MATLAB - Correction based on multiple condition. Help to speed up a function











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I need help to speed up a function I created to correct x and y elements based on the position of obstacles.



I explain better my function with one example only in the x direction (this means 1D example). Let's suppose I have the values of x and Geometry as



x = [1.8;3.4;4.2;6.3;6.4;8.6;9.1];
Geometry = [0;1;1;0;1;0;1;1;0;0]; % the value 1 means obstacle


So if I do find(Geometry) i'll get the position of the obstacles. Then the function corrects the position based on the x elements that are greater than the position of the obstacles. So for example x(1) is not corrected because its position is not greater than any obstacle. But x(2) will be corrected by two Geometry elements so Corrected_x(2) = 3.4 - 2. The answer of this example is:



Corrected_x = [1.8;1.4;2.2;3.3;3.4;3.6;4.1];


Note that the last element of x is corrected by 5 because the value is greater than the position of the 5 obstacles present in Geometry.



So far I have written a function that does this in 2D and gives me the result I want but it takes too much time. The problem is that Geometry is MxN matrix that I need to go through every row and column since the obstacle are in different positions in each row or column.
The function takes too much time because of the loops going through every row and column, but also because x and y are large arrays.



Any ideas to speed this up?



function [Corrected_x, Corrected_y]= PositionCorrection(x, y, Geometry)    

yRange = floor(min(y)):ceil(max(y));
xRange = floor(min(x)):ceil(max(x));
nY = length(yRange); % nRows
nX = length(xRange); % nCols

%% Save in cell arrays the x and y elements within the ranges
xPositionCorrected = cell(nY-1,1);
yPositionCorrected = cell(nX-1,1);
% Remember the order of the elements
Rememberpositionx = cell(nY-1,1);
Rememberpositiony = cell(nX-1,1);

%% Loop of search and correction of x elements
for iY = 1:nY-1
% x elements within yRange
xinBinY = x(y(:)>yRange(iY) & y(:)<yRange(iY+1));
[~,xObstacle] = find(Geometry(iY,:)); % xPosition of the obstacle
[~,Rememberpositionx{iY,1}] = ismember(xinBinY,x);
% Correction
if isempty(xinBinY)
xPositionCorrected{iY,1} = ;
elseif isempty(xObstacle)
xPositionCorrected{iY,1} = xinBinY;
else
matrixCorrectedResults = bsxfun(@gt, xinBinY,xObstacle);
xCorrection = sum(matrixCorrectedResults,2);
xPositionCorrected{iY,1} = xinBinY-xCorrection;
end

end

%% Loop of search and correction of y elements
for iX = 1:nX-1
% y elements within xRange
yinBinX = y(x(:)>xRange(iX) & x(:)<xRange(iX+1));
[yObstacle,~] = find(Geometry(:,iX));
[~,Rememberpositiony{iX,1}] = ismember(yinBinX,y);
if isempty(yinBinX)
yPositionCorrected{iX,1} = ;
elseif isempty(yObstacle)
yPositionCorrected{iX,1} = yinBinX;
else
matrixCorrectedResults = bsxfun(@gt, yinBinX,yObstacle.');
yCorrection = sum(matrixCorrectedResults,2);
yPositionCorrected{iX,1} = yinBinX-yCorrection;
end
end
%% Get the original order
xPositionCorrected(all(cellfun('isempty', xPositionCorrected),2),:) = ;
Rememberpositionx(all(cellfun('isempty',Rememberpositionx),2),:) = ;

yPositionCorrected(all(cellfun('isempty', yPositionCorrected),2),:) = ;
Rememberpositiony(all(cellfun('isempty',Rememberpositiony),2),:) = ;

% x
Lx = 1:length(x);
Corrected_x = cell2mat(xPositionCorrected);
idx = cell2mat(Rememberpositionx);
[~,idtemp] = ismember(Lx,idx);
Corrected_x = Corrected_x(idtemp);

% y
Corrected_y = cell2mat(yPositionCorrected);
idy = cell2mat(Rememberpositiony);
[~,idtemp] = ismember(Lx,idy);
Corrected_y = Corrected_y(idtemp);
end









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  • Indenting your code will format it properly, rather than adding backticks before and after every single line!
    – Wolfie
    Nov 19 at 12:28















up vote
0
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favorite












I need help to speed up a function I created to correct x and y elements based on the position of obstacles.



