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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:
Related
I am trying to create a vector with the average of every 48 elements in eddyCO2.
Tweedle = eddyCO2(1:47:end);
Tweedle(1) = mean(eddyCO2(1):eddyCO2(48));
for i = 2:length(Tweedle)
Tweedle(i) = mean(eddyCO2((i-1)*48):eddyCO2(i*48)); (ERROR: Index exceeds matrix dimensions)
end
I've tried reshaping and the only thing that seems to work is entering values manually but the size is too large to work without a loop. Why is this error appearing?
Does the following modified version of your script do what you intend? Hope it helps.
Tweedle = eddyCO2(1:48:end);
sz = length(eddyCO2);
for i = 1:length(Tweedle)-1
Tweedle(i) = mean(eddyCO2((i-1)*48+1):eddyCO2(i*48)); % averages elements 1-48, 49-96, etc.
end
Tweedle(i+1) = mean(eddyCO2(i*48+1):sz); % averages remaining items at end of vector
I wrote this matlab code in order to concatenate the results of the integration of all the columns of a matrix extracted form a multi matrix array.
"datimf" is a matrix composed by 100 matrices, each of 224*640, vertically concatenated.
In the first loop i select every single matrix.
In the second loop i integrate every single column of the selected matrix
obtaining a row of 640 elements.
The third loop must concatenate vertically all the lines previously calculated.
Anyway i got always a problem with the third loop. Where is the error?
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=0:224:(22400-224)
for j = 1:640
for k = 1:100
singleframe(:,:) = datimf([i+1:(i+223)+1],:);
int_frame_all(:,j) = trapz(singleframe(:,j));
conc(:,k) = vertcat(int_frame_all);
end
end
end
An alternate way to do this without using any explicit loops (edited in response to rayryeng's comment below. It's also worth noting that using cellfun may not be more efficient than explicitly looping.):
nmats = 100;
nrows = 224;
ncols = 640;
datimf = rand(nmats*nrows, ncols);
% convert to an nmats x 1 cell array containing each matrix
cellOfMats = mat2cell(datimf, ones(1, nmats)*nrows, ncols);
% Apply trapz to the contents of each cell
cellOfIntegrals = cellfun(#trapz, cellOfMats, 'UniformOutput', false);
% concatenate the results
conc = cat(1, cellOfIntegrals{:});
Taking inspiration from user2305193's answer, here's an even better "loop-free" solution, based on reshaping the matrix and applying trapz along the appropriate dimension:
datReshaped = reshape(datimf, nrows, nmats, ncols);
solution = squeeze(trapz(datReshaped, 1));
% verify solutions are equivalent:
all(solution(:) == conc(:)) % ans = true
I think I understand what you want. The third loop is unnecessary as both the inner and outer loops are 100 elements long. Also the way you have it you are assigning singleframe lots more times than necessary since it does not depend on the inner loops j or k. You were also trying to add int_frame_all to conc before int_frame_all was finished being populated.
On top of that the j loop isn't required either since trapz can operate on the entire matrix at once anyway.
I think this is closer to what you intended:
datimf = rand(224*100,640);
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=1:100
idx = (i-1)*224+1;
singleframe(:,:) = datimf(idx:idx+223,:);
% for j = 1:640
% int_frame_all(:,j) = trapz(singleframe(:,j));
% end
% The loop is uncessary as trapz can operate on the entire matrix at once.
int_frame_all = trapz(singleframe,1);
%I think this is what you really want...
conc(i,:) = int_frame_all;
end
It looks like you're processing frames in a video.
The most efficent approach in my experience would be to reshape datimf to be 3-dimensional. This can easily be achieved with the reshape command.
something along the line of vid=reshape(datimf,224,640,[]); should get you far in this regard, where the 3rd dimension is time. vid(:,:,1) then would display the first frame of the video.
I would like to divide a vector in many vectors and put all of them in a matrix. I got this error "Subscripted assignment dimension mismatch."
STEP = zeros(50,1);
STEPS = zeros(50,length(locate));
for i = 1:(length(locate)-1)
STEP = filtered(locate(i):locate(i+1));
STEPS(:,i) = STEP;
end
I take the value of "filtered" from (1:50) at the first time for example and I would like to stock it in the first row of a matrix, then for iterations 2, I take value of "filtered from(50:70) for example and I stock it in row 2 in the matrix, and this until the end of the loop..
If someone has an idea, I don't get it! Thank you!
