How to find a value in a matrix

3 次查看（过去 30 天）
Honey
Honey 2021-11-27
评论： Honey 2021-11-28
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

请先登录，再回答此问题。

采纳的回答

DGM
DGM 2021-11-27
Ran in:
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 个评论
显示 3更早的评论
DGM
DGM 2021-11-27
Ran in:
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 2021-11-28
Great ! It works and now with 'find()', I can have the pixel numbers too

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by