When indexing matrices in MATLAB, can I specify only the first or last n dimensions, and have all others dimensions "selected automatically"?
For example, I am writing a function which takes in an image, and displays it with imshow, which can either display a 3-D color image (e.g 1024×768×3) or a 2-D monochrome array (e.g 1024x768).
My function does not care about how many color channels the image has, imshow will take care of that. All I want to do is pass parameters to select a single region:
imshow(frame(x1:x2, y1:y2, :))
What do I put in place of the last colon to say "include all the others dimensions"?
You can use comma-separated-list expansion together with the ':' indexing.
Suppose your input is:
A = rand([7,4,2,3]);
To retrieve only first 2:
cln = {':', ':'};
A(cln{:})
To retrieve the last 3:
cln = {1, ':', ':', ':'};
A(cln{:})
Which can be generalized with:
sten = 2:3; % Which dims to retrieve
cln(1:ndims(A)) = {1};
cln(sten) = {':'};
A(cln{:})
Following from Oleg's answer, here is a function that will work if you are selecting from several of the first dimensions. If other dimensions are needed, I think you can see how to modify.
function [dat] = getblock2(dat, varargin)
%[dat] = getblock(dat, varargin) select subarray and retain all others
% unchanged
%dat2 = getblock(dat, [1,2], [3,5]) is equivalent to
% dat2 = dat(1:2, 3:5, :, :, :) etc.
%Peter Burns 4 June 2013
arg1(1:ndims(dat)) = {':,'};
v = cell2mat(varargin);
nv = length(v)/2;
v = reshape(v,2,nv)';
for ii=1:nv
arg1{ii} = [num2str(v(ii,1)),':',num2str(v(ii,2)),','];
end
arg2 = cell2mat(arg1);
arg2 = ['dat(',arg2(1:end-1),')'];
dat = eval(arg2);
Related
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.
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:
I would like to name variable (type double) in the following way:
k0 = D(1,1);
k1 = D(2,2);
k2 = D(3,3);
k3 = D(4,4);
k4 = D(5,5);
k5 = D(6,6);
k6 = D(7,7);
k7 = D(8,8);
...
up to k99 automatically using for loop. So I see that I should use array or cell instead of double variable using eval as it is slow. But if I should use array or cell instead of double variable, I have to start at k{1} or k(1), which loses the meaning as I want exactly that k0 refers to D(1,1), i.e. the number in my variable is 1 less. How do I create meaningful cell name like k{0}?
Also, say I have an array A. There are also some times i need meaningful variable name, such as
c111 = A(1)*A(1)*A(1)
c222 = A(2)*A(2)*A(2)
c333 = A(3)*A(3)*A(3)
How can I create c{111} efficiently using for loop?
Use structures:
D = rand(21);
c = 1;
for k = -10:10
if k<0
s.(['k_' num2str(abs(k))]) = D(c,c);
else
s.(['k' num2str(k)]) = D(c,c);
end
c = c+1;
end
This will give you a structure like:
s =
k_10: 0.51785
k_9: 0.90121
k_8: 0.40746
k_7: 0.092989
.
.
k_1: 0.75522
k0: 0.55257
k1: 0.28708
.
.
k9: 0.94182
k10: 0.2124
and don't use eval...
Answer to 1st Question:-
D=randn(100); % A matrix of random elements of size 8x8
for m=0:99
assignin('base', ['k' num2str(m)], D(m+1,m+1))
end
Answer to 2nd Question:-
A=randn(1,3); % An array of 3 random elements
for n=1:3
assignin('base', ['c' num2str(111*n)], A(n)^3)
end
Comments:-
You've stated that you need variables like k0,k1,k2,... and c111,c222,c333 but you're asking how to create k{0}, k{1},k{2},... and c{111},c{222},c{333}. As far as your need is concerned, I have given answer to it. Regarding the latter, k{0} is never possible and c{111},c{222},c{333},... don't make good sense without using any of the first 0:100 values and then 112:221 values and so on. Although you can do it using:
A=rand(1,3); % An array of 3 random elements
c{333} = 0 ; % Pre-allocation
for p=1:3 % Since you want to use a 'for loop'
c{111*p} = A(p)^3;
end
And regarding the requirement that you made in the comment in these words "I also have some variable using negative index", you can never have variables in the negative index. If you mean you want to create variables with names like k-1, k-2,... etc, it is not possible. An alternate way is to use k_1, k_2,... etc but then as you said in the question "k0 refers to D(1,1), i.e. the number in my variable is 1 less". It means k_1 will refer to D(0,0) and so on which is again an invalid thing for MATLAB.
Recommendation:-
You really need to modify your code.
I've created some Matlab code which anyone helping can run and see the problem.
When I run the following code, for each data point on my plot I seem to get all 15 labels instead of only 1 specific label.