I explain better my function with one example only in the x direction (this means 1D example). Let's suppose I have the values of x and Geometry as



x = [1.8;3.4;4.2;6.3;6.4;8.6;9.1];
Geometry = [0;1;1;0;1;0;1;1;0;0]; % the value 1 means obstacle


So if I do find(Geometry) i'll get the position of the obstacles. Then the function corrects the position based on the x elements that are greater than the position of the obstacles. So for example x(1) is not corrected because its position is not greater than any obstacle. But x(2) will be corrected by two Geometry elements so Corrected_x(2) = 3.4 - 2. The answer of this example is:



Corrected_x = [1.8;1.4;2.2;3.3;3.4;3.6;4.1];


Note that the last element of x is corrected by 5 because the value is greater than the position of the 5 obstacles present in Geometry.



So far I have written a function that does this in 2D and gives me the result I want but it takes too much time. The problem is that Geometry is MxN matrix that I need to go through every row and column since the obstacle are in different positions in each row or column.
The function takes too much time because of the loops going through every row and column, but also because x and y are large arrays.



Any ideas to speed this up?



function [Corrected_x, Corrected_y]= PositionCorrection(x, y, Geometry)    

yRange = floor(min(y)):ceil(max(y));
xRange = floor(min(x)):ceil(max(x));
nY = length(yRange); % nRows
nX = length(xRange); % nCols

%% Save in cell arrays the x and y elements within the ranges
xPositionCorrected = cell(nY-1,1);
yPositionCorrected = cell(nX-1,1);
% Remember the order of the elements
Rememberpositionx = cell(nY-1,1);
Rememberpositiony = cell(nX-1,1);

%% Loop of search and correction of x elements
for iY = 1:nY-1
% x elements within yRange
xinBinY = x(y(:)>yRange(iY) & y(:)<yRange(iY+1));
[~,xObstacle] = find(Geometry(iY,:)); % xPosition of the obstacle
[~,Rememberpositionx{iY,1}] = ismember(xinBinY,x);
% Correction
if isempty(xinBinY)
xPositionCorrected{iY,1} = ;
elseif isempty(xObstacle)
xPositionCorrected{iY,1} = xinBinY;
else
matrixCorrectedResults = bsxfun(@gt, xinBinY,xObstacle);
xCorrection = sum(matrixCorrectedResults,2);
xPositionCorrected{iY,1} = xinBinY-xCorrection;
end

end

%% Loop of search and correction of y elements
for iX = 1:nX-1
% y elements within xRange
yinBinX = y(x(:)>xRange(iX) & x(:)<xRange(iX+1));
[yObstacle,~] = find(Geometry(:,iX));
[~,Rememberpositiony{iX,1}] = ismember(yinBinX,y);
if isempty(yinBinX)
yPositionCorrected{iX,1} = ;
elseif isempty(yObstacle)
yPositionCorrected{iX,1} = yinBinX;
else
matrixCorrectedResults = bsxfun(@gt, yinBinX,yObstacle.');
yCorrection = sum(matrixCorrectedResults,2);
yPositionCorrected{iX,1} = yinBinX-yCorrection;
end
end
%% Get the original order
xPositionCorrected(all(cellfun('isempty', xPositionCorrected),2),:) = ;
Rememberpositionx(all(cellfun('isempty',Rememberpositionx),2),:) = ;

yPositionCorrected(all(cellfun('isempty', yPositionCorrected),2),:) = ;
Rememberpositiony(all(cellfun('isempty',Rememberpositiony),2),:) = ;

% x
Lx = 1:length(x);
Corrected_x = cell2mat(xPositionCorrected);
idx = cell2mat(Rememberpositionx);
[~,idtemp] = ismember(Lx,idx);
Corrected_x = Corrected_x(idtemp);

% y
Corrected_y = cell2mat(yPositionCorrected);
idy = cell2mat(Rememberpositiony);
[~,idtemp] = ismember(Lx,idy);
Corrected_y = Corrected_y(idtemp);
end









share|improve this question
























  • Indenting your code will format it properly, rather than adding backticks before and after every single line!
    – Wolfie
    Nov 19 at 12:28













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need help to speed up a function I created to correct x and y elements based on the position of obstacles.



