I am trying to concatenate matrices station_1, station_2,.....station_10 from my matlab workspace and trying to concatenate all stations automatically using a loop and not calling them one by one like this
cat(1,station_1,station_2,station_3,station_4... ,station_5,station_6,station_7,station_8... ,station_9,station_10 )
Any ideas?
the code below is what i was trying to improve
for jj= 1 : 10 T= cat(1,eval(['station_', num2str(jj)])); MegaMat = cat(1,T) end
Reading your code I think at the end of your loop you will have T = station_10.
If you want to concatenate all of them you would do
T = []
for jj= 1:10
T = cat(1, T, eval(['station_', num2str(jj)]));
end
MegaMat = T;
Using eval is not a good practice. Instead of creating station_1 to station_10 you could create a cell array
station{1} = ...
station{2} = ...
Then you could iterate like
T = []
for jj = 1:length(station)
T = cat(1, T, station{jj});
end
If the number of arrays is big this will be slow due to memory reallocation and copy. In that case is more efficient to initialize T as a matrix of the final dimension and write slices.
Appendix:
There is an interesting notation trick pointed by #Cris Luengo in the comments, that is when you have a cell array station, use the notation [station{:}], I have to admit, this notation is new to me. The only caveat is that if you set the items station{i} = ... then you will have the matrices concatenated horizontally rather than vertically.
The answer from #Mateo V, is also good, probably with leas overhead since it calls eval only once. That approach can be refined giving a one linear solution, and to be honest It felt not very unreadable.
MegaMat = eval(['cat(1', num2str(1:10, ', station_%d'), ')']);
No need for loops:
str = num2str(1:10, 'station_%i,'); % returns string 'station_1, station_2, ..., station_10,'
str = str(1:end-1); % remove last comma
eval(['MegaMat = cat(1, ', str, ');'])
I have a variable pth which is a cell array of dimension 1xn where n is a user input. Each of the elements in pth is itself a cell array and length(pth{k}) for k=1:n is variable (result of another function). Each element pth{k}{kk} where k=1:n and kk=1:length(pth{k}) is a 1D vector of integers/node numbers of again variable length. So to summarise, I have a variable number of variable-length vectors organised in a avriable number of cell arrays.
I would like to try and find all possible intersections when you take a vector at random from pth{1}, pth{2}, {pth{3}, etc... There are various functions on the File Exchange that seem to do that, for example this one or this one. The problem I have is you need to call the function this way:
mintersect(v1,v2,v3,...)
and I can't write all the inputs in the general case because I don't know explicitly how many there are (this would be n above). Ideally, I would like to do some thing like this;
mintersect(pth{1}{1},pth{2}{1},pth{3}{1},...,pth{n}{1})
mintersect(pth{1}{1},pth{2}{2},pth{3}{1},...,pth{n}{1})
mintersect(pth{1}{1},pth{2}{3},pth{3}{1},...,pth{n}{1})
etc...
mintersect(pth{1}{1},pth{2}{length(pth{2})},pth{3}{1},...,pth{n}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{2},...,pth{n}{1})
etc...
keep going through all the possible combinations, but I can't write this in code. This function from the File Exchange looks like a good way to find all possible combinations but again I have the same problem with the function call with the variable number of inputs:
allcomb(1:length(pth{1}),1:length(pth{2}),...,1:length(pth{n}))
Does anybody know how to work around this issue of function calls with variable number of input arguments when you can't physically specify all the input arguments because their number is variable? This applies equally to MATLAB and Octave, hence the two tags. Any other suggestion on how to find all possible combinations/intersections when taking a vector at random from each pth{k} welcome!