So how do I get the Matlab data point labels correct for the following code?
Based on the the suggestions, I did the following:
I replaced these two lines of code:
labels = num2str(test_vector_label,'F%d');
labels_cell = cellstr(labels);
With this line of code as suggested:
labels_cell = strread(num2str(test_vector_label),'%s');
Now there are two follow-up questions:
1) A warning appears stating that I should use textscan instead of strread:
labels_cell = textscan(num2str(test_vector_label),'%s');
Then when I use textscan as in the above line of code above, I get an error?
"Error using text
Cell array of strings may only contain string and numeric
matrices"
"Error in Code_Test (line 46)
text(x_val,y_val,labels_cell,'horizontal','left',
'vertical','bottom')"
2) How do I put a letter in front of the number labels? For example, in the original code I had put letter F followed by a number?
%--------------Randomly select training and testing data.-----------
num_data = 35;
data_idx = 1:35;
train_data_idx_tmp = randsample(num_data,20)
train_dataRand_idx = sort(train_data_idx_tmp)
% Lia = ismember(A,B) returns an array the same size as A, containing 1 (true)
% where the elements of A are found in B, and 0 (false) elsewhere.
test_data_idx_tmp = ismember(data_idx,train_dataRand_idx)
test_dataRand_idx = data_idx(~test_data_idx_tmp)'
% Check to see if training and test data index are exclusive.
check_train_test_idx = ismember(train_dataRand_idx,test_dataRand_idx)
%--------------------------------------------------------------------------
% Testing stage.
test_vector = test_dataRand_idx; %Select randomly obtained testing data.
% Training stage.
train_vector = train_dataRand_idx; %Select randomly obtained training
x_val = [1:15];
y_val = 2*[1:15];
plot(x_val,y_val,'or','MarkerFaceColor','r')
grid on
%Put specific data point labels on plots.
test_vector_label = test_vector';
labels = num2str(test_vector_label,'F%d');
labels_cell = cellstr(labels);
text(x_val,y_val,labels_cell,'horizontal','left', 'vertical','bottom')
Your variable labels_cell is a 1x1 string cell not an array of strings. Replace
labels = num2str(test_vector_label,'F%d');
labels_cell = cellstr(labels);
with
labels_cell = strread(num2str(test_vector_label),'%s');
I want to create a MATLAB function to import data from files in another directory and fit them to a given model, but because the data need to be filtered (there's "thrash" data in different places in the files, eg. measurements of nothing before the analyzed motion starts).
So the vectors that contain the data used to fit end up having different lengths and so I can't return them in a matrix (eg. x in my function below). How can I solve this?
I have a lot of datafiles so I don't want to use a "manual" method. My function is below. All and suggestions are welcome.
datafit.m
function [p, x, y_c, y_func] = datafit(pattern, xcol, ycol, xfilter, calib, p_calib, func, p_0, nhl)
datafiles = dir(pattern);
path = fileparts(pattern);
p = NaN(length(datafiles));
y_func = [];
for i = 1:length(datafiles)
exist(strcat(path, '/', datafiles(i).name));
filename = datafiles(i).name;
data = importdata(strcat(path, '/', datafiles(i).name), '\t', nhl);
filedata = data.data/1e3;
xdata = filedata(:,xcol);
ydata = filedata(:,ycol);
filter = filedata(:,xcol) > xfilter(i);
x(i,:) = xdata(filter);
y(i,:) = ydata(filter);
y_c(i,:) = calib(y(i,:), p_calib);
error = #(par) sum(power(y_c(i,:) - func(x(i,:), par),2));
p(i,:) = fminsearch(error, p_0);
y_func = [y_func; func(x(i,:), p(i,:))];
end
end
sample data: http://hastebin.com/mokocixeda.md
There are two strategies I can think of:
I would return the data in a vector of cells instead, where the individual cells store vectors of different lengths. You can access data the same way as arrays, but use curly braces: Say c{1}=[1 2 3], c{2}=[1 2 10 8 5] c{3} = [ ].
You can also filter the trash data upon reading a line, if that makes your vectors have the same length.
If memory is not an major issue, try filling up the vectors with distinct values, such as NaN or Inf - anything, that is not found in your measurements based on their physical context. You might need to identify the longest data-set before you allocate memory for your matrices (*). This way, you can use equally sized matrices and easily ignore the "empty data" later on.
(*) Idea ... allocate memory based on the size of the largest file first. Fill it up with e.g. NaN's
matrix = zeros(length(datafiles), longest_file_line_number) .* NaN;
Then run your function. Determine the length of the longest consecutive set of data.
new_max = length(xdata(filter));
if new_max > old_max
old_max = new_max;
end
matrix(i, length(xdata(filter))) = xdata(filter);
Crop your matrix accordingly, before the function returns it ...
matrix = matrix(:, 1:old_max);