As mentioned in the comments, to make it work you can edit the loopy code at the end with -
STEPS(1:numel(STEP),i) = STEP;
Also, output array STEPS doesn't seem to use the last column. So, the initialization could use one less column, like so -
STEPS = zeros(50,length(locate)-1);
All is good with the loopy code, but in the long run with a high level language like MATLAB, you might want to look for faster codes and one way to achieve that would be vectorized codes. So, let me suggest a vectorized solution using bsxfun's masking capability to process such ragged-arrays. The implementation to cover generic elements in locate would look something like this -
% Get differentiation, which represent the interval lengths for each col
diffs = diff(locate)+1;
% Initialize output array
out = zeros(max(diffs),length(locate)-1);
% Get elements from filtered array for setting into o/p array
vals = filtered(sort([locate(1):locate(end) locate(2:end-1)]));
% Use bsxfun to create a mask that are to be set in o/p array and set thereafter
out(bsxfun(#ge,diffs,(1:max(diffs)).')) = vals;
Sample run for verification -
>> % Inputs
locate = [6,50,70,82];
filtered = randi(9,1,120);
% Get extent of output array for number of rows
N = max(diff(locate))+1;
>> % Original code with corrections
STEP = zeros(N,1);
STEPS = zeros(N,length(locate)-1);
for i = 1:(length(locate)-1)
STEP = filtered(locate(i):locate(i+1));
STEPS(1:numel(STEP),i) = STEP;
end
>> % Proposed code
diffs = diff(locate)+1;
out = zeros(max(diffs),length(locate)-1);
vals = filtered(sort([locate(1):locate(end) locate(2:end-1)]));
out(bsxfun(#ge,diffs,(1:max(diffs)).')) = vals;
>> max_error = max(abs(out(:)-STEPS(:)))
max_error =
0
What is the best way to do random sample with replacement from dataset? I am using 316 * 34 as my dataset. I want to segment the data into three buckets but with replacement. Should I use the randperm because I need to make sure I keep the index intact where that index would be handy in identifying the label data. I am new to matlab I saw there are couple of random sample methods but they didn't look like its doing what I am looking for, its strange to think that something like doesn't exist in matlab, but I did the follwoing:
My issue is when I do this row_idx = round(rand(1)*316) sometimes I get zero, that leads to two questions
what should I do to avoid zeor?
What is the best way to do the random sample with replacement.
shuffle_X = X(randperm(size(X,1)),:);
lengthOf_shuffle_X = length(shuffle_X)
number_of_rows_per_bucket = round(lengthOf_shuffle_X / 3)
bucket_cell = cell(3,1)
bag_matrix = []
for k = 1:length(bucket_cell)
for i = 1:number_of_rows_per_bucket
row_idx = round(rand(1)*316)
bag_matrix(i,:) = shuffle_X(row_idx,:)
end
bucket_cell{k} = bag_matrix
end
I could do following:
if row_idx == 0
row_idx = round(rand(1)*316)
assuming random number will never give two zeros values in two consecutive rounds.
randi is a good way to get integer indices for sampling with replacement. Assuming you want to fill three buckets with an equal number of samples, then you can write
data = rand(316,34); %# create some dummy data
number_of_data = size(data,1);
number_of_rows_per_bucket = 50;
bucket_cell = cell(1,3);
idx = randi([1,number_of_data],[number_of_rows_per_bucket,3]);
for iBucket = 1:3
bucket_cell{iBucket} = data(idx(:,iBucket),:);
end
To the question: if you use randperm it will give you a draw order without replacement, since you can draw any item once.
If you use randi it draws you with replacement, that is you draw an item possibly many times.
If you want to "segment" a dataset, that usually means you split the dataset into three distinct sets. For that you use draw without replacement (you don't put the items back; use randperm). If you'd do it with replacement (using randi), it will be incredibly slow, since after some time the chance that you draw an item which you have not before is very low.
(Details in coupon collector ).
If you need a segmentation that is a split, you can just go over the elements and independently decide where to put it. (That is you choose a bucket for each item with replacement -- that is you put any chosen bucket back into the game.)
For that:
% if your data items are vectors say data = [1 1; 2 2; 3 3; 4 4]
num_data = length(data);
bucket_labels = randi(3,[1,num_data]); % draw a bucket label for each item, independently.
for i=1:3
bucket{i} = data(bucket_labels==i,:);
end
%if your data items are scalars say data = [1 2 3 4 5]
num_data = length(data);
bucket_labels = randi(3,[1,num_data]);
for i=1:3
bucket{i} = data(bucket_labels==i);
end
there we go.
I have a data file m.txt that looks something like this (with a lot more points):
286.842995
3.444398
3.707202
338.227797
3.597597
283.740414
3.514729
3.512116
3.744235
3.365461
3.384880
Some of the values (like 338.227797) are very different from the values I generally expect (smaller numbers).
So, I am thinking that
I will remove all the points that lie outside the 3-sigma range. How can I do that in MATLAB?
Also, the bigger problem is that this file has a separate file t.txt associated with it which stores the corresponding time values for these numbers. So, I'll have to remove the corresponding time values from the t.txt file also.
I am still learning MATLAB, and I know there would be some good way of doing this (better than storing indices of the elements that were removed from m.txt and then removing those elements from the t.txt file)
#Amro is close, but the FIND is unnecessary (look up logical subscripting) and you need to include the mean for a true +/-3 sigma range. I would go with the following:
%# load files
m = load('m.txt');
t = load('t.txt');
%# find values within range
z = 3;
meanM = mean(m);
sigmaM = std(m);
I = abs(m - meanM) <= z * sigmaM;
%# keep values within range
m = m(I);
t = t(I);
%# load files
m = load('m.txt');
t = load('t.txt');
%# find outliers indices
z = 3;
idx = find( abs(m-mean(m)) > z*std(m) );
%# remove them from both data and time values
m(idx) = [];
t(idx) = [];