I explain better my function with one example only in the x direction (this means 1D example). Let's suppose I have the values of x and Geometry as



x = [1.8;3.4;4.2;6.3;6.4;8.6;9.1];
Geometry = [0;1;1;0;1;0;1;1;0;0]; % the value 1 means obstacle


So if I do find(Geometry) i'll get the position of the obstacles. Then the function corrects the position based on the x elements that are greater than the position of the obstacles. So for example x(1) is not corrected because its position is not greater than any obstacle. But x(2) will be corrected by two Geometry elements so Corrected_x(2) = 3.4 - 2. The answer of this example is:



Corrected_x = [1.8;1.4;2.2;3.3;3.4;3.6;4.1];


Note that the last element of x is corrected by 5 because the value is greater than the position of the 5 obstacles present in Geometry.



So far I have written a function that does this in 2D and gives me the result I want but it takes too much time. The problem is that Geometry is MxN matrix that I need to go through every row and column since the obstacle are in different positions in each row or column.
The function takes too much time because of the loops going through every row and column, but also because x and y are large arrays.



Any ideas to speed this up?



function [Corrected_x, Corrected_y]= PositionCorrection(x, y, Geometry)    

yRange = floor(min(y)):ceil(max(y));
xRange = floor(min(x)):ceil(max(x));
nY = length(yRange); % nRows
nX = length(xRange); % nCols

%% Save in cell arrays the x and y elements within the ranges
xPositionCorrected = cell(nY-1,1);
yPositionCorrected = cell(nX-1,1);
% Remember the order of the elements
Rememberpositionx = cell(nY-1,1);
Rememberpositiony = cell(nX-1,1);

%% Loop of search and correction of x elements
for iY = 1:nY-1
% x elements within yRange
xinBinY = x(y(:)>yRange(iY) & y(:)<yRange(iY+1));
[~,xObstacle] = find(Geometry(iY,:)); % xPosition of the obstacle
[~,Rememberpositionx{iY,1}] = ismember(xinBinY,x);
% Correction
if isempty(xinBinY)
xPositionCorrected{iY,1} = ;
elseif isempty(xObstacle)
xPositionCorrected{iY,1} = xinBinY;
else
matrixCorrectedResults = bsxfun(@gt, xinBinY,xObstacle);
xCorrection = sum(matrixCorrectedResults,2);
xPositionCorrected{iY,1} = xinBinY-xCorrection;
end

end

%% Loop of search and correction of y elements
for iX = 1:nX-1
% y elements within xRange
yinBinX = y(x(:)>xRange(iX) & x(:)<xRange(iX+1));
[yObstacle,~] = find(Geometry(:,iX));
[~,Rememberpositiony{iX,1}] = ismember(yinBinX,y);
if isempty(yinBinX)
yPositionCorrected{iX,1} = ;
elseif isempty(yObstacle)
yPositionCorrected{iX,1} = yinBinX;
else
matrixCorrectedResults = bsxfun(@gt, yinBinX,yObstacle.');
yCorrection = sum(matrixCorrectedResults,2);
yPositionCorrected{iX,1} = yinBinX-yCorrection;
end
end
%% Get the original order
xPositionCorrected(all(cellfun('isempty', xPositionCorrected),2),:) = ;
Rememberpositionx(all(cellfun('isempty',Rememberpositionx),2),:) = ;

yPositionCorrected(all(cellfun('isempty', yPositionCorrected),2),:) = ;
Rememberpositiony(all(cellfun('isempty',Rememberpositiony),2),:) = ;

% x
Lx = 1:length(x);
Corrected_x = cell2mat(xPositionCorrected);
idx = cell2mat(Rememberpositionx);
[~,idtemp] = ismember(Lx,idx);
Corrected_x = Corrected_x(idtemp);

% y
Corrected_y = cell2mat(yPositionCorrected);
idy = cell2mat(Rememberpositiony);
[~,idtemp] = ismember(Lx,idy);
Corrected_y = Corrected_y(idtemp);
end









share|improve this question















I need help to speed up a function I created to correct x and y elements based on the position of obstacles.