EDIT 27/05/20
Thanks to Mad Physicist's answer, I have ended up using the following which works:
disp('Computing intersections for all possible paths...')
grids = cellfun(#(x) 1:numel(x), pth, 'UniformOutput', false);
idx = cell(1, numel(pth));
[idx{:}] = ndgrid(grids{:});
idx = cellfun(#(x) x(:), idx, 'UniformOutput', false);
idx = cat(2, idx{:});
valid_comb = [];
k = 1;
for ii = idx'
indices = reshape(num2cell(ii), size(pth));
selection = cellfun(#(p,k) p{k}, pth, indices, 'UniformOutput', false);
if my_intersect(selection{:})
valid_comb = [valid_comb k];
endif
k = k+1;
end
My own version is similar but uses a for loop instead of the comma-separated list:
disp('Computing intersections for all possible paths...')
grids = cellfun(#(x) 1:numel(x), pth, 'UniformOutput', false);
idx = cell(1, numel(pth));
[idx{:}] = ndgrid(grids{:});
idx = cellfun(#(x) x(:), idx, 'UniformOutput', false);
idx = cat(2, idx{:});
[n_comb,~] = size(idx);
temp = cell(n_pipes,1);
valid_comb = [];
k = 1;
for k = 1:n_comb
for kk = 1:n_pipes
temp{kk} = pth{kk}{idx(k,kk)};
end
if my_intersect(temp{:})
valid_comb = [valid_comb k];
end
end
In both cases, valid_comb has the indices of the valid combinations, which I can then retrieve using something like:
valid_idx = idx(valid_comb(1),:);
for k = 1:n_pipes
pth{k}{valid_idx(k)} % do something with this
end
When I benchmarked the two approaches with some sample data (pth being 4x1 and the 4 elements of pth being 2x1, 9x1, 8x1 and 69x1), I got the following results:
>> benchmark
Elapsed time is 51.9075 seconds.
valid_comb = 7112
Elapsed time is 66.6693 seconds.
valid_comb = 7112
So Mad Physicist's approach was about 15s faster.
I also misunderstood what mintersect did, which isn't what I wanted. I wanted to find a combination where no element present in two or more vectors, so I ended writing my version of mintersect:
function valid_comb = my_intersect(varargin)
% Returns true if a valid combination i.e. no combination of any 2 vectors
% have any elements in common
comb_idx = combnk(1:nargin,2);
[nr,nc] = size(comb_idx);
valid_comb = true;
k = 1;
% Use a while loop so that as soon as an intersection is found, the execution stops
while valid_comb && (k<=nr)
temp = intersect(varargin{comb_idx(k,1)},varargin{comb_idx(k,2)});
valid_comb = isempty(temp) && valid_comb;
k = k+1;
end
end
Couple of helpful points to construct a solution:
This post shows you how to construct a Cartesian product between arbitrary arrays using ndgrid.
cellfun accepts multiple cell arrays simultaneously, which you can use to index specific elements.
You can capture a variable number of arguments from a function using cell arrays, as shown here.
So let's get the inputs to ndgrid from your outermost array:
grids = cellfun(#(x) 1:numel(x), pth, 'UniformOutput', false);
Now you can create an index that contains the product of the grids:
index = cell(1, numel(pth));
[index{:}] = ndgrid(grids{:});
You want to make all the grids into column vectors and concatenate them sideways. The rows of that matrix will represent the Cartesian indices to select the elements of pth at each iteration:
index = cellfun(#(x) x(:), index, 'UniformOutput', false);
index = cat(2, index{:});
If you turn a row of index into a cell array, you can run it in lockstep over pth to select the correct elements and call mintersect on the result.
for i = index'
indices = num2cell(i');
selection = cellfun(#(p, i) p{i}, pth, indices, 'UniformOutput', false);
mintersect(selection{:});
end
This is written under the assumption that pth is a row array. If that is not the case, you can change the first line of the loop to indices = reshape(num2cell(i), size(pth)); for the general case, and simply indices = num2cell(i); for the column case. The key is that the cell from of indices must be the same shape as pth to iterate over it in lockstep. It is already generated to have the same number of elements.
I believe this does the trick. Calls mintersect on all possible combinations of vectors in pth{k}{kk} for k=1:n and kk=1:length(pth{k}).
Using eval and messing around with sprintf/compose a bit. Note that typically the use of eval is very much discouraged. Can add more comments if this is what you need.
