Can't seem to properly debug "In an assignment A(:) = B..." error

1 visualización (últimos 30 días)
user
user el 19 de Mzo. de 2018
Comentada: Walter Roberson el 19 de Mzo. de 2018
I am doing some image analysis and getting the following error message:
In an assignment A(:) = B, the number of elements in A and B must be the same.
Error in FINAL (line 37)
bottomLine(col) = find(croppedImage(:, col), 1, 'last');
I have made sure that my column vector (col) is the proper length, but it doesn't seem to be working. For some context, the purpose of this set of code is to analyze a video file of a droplet on a surface. As the droplet freezes, I see the freezing line move up.
When I evaluate my "problem" line of code (line 37) all of my variables are of acceptable length, the columns vector is 236 and so is the columns of the croppedImage matrix. The for loops seems to stop at column number 148 every time which I don't understand.
Here is my code:
nstart = 260;
nend = 700;
Height = (90:370); %Temporary
Width1 = (94:796); %Temporary
Width2 = (1060:1775); %Temporary
load('Crop1.mat')
%%Processing for Droplet 1
numberOfCells = length(Crop1);
freezingLine1 = zeros(1, numberOfCells);
L1 = round((length(Width1)/3)):round((2*length(Width1)/3));
for k = 1:numberOfCells
thisCell = Crop1{k};
if ~isempty(thisCell)
subplot(2, 2, 1);
imshow(thisCell);
axis on;
caption = sprintf('Frame %d of %d', k, numberOfCells);
title(caption, 'FontSize', fontSize);
% Extract columns 200-300
croppedImage = thisCell(1:length(Height), L1);
subplot(2, 2, 2);
imshow(croppedImage);
axis on;
title(caption, 'FontSize', fontSize);
% Get the bottom most pixel
[rows, columns] = size(croppedImage);
bottomLine = zeros(1, columns);
for col = 1 : columns
bottomLine(col) = find(croppedImage(:, col), 1, 'last');
end
subplot(2, 2, 3);
plot(bottomLine, 'b-', 'LineWidth', 2);
ylim([1, rows]);
xlim([1, columns]);
title('Bottom Pixels', 'FontSize', fontSize);
xlabel('Column', 'FontSize', fontSize);
ylabel('Row', 'FontSize', fontSize);
grid on;
% Throw out outliers by looking at the median absolute deviation.
medianValue = median(bottomLine);
mad = bottomLine - medianValue;
goodIndexes = mad <= 3; % Good locations have a mad of less than or equal to 3.
% Find the mean of the good indexes.
freezingLine1(k) = mean(bottomLine(goodIndexes));
subplot(2, 2, 4);
plot(freezingLine1, 'b-', 'LineWidth', 2);
grid on;
title('Freezing Line Row1', 'FontSize', fontSize);
xlabel('Frame', 'FontSize', fontSize);
ylabel('Row', 'FontSize', fontSize);
drawnow;
end
end
and attached is a .mat file that needs to be loaded to run the code.

Respuesta aceptada

Ahmet Cecen
Ahmet Cecen el 19 de Mzo. de 2018
Editada: Ahmet Cecen el 19 de Mzo. de 2018
I am guessing this error is due to null matrices appearing as a result of this call:
find(croppedImage(:, col), 1, 'last');
Try this instead:
for col = 1 : 236
try
bottomLine(col) = find(croppedImage(:, col), 1, 'last');
catch
bottomLine(col) = NaN;
end
end
This would mean some columns in the image simply doesn't have a non-zero value. I am dealing with them as NaNs here but you can assign them whatever you want.
  3 comentarios
Ahmet Cecen
Ahmet Cecen el 19 de Mzo. de 2018
Yes, when you try to assign an empty value to an indexed position, MATLAB throws an error saying that this is an invalid assignment. See this case:
a = [ 1 2 3]
b = []
a(2) = b
will give an error. I suspect this is a safety feature due to the following behavior where the array element is instead destroyed:
a = [ 1 2 3]
a(2) = []
Try will "try" to execute a statement within the try block and if there is an error "thrown" during that execution, it will "catch" that error and execute the "catch" block instead. Your understanding is correct.
Walter Roberson
Walter Roberson el 19 de Mzo. de 2018
I recommend instead
for col = 1 : 236
loc = find(croppedImage(:,col), 1, 'last');
if isempty(loc)
bottomLine(col) = NaN;
else
bottomLine(col) = loc;
end
end

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Logical en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by