In a matrix mat I would like to set an individual number of rows to zero according a cell array startRowsthat holds row numbers. I do have a working solution, however, I am wondering whether I can avoid the loop and replace it with some more sophisticated code?
mat = ones(100,5);
startRows = {10,10,30,10,40};
% How to avoid this loop?
for c = 1 : 5
mat(1:startRows{c}-1, c) = 0;
end
You can use Matlab's auto expansion of mismatched array sizes in recent versions. For older version of Matlab you will need to use bsxfun in the last line
mat = ones(100,5);
startRows = {10,10,30,10,40};
startRowsMatrix = [startRows{:}]
mat((1:100)' < [startRows{:}]) = 0;
Related
Setup
I have an array of captured data. The data may be captured on just 1 device or up to a dozen devices, with each device being a column in the array. I have a prior statement which I execute on the array to then turn it into a logical array to find particular points of interest in the data. Due to the nature of the data, there are many 0's and only a few 1's. I need to return an array with the indices of the 1's so I can go back and capture the data between those points (see update below).
find is an obvious choice for a function - however, the result I need, needs to have 1 column for each device. Normally find will do a linear index regardless of the dimensions of the array.
The devices follow a pattern - but aren't exactly the same. So, complicating this is the fact that the number of 1's in each column is close to, but not guaranteed to be exactly the same depending on the exact timing the data capture is stopped (they are most often different from each other by 1 element, but could be different by more).
MATLAB CODE ATTEMPTS
Because of that difference, I can't use the following simple code:
for p = 1:np
indices( :, p ) = find( device.data.cross( :, p ) );
end
Notes:
np is the number of columns in the data = number of devices captured.
devices is a class representing the collection of devices
data is a TimeTable containing captured data on all the devices
cross is a column in the data TimeTable which contains the logical array
Even this simple code is inefficient and generates the Code Analyzer warning:
The variable 'indices' appears to change size on every loop
iteration (within a script). Consider preallocating for speed.
As expected, it doesn't work as I get an error similar to the following:
Unable to perform assignment because the size of the left side is
448-by-1 and the size of the right side is 449-by-1.
I know why I get this error - each column in an array in MATLAB has to have the same number of rows, so I can't make the assignment if the row size doesn't match. I need to pad the "short" columns somehow. In this case, repeating the last index will work for my later operations without causing an error.
I can't figure out a "good" way to do this. I can't pre-populate the array rows because I don't know how many rows there will be until I've done the find operation.
I can change the code as follows:
indices = [];
for p = 1:np
tempindices = find( devices.data.cross(:, p) );
sizediff = size( tempindices, 1 ) - size( indices, 1 );
if p > 1
if sizediff > 0
padding = repmat(indices(end, 1:(p - 1)), sizediff, 1);
indices = [indices; padding];
elseif sizediff < 0
padding = repmat(tempindices(end), abs(sizediff), 1);
tempindices = [tempindices; padding];
end
end
indices(:,p) = tempindices;
end
Note: padarray would have been useful here, but I don't have the Image Processing Toolbox so I cannot use it.
This code works, but it is very inefficient, it creates multiple otherwise unneeded variables in the workspace and generates multiple "appears to change size on every loop iteration" warnings in Code Analyzer. Is there a more efficient way to do this?
Update / Additional Information:
Some more context is needed for my issue. Given that devices.data.cross is a logical array, to just "pick" the data I want from other columns in my table (as I originally described my problem) I could select a column from devices.data.cross and pass that logical column as a subscript to get that data. I do that where it works. However, for some of the columns I need to select "chunks" of the data between the indices and that's where (I think) I need the indices. Or, at least I don't know of another way to do it.
Here is example code of how I use the indices:
for p = 1:np
for i = 2:num_indices
these_indices = indices(i-1, p):( indices(i, p) - 1 );
rmsvoltage = sqrt( mean( devices.data.voltage(these_indices).^2 ) );
end
end
This is just one routine I do on the "chunks" of data. I also have a couple of functions where these chunks of data are passed for processing.
When I understood your problem correctly, the code below should work. I'm using the approach that Cris Luengo suggested in a comment under your question.
Key element is [rowIdcs, colIdcs] = find( cros ); which gives you the subscripts of positions in cros having a value of one. Please find further comments inline.
% Create some data for testing
volt = randn(10,10);
cros = randi(10,10,10) > 9;
% Get rowIdcs and colIdcs, which have both a size of Nx1,
% with N denoting the number of ones in the mask.
% rowIdcs and colIdcs are the subscripts of the ones in the mask.