% generate some data
n = 5;
pth = cell(1,n);
for k = 1:n
pth{k} = cell(1,randi([1 10]));
for kk = 1:numel(pth{k})
pth{k}{kk} = randi([1 100], randi([1 10]), 1);
end
end
% get all combs
str_to_eval = compose('1:length(pth{%i})', 1:numel(pth));
str_to_eval = strjoin(str_to_eval,',');
str_to_eval = sprintf('allcomb(%s)',str_to_eval);
% use eval to get all combinations for a given pth
all_combs = eval(str_to_eval);
% and make strings to eval in intersect
comp = num2cell(1:numel(pth));
comp = [comp ;repmat({'%i'}, 1, numel(pth))];
str_pattern = sprintf('pth{%i}{%s},', comp{:});
str_pattern = str_pattern(1:end-1); % get rid of last ,
strings_to_eval = cell(length(all_combs),1);
for k = 1:size(all_combs,1)
strings_to_eval{k} = sprintf(str_pattern, all_combs(k,:));
end
% and run eval on all those strings
result = cell(length(all_combs),1);
for k = 1:size(all_combs,1)
result{k} = eval(['mintersect(' strings_to_eval{k} ')']);
%fprintf(['mintersect(' strings_to_eval{k} ')\n']); % for debugging
end
For a randomly generated pth, the code produces the following strings to evaluate (where some pth{k} have only one cell for illustration):
mintersect(pth{1}{1},pth{2}{1},pth{3}{1},pth{4}{1},pth{5}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{1},pth{4}{2},pth{5}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{1},pth{4}{3},pth{5}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{2},pth{4}{1},pth{5}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{2},pth{4}{2},pth{5}{1})
mintersect(pth{1}{1},pth{2}{1},pth{3}{2},pth{4}{3},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{1},pth{4}{1},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{1},pth{4}{2},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{1},pth{4}{3},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{2},pth{4}{1},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{2},pth{4}{2},pth{5}{1})
mintersect(pth{1}{2},pth{2}{1},pth{3}{2},pth{4}{3},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{1},pth{4}{1},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{1},pth{4}{2},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{1},pth{4}{3},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{2},pth{4}{1},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{2},pth{4}{2},pth{5}{1})
mintersect(pth{1}{3},pth{2}{1},pth{3}{2},pth{4}{3},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{1},pth{4}{1},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{1},pth{4}{2},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{1},pth{4}{3},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{2},pth{4}{1},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{2},pth{4}{2},pth{5}{1})
mintersect(pth{1}{4},pth{2}{1},pth{3}{2},pth{4}{3},pth{5}{1})
As Madphysicist pointed out, I misunderstood the initial structure of your initial cell array, however the point stands. The way to pass an unknown number of arguments to a function is via comma-separated-list generation, and your function needs to support it by being declared with varargin. Updated example below.
Create a helper function to collect a random subcell from each main cell:
% in getRandomVectors.m
function Out = getRandomVectors(C) % C: a double-jagged array, as described
N = length(C);
Out = cell(1, N);
for i = 1 : length(C)
Out{i} = C{i}{randi( length(C{i}) )};
end
end
Then assuming you already have an mintersect function defined something like this:
% in mintersect.m
function Intersections = mintersect( varargin )
Vectors = varargin;
N = length( Vectors );
for i = 1 : N; for j = 1 : N
Intersections{i,j} = intersect( Vectors{i}, Vectors{j} );
end; end
end
Then call this like so:
C = { { 1:5, 2:4, 3:7 }, {1:8}, {2:4, 3:9, 2:8} }; % example double-jagged array
In = getRandomVectors(C); % In is a cell array of randomly selected vectors
Out = mintersect( In{:} ); % Note the csl-generator syntax
PS. I note that your definition of mintersect differs from those linked. It may just be you didn't describe what you want too well, in which case my mintersect function is not what you want. What mine does is produce all possible intersections for the vectors provided. The one you linked to produces a single intersection which is common to all vectors provided. Use whichever suits you best. The underlying rationale for using it is the same though.
PS. It is also not entirely clear from your description whether what you're after is a random vector k for each n, or the entire space of possible vectors over all n and k. The above solution does the former. If you want the latter, see MadPhysicist's solution on how to create a cartesian product of all possible indices instead.
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 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