 0.1111 0.1111 0.1111 0.1111 0.1111 0.1111 0.1111 0.1111 0.1111 0.1111 0.2222 0.2222 0.2222 0.2222 0.2222 0.2222 0.2222 0.2222 0.2222 0.2222 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.4444 0.4444 0.4444 0.4444 0.4444 0.4444 0.4444 0.4444 0.4444 0.4444 0.5556 0.5556 0.5556 0.5556 0.5556 0.5556 0.5556 0.5556 0.5556 0.5556 0.6667 0.6667 0.6667 0.6667 0.6667 0.6667 0.6667 0.6667 0.6667 0.6667 0.7778 0.7778 0.7778 0.7778 0.7778 0.7778 0.7778 0.7778 0.7778 0.7778 0.8889 0.8889 0.8889 0.8889 0.8889 0.8889 0.8889 0.8889 0.8889 0.8889 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
% the point you're trying to locate
targetpt = [0.3333 0.5555]; % [x y]
tol = 0.001; % tolerance
% the row and column where the point is found
[tprow tpcol] = find(abs(x-targetpt(1))<tol & abs(y-targetpt(2))<tol)
tprow = 6
tpcol = 4
Don't expect simple equality tests to work with floating point numbers like this. You'll have to test matches to within some defined tolerance.
  5 comentarios
DGM
DGM el 27 de Nov. de 2021
Ah yeah. Those aren't really close to the mesh at all. It depends on what you want to do. If you want to keep working on the fixed mesh and just find the nearest vertices, you can do something like:
X = xlsread('X,Y.xlsx','X');
Y = xlsread('X,Y.xlsx','Y');
yx = xlsread('querypoints.xlsx');
% X and Y are meshgrids; don't really need all that
Xv = X(:,1).';
Yv = Y(1,:);
% find the subscripts of the nearest match
[~,idxx] = min(abs(Xv-yx(:,2)),[],2); % index along dim 1
[~,idxy] = min(abs(Yv-yx(:,1)),[],2); % index along dim 2
% bear in mind the vector orientation
% here, "x" and "y" refer to the names from the spreadsheet
% not the array dimensions
% for example, look up the closest match for a selected query point
k = 1; % pick a point
yx(k,:)
ans = 1×2
53.6114 36.6867
X(idxx(k),idxy(k))
ans = 36.6550
Y(idxx(k),idxy(k))
ans = 53.6510
I had to rename the second file only because the web-version refused to recognize the filename. Otherwise it worked fine on desktop.
Honey
Honey el 28 de Nov. de 2021
Great ! It works and now with 'find()', I can have the pixel numbers too

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Resizing and Reshaping Matrices en Help Center y File Exchange.

Etiquetas

Productos


Versión

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by