I explain better my function with one example only in the x direction (this means 1D example). Let's suppose I have the values of x and Geometry as



x = [1.8;3.4;4.2;6.3;6.4;8.6;9.1];
Geometry = [0;1;1;0;1;0;1;1;0;0]; % the value 1 means obstacle


So if I do find(Geometry) i'll get the position of the obstacles. Then the function corrects the position based on the x elements that are greater than the position of the obstacles. So for example x(1) is not corrected because its position is not greater than any obstacle. But x(2) will be corrected by two Geometry elements so Corrected_x(2) = 3.4 - 2. The answer of this example is:



Corrected_x = [1.8;1.4;2.2;3.3;3.4;3.6;4.1];


Note that the last element of x is corrected by 5 because the value is greater than the position of the 5 obstacles present in Geometry.



So far I have written a function that does this in 2D and gives me the result I want but it takes too much time. The problem is that Geometry is MxN matrix that I need to go through every row and column since the obstacle are in different positions in each row or column.
The function takes too much time because of the loops going through every row and column, but also because x and y are large arrays.



Any ideas to speed this up?



function [Corrected_x, Corrected_y]= PositionCorrection(x, y, Geometry)    

yRange = floor(min(y)):ceil(max(y));
xRange = floor(min(x)):ceil(max(x));
nY = length(yRange); % nRows
nX = length(xRange); % nCols

%% Save in cell arrays the x and y elements within the ranges
xPositionCorrected = cell(nY-1,1);
yPositionCorrected = cell(nX-1,1);
% Remember the order of the elements
Rememberpositionx = cell(nY-1,1);
Rememberpositiony = cell(nX-1,1);

%% Loop of search and correction of x elements
for iY = 1:nY-1
% x elements within yRange
xinBinY = x(y(:)>yRange(iY) & y(:)<yRange(iY+1));
[~,xObstacle] = find(Geometry(iY,:)); % xPosition of the obstacle
[~,Rememberpositionx{iY,1}] = ismember(xinBinY,x);
% Correction
if isempty(xinBinY)
xPositionCorrected{iY,1} = ;
elseif isempty(xObstacle)
xPositionCorrected{iY,1} = xinBinY;
else
matrixCorrectedResults = bsxfun(@gt, xinBinY,xObstacle);
xCorrection = sum(matrixCorrectedResults,2);
xPositionCorrected{iY,1} = xinBinY-xCorrection;
end

end

%% Loop of search and correction of y elements
for iX = 1:nX-1
% y elements within xRange
yinBinX = y(x(:)>xRange(iX) & x(:)<xRange(iX+1));
[yObstacle,~] = find(Geometry(:,iX));
[~,Rememberpositiony{iX,1}] = ismember(yinBinX,y);
if isempty(yinBinX)
yPositionCorrected{iX,1} = ;
elseif isempty(yObstacle)
yPositionCorrected{iX,1} = yinBinX;
else
matrixCorrectedResults = bsxfun(@gt, yinBinX,yObstacle.');
yCorrection = sum(matrixCorrectedResults,2);
yPositionCorrected{iX,1} = yinBinX-yCorrection;
end
end
%% Get the original order
xPositionCorrected(all(cellfun('isempty', xPositionCorrected),2),:) = ;
Rememberpositionx(all(cellfun('isempty',Rememberpositionx),2),:) = ;

yPositionCorrected(all(cellfun('isempty', yPositionCorrected),2),:) = ;
Rememberpositiony(all(cellfun('isempty',Rememberpositiony),2),:) = ;

% x
Lx = 1:length(x);
Corrected_x = cell2mat(xPositionCorrected);
idx = cell2mat(Rememberpositionx);
[~,idtemp] = ismember(Lx,idx);
Corrected_x = Corrected_x(idtemp);

% y
Corrected_y = cell2mat(yPositionCorrected);
idy = cell2mat(Rememberpositiony);
[~,idtemp] = ismember(Lx,idy);
Corrected_y = Corrected_y(idtemp);
end






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edited Nov 19 at 12:50









rinkert

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asked Nov 19 at 12:20









S. Perez

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  • Indenting your code will format it properly, rather than adding backticks before and after every single line!
    – Wolfie
    Nov 19 at 12:28


















  • Indenting your code will format it properly, rather than adding backticks before and after every single line!
    – Wolfie
    Nov 19 at 12:28
















Indenting your code will format it properly, rather than adding backticks before and after every single line!
– Wolfie
Nov 19 at 12:28




Indenting your code will format it properly, rather than adding backticks before and after every single line!
– Wolfie
Nov 19 at 12:28

















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