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 am trying to concatenate several structs. What I take from each struct depends on a function that requires a for loop. Here is my simplified array:
t = 1;
for t = 1:5 %this isn't the for loop I am asking about
a(t).data = t^2; %it just creates a simple struct with 5 data entries
end
Here I am doing concatenation manually:
A = [a(1:2).data a(1:3).data a(1:4).data a(1:5).data] %concatenation function
As you can see, the range (1:2), (1:3), (1:4), and (1:5) can be looped, which I attempt to do like this:
t = 2;
A = [for t = 2:5
a(1:t).data
end]
This results in an error "Illegal use of reserved keyword "for"."
How can I do a for loop within the concatenate function? Can I do loops within other functions in Matlab? Is there another way to do it, other than copy/pasting the line and changing 1 number manually?
You were close to getting it right! This will do what you want.
A = []; %% note: no need to initialize t, the for-loop takes care of that
for t = 2:5
A = [A a(1:t).data]
end
This seems strange though...you are concatenating the same elements over and over...in this example, you get the result:
A =
1 4 1 4 9 1 4 9 16 1 4 9 16 25
If what you really need is just the .data elements concatenated into a single array, then that is very simple:
A = [a.data]
A couple of notes about this: why are the brackets necessary? Because the expressions
a.data, a(1:t).data
don't return all the numbers in a single array, like many functions do. They return a separate answer for each element of the structure array. You can test this like so:
>> [b,c,d,e,f] = a.data
b =
1
c =
4
d =
9
e =
16
f =
25
Five different answers there. But MATLAB gives you a cheat -- the square brackets! Put an expression like a.data inside square brackets, and all of a sudden those separate answers are compressed into a single array. It's magic!
Another note: for very large arrays, the for-loop version here will be very slow. It would be better to allocate the memory for A ahead of time. In the for-loop here, MATLAB is dynamically resizing the array each time through, and that can be very slow if your for-loop has 1 million iterations. If it's less than 1000 or so, you won't notice it at all.
Finally, the reason that HBHB could not run your struct creating code at the top is that it doesn't work unless a is already defined in your workspace. If you initialize a like this:
%% t = 1; %% by the way, you don't need this, the t value is overwritten by the loop below
a = []; %% always initialize!
for t = 1:5 %this isn't the for loop I am asking about
a(t).data = t^2; %it just creates a simple struct with 5 data entries
end
then it runs for anyone the first time.
As an appendix to gariepy's answer:
The matrix concatenation
A = [A k];
as a way of appending to it is actually pretty slow. You end up reassigning N elements every time you concatenate to an N size vector. If all you're doing is adding elements to the end of it, it is better to use the following syntax
A(end+1) = k;
In MATLAB this is optimized such that on average you only need to reassign about 80% of the elements in a matrix. This might not seam much, but for 10k elements this adds up to ~ an order of magnitude of difference in time (at least for me).
Bare in mind that this works only in MATLAB 2012b and higher as described in this thead: Octave/Matlab: Adding new elements to a vector
This is the code I used. tic/toc syntax is not the most accurate method for profiling in MATLAB, but it illustrates the point.
close all; clear all; clc;
t_cnc = []; t_app = [];
N = 1000;
for n = 1:N;
% Concatenate
tic;
A = [];
for k = 1:n;
A = [A k];
end
t_cnc(end+1) = toc;
% Append
tic;
A = [];
for k = 1:n;
A(end+1) = k;
end
t_app(end+1) = toc;
end
t_cnc = t_cnc*1000; t_app = t_app*1000; % Convert to ms
% Fit a straight line on a log scale
P1 = polyfit(log(1:N),log(t_cnc),1); P_cnc = #(x) exp(P1(2)).*x.^P1(1);
P2 = polyfit(log(1:N),log(t_app),1); P_app = #(x) exp(P2(2)).*x.^P2(1);
% Plot and save
loglog(1:N,t_cnc,'.',1:N,P_cnc(1:N),'k--',...
1:N,t_app,'.',1:N,P_app(1:N),'k--');
grid on;
xlabel('log(N)');
ylabel('log(Elapsed time / ms)');
title('Concatenate vs. Append in MATLAB 2014b');
legend('A = [A k]',['O(N^{',num2str(P1(1)),'})'],...
'A(end+1) = k',['O(N^{',num2str(P2(1)),'})'],...
'Location','northwest');
saveas(gcf,'Cnc_vs_App_test.png');
I have a 3 nested for loop that produces a vector, but each time it runs through the loops the vector that is produced changes size. I need to save each of these vectors at the ended of the for loops, so I was thinking of using mat2cell and store them in a cell. But I don't know the code get out a cell vector that will contain each of these different sized vectors.
I will give an example code
for ip = n_low:n_up
for jx = x_low:x_up
for jphi = phi_lowx:phi_upx
lx = find_path(ip,jx,jphi,0,1);
.
.
.
A_r = volumeintegrate(integr_final_r , r , z , phi);
end
end
end
Obviously, you don't know what these variables are or the numbers but I think its not needed to solve the problem. Anyways A_r is what is spit out at the end of the loops, but A_r varies in size as the loop repeats itself. I need to save every A_r vector.
Add a counter and save to a cell element: for example:
counter=0
for ...
for ...
for ...
counter=counter+1;
A_r{counter} = volumeintegrate(integr_final_r , r , z , phi);
then to extract the n-th vector just write A_r{n}
Just create a cell array:
A_r = cell(Ni, Nj, Nk)
Then create loops - note I am indexing over well behaved integers which I will use as index into my cell array, then computing the actual value for the variables you need by looking at the array iVec etc:
iVec = n_low:n_up;
Ni = numel(iVec);
jVec = x_low:x_up;
Nj = numel(jVec);
kVec = phi_lowx:phi_upx;
Nk = numel(kVec);
A_r = cell(Ni, Nj, Nk);
for ii = 1:Ni
ip = iVec(ii);
for jj = 1:Nj
jx = jVec(jj);
for kk = 1:Nk
jphi = kVec(kk);
lx = find_path(ip,jx,jphi,0,1);
....
A_r{ii,jj,kk} = volumeintegrate(integr_final_r , r , z , phi);;
end
end
end
You can now access each array in the same way that it was assigned:
A_r{ii,jj,kk}