[rowIdcs, colIdcs] = find( cros );
% The number of chunks is equal to number N of ones found in the mask;
numChunks = numel( rowIdcs );
% Initilize a vector for the rms
rms = zeros( numChunks, 1 );
% Loop over the chunks
for k = 1 : numChunks
curRow = rowIdcs(k);
curCol = colIdcs(k);
% Get indices of range over neighbouring rows
chunkRowIdcs = curRow + [-1 0 1]; %i.e. these_indices in your example
% Remove indices that are out of range
chunkRowIdcs( chunkRowIdcs < 1 | chunkRowIdcs > size(volt,1) ) = [];
% Get voltages covered by chunk
chunkVoltages = volt( chunkRowIdcs, curCol );
% Get RMS over voltages
rms(k) = sqrt( mean( chunkVoltages(:).^2 ));
end
Using Matlab R2019a, is there any way to avoid the for-loop in the following code in spite of the dimensions containing different element so that each element has to be checked? M is a vector with indices, and Inpts.payout is a 5D array with numerical data.
for m = 1:length(M)-1
for power = 1:noScenarios
for production = 1:noScenarios
for inflation = 1:noScenarios
for interest = 1:noScenarios
if Inpts.payout(M(m),power,production,inflation,interest)<0
Inpts.payout(M(m+1),power,production,inflation,interest)=...
Inpts.payout(M(m+1),power,production,inflation,interest)...
+Inpts.payout(M(m),power,production,inflation,interest);
Inpts.payout(M(m),power,production,inflation,interest)=0;
end
end
end
end
end
end
It is quite simple to remove the inner 4 loops. This will be more efficient unless you have a huge matrix Inpts.payout, as a new indexing matrix must be generated.
The following code extracts the two relevant 'planes' from the input data, does the logic on them, then writes them back:
for m = 1:length(M)-1
payout_m = Inpts.payout(M(m),:,:,:,:);
payout_m1 = Inpts.payout(M(m+1),:,:,:,:);
indx = payout_m < 0;
payout_m1(indx) = payout_m1(indx) + payout_m(indx);
payout_m(indx) = 0;
Inpts.payout(M(m),:,:,:,:) = payout_m;
Inpts.payout(M(m+1),:,:,:,:) = payout_m1;
end
It is possible to avoid extracting the 'planes' and writing them back by working directly with the input data matrix. However, this yields more complex code.
However, we can easily avoid some indexing operations this way:
payout_m = Inpts.payout(M(1),:,:,:,:);
for m = 1:length(M)-1
payout_m1 = Inpts.payout(M(m+1),:,:,:,:);
indx = payout_m < 0;
payout_m1(indx) = payout_m1(indx) + payout_m(indx);
payout_m(indx) = 0;
Inpts.payout(M(m),:,:,:,:) = payout_m;
payout_m = payout_m1;
end
Inpts.payout(M(m+1),:,:,:,:) = payout_m1;
It seems like there is not a way to avoid this. I am assuming that each for lop independently changes a variable parameter used in the main calculation. Thus, it is required to have this many for loops. My only suggestion is to turn your nested loops into a function if you're concerned about appearance. Not sure if this will help run-time.
This question already has answers here:
Get the indices of the n largest elements in a matrix
(4 answers)
Closed 6 years ago.
When using a binary image with several lines I know that this code displays the longest line:
lineStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = max([lineStats.Area]);
longestLine = zeros(size(imsk));
longestLine(lineStats(index).PixelIdxList)=1;
figure
imshow(longestLine)
Is there a way to display the second longest line? I need to display a line that is a little shorter than the longest line in order to connect them.
EDIT: Is there a way to display both lines on the binary image figure?
Thank you.
I would set the longest line to zero and use max again, after I copy the original vector.
lineStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = max([lineStats.Area]);
lineAreas = [lineStats.Area]; %copy all lineStats.Area values into a new vector
lineAreas(index) = NaN; %remove the longest line by setting it to not-a-number
[length2, index2] = max(lineAreas);
EDIT: Response to new question
sort may be a more straight forward approach for multiples, but you can still use max.
lineAreas = [lineStats.Area]; %copy all lineStats.Area values into a new vector
% add a for loop that iteratively stores the desired indices
nLines = 3;
index = zeros(1,nLines);
for iLines = 1:nLines
[length, index(iLines)] = max(lineAreas);
lineAreas(index) = NaN; %remove the longest line by setting it to not-a-number
end
longestLine = zeros(size(imsk));
% I cannot be certain this will work since your example is not reproducible
longestLine([lineStats(index).PixelIdxList]) = 1;
figure
imshow(longestLine)
Instead of using max use sort in descending order and take the second element. Like max, sort also provides the indexes of the returned values, so the two functions are pretty compatible.
eStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = sort([lineStats.Area], 'descend');
longestLine = zeros(size(imsk));
longestLine(lineStats(index(2)).PixelIdxList)=1; % here take the second largest
figure
imshow(longestLine)
As an alternative with focus on performance and ease of use, here's one approach using bwlabel instead of regionprops -
[L, num] = bwlabel(imsk, 8);
count_pixels_per_obj = sum(bsxfun(#eq,L(:),1:num));
[~,sidx] = sort(count_pixels_per_obj,'descend');
N = 3; % Shows N biggest objects/lines
figure,imshow(ismember(L,sidx(1:N))),title([num2str(N) ' biggest blobs'])
On the performance aspect, here's one post that does some benchmarking on snowflakes and coins images from MATLAB's image gallery.
Sample run -
imsk = im2bw(imread('coins.png')); %%// Coins photo from MATLAB Library
N = 2:
N = 3:
This is my code in Matlab: How could I get all values of all 5 images saved? This code only returns the last image! I tried using IM(l) but it gives me an error: In an assignment A(I) = B, the number of elements in B and I must be the same.
Amount_measurements = 5;
IM=zeros(2097152,1);
l=1;
for l=(1:Amount_measurements)
if l < 9
%index = double(0)+double(0)+double(l+1);
index = strcat(num2str(double(0)),num2str(double(0)),num2str(double(l+1)));
elseif l < 99
index = double(0)+double(l+1);
else
index = double(l+1);
end
file_name1='trial.nii.gz';
%disp(file_name1);
jesu=load_nii(file_name1);
[x,y,z] = meshgrid(1:256,1:256,1:256);
[lx,ly,lz] = meshgrid(1:2:256,1:2:256,1:2:256);
newImage = interp3(x,y,z,jesu.img,lx,ly,lz);
IM= newImage(:);
end
I want the values newImage(:) to be stored as IM1=newImage(:) IM2=newImage(:) IM3=newImage(:) IM4=newImage(:) so on... How could I go about with it?
Since you mentioned wanting a variable-length version of IM1=newImage(:) IM2=newImage(:) IM3=newImage(:) IM4=newImage(:), you're looking for a cell array. Try
IM{l} = newImage;
instead of
IM(l) = newImage(:);
The important difference is the use of braces rather than parentheses. Use a right-hand side ofnewImage(:) if you want to reshape into a vector, just newImage if you want to preserve it as a matrix.
By using IM(l) you're trying to add an entire column vector (newImage(:)) to a single element (the l-th element) in the array IM, that's why Matlab throws the error.
You should consider concatenation: since newImage(:) is a column-vector, replace
IM= newImage(:);
with
IM=[IM newImage(:)];
but at the top of the script you should also initialize IM as
IM=[];
At the end of the loop, the resulting IM will have Amount_measurements columns where 1 column = 1 newImage(:).
Note #1: this will only work if newImage(:) always has the same length.
Note #2: if you know a priori how long the vector newImage(:) is and, again, by assuming that its length never changes, you should consider preallocating the IM matrix by replacing IM=[]; with IM=zeros(X,Amount_measurements); where X is the number of elements in newImage(:). Finally, regarding the concatenation stage, you should replace IM=[IM newImage(:)]; with IM(:,l)=newImage(:).
Note #3: as instead, if the size of newImage(:) can change you cannot rely on preallocation and matrices, but you must use cell arrays: the last instruction in your loop should be IM{l}=newImage(:);.
I have the following piece of code:
for query = queryFiles
queryImage = imread(strcat('Queries/', query));
queryImage = im2single(rgb2gray(queryImage));
[qf,qd] = vl_covdet(queryImage, opts{:}) ;
for databaseEntry = databaseFiles
entryImage = imread(databaseEntry.name);
entryImage = im2single(rgb2gray(entryImage));
[df,dd] = vl_covdet(entryImage, opts{:}) ;
[matches, H] = matchFeatures(qf,qf,df,dd) ;
result = [result; query, databaseEntry, length(matches)];
end
end
It is my understanding that it should work as a Java/C++ for(query:queryFiles), however the query appears to be a copy of the queryFiles. How do I iterate through this vector normally?
I managed to sort the problem out. It was mainly to my MATLAB ignorance. I wasn't aware of cell arrays and that's the reason I had this problem. That and the required transposition.
From your code it appears that queryFiles is a numeric vector. Maybe it's a column vector? In that case you should convert it into a row:
for query = queryFiles.'
This is because the for loop in Matlab picks a column at each iteration. If your vector is a single column, it picks the whole vector in just one iteration.
In MATLAB, the for construct expects a row vector as input:
for ii = 1:5
will work (loops 5 times with ii = 1, 2, ...)
x = 1:5;
for ii = x
works the same way
However, when you have something other than a row vector, you would simply get a copy (or a column of data at a time).
To help you better, you need to tell us what the data type of queryFiles is. I am guessing it might be a cell array of strings since you are concatenating with a file path (look at fullfile function for the "right" way to do this). If so, then a "safe" approach is:
for ii = 1:numel(queryFiles)
query = queryFiles{ii}; % or queryFiles(ii)
It is often helpful to know what loop number you are in, and in this case ii provides that count for you. This approach is robust even when you don't know ahead of time what the shape of queryFiles is.
Here is how you can loop over all elements in queryFiles, this works for scalars, row vectors, column vectors and even high dimensional matrices:
for query = queryFiles(:)'
% Do stuff
end
Is queryFiles a cell array? The safest way to do this is to use an index:
for i = 1:numel(queryFiles)
query = queryFiles{i};
...
end