Is there a way to create an array of sets in Matlab.
Eg: I have:
a = ones(10,1);
b = zeros(10,1);
I need c such that c = [(1,0); (1,0); ...], i.e. each set in c has first element from a and 2nd element from b with the corresponding index.
Also is there some way I can check if an unknown set (x,y) is in c.
Can you all please help me out? I am a Matlab noob. Thanks!
There are not sets in your understanding in MATLAB (I assume that you are thinking of tuples on Python...) But there are cells in MATLAB. That is a data type that can store pretty much anything (you may think of pointers if you are familiar with the concept). It is indicated by using { }.
Knowing this, you could come up with a cell of arrays and check them using cellfun
% create a cell of numeric arrays
C = {[1,0],[0,2],[99,-1]}
% check which input is equal to the array [1,0]
lg = cellfun(#(x)isequal(x,[1,0]),C)
Note that you access the address of a cell with () and the content of a cell with {}. [] always indicate arrays of something. We come to this in a moment.
OK, this was the part that you asked for; now there is a bonus:
That you use the term set makes me feel that they always have the same size. So why not create an array of arrays (or better an array of vectors, which is a matrix) and check this matrix column-wise?
% array of vectors (there is a way with less brackets but this way is clearer):
M = [[1;0],[0;2],[99;-1]]
% check at which column (direction ",1") all rows are equal to the proposed vector:
lg = all(M == [0;2],1)
This way is a bit clearer, better in terms of memory and faster.
Note that both variables lg are arrays of logicals. You can use them directly to index the original variable, i.e. M(:,lg) and C{lg} returns the set that you are looking for.
If you would like to get logical value regarding if p is in C, maybe you can try the code below
any(sum((C-p).^2,2)==0)
or
any(all(C-p==0,2))
Example
C = [1,2;
3,-1;
1,1;
-2,5];
p1 = [1,2];
p2 = [1,-2];
>> any(sum((C-p1).^2,2)==0) # indicating p1 is in C
ans = 1
>> any(sum((C-p2).^2,2)==0) # indicating p2 is not in C
ans = 0
>> any(all(C-p1==0,2))
ans = 1
>> any(all(C-p2==0,2))
ans = 0
Related
I'm pretty new to matlab, so I'm guessing there is some shortcut way to do this but I cant seem to find it
results = eqs\soltns;
A = results(1);
B = results(2);
C = results(3);
D = results(4);
E = results(5);
F = results(6);
soltns is a 6x1 vector and eqs is a 6x6 matrix, and I want the results of the operation in their own separate variables. It didn't let me save it like
[A, B, C, D, E, F] = eqs\soltns;
Which I feel like would make sense, but it doesn't work.
Up to now, I have never come across a MATLAB function doing this directly (but maybe I'm missing something?). So, my solution would be to write a function distribute on my own.
E.g. as follows:
result = [ 1 2 3 4 5 6 ];
[A,B,C,D,E,F] = distribute( result );
function varargout = distribute( vals )
assert( nargout <= numel( vals ), 'To many output arguments' )
varargout = arrayfun( #(X) {X}, vals(:) );
end
Explanation:
nargout is special variable in MATLAB function calls. Its value is equal to the number of output parameters that distribute is called with. So, the check nargout <= numel( vals ) evaluates if enough elements are given in vals to distribute them to the output variables and raises an assertion otherwise.
arrayfun( #(X) {X}, vals(:) ) converts vals to a cell array. The conversion is necessary as varargout is also a special variable in MATLAB's function calls, which must be a cell array.
The special thing about varargout is that MATLAB assigns the individual cells of varargout to the individual output parameters, i.e. in the above call to [A,B,C,D,E,F] as desired.
Note:
In general, I think such expanding of variables is seldom useful. MATLAB is optimized for processing of arrays, separating them to individual variables often only complicates things.
Note 2:
If result is a cell array, i.e. result = {1,2,3,4,5,6}, MATLAB actually allows to split its cells by [A,B,C,D,E,F] = result{:};
One way as long as you know the size of results in advance:
results = num2cell(eqs\soltns);
[A,B,C,D,E,F] = results{:};
This has to be done in two steps because MATLAB does not allow for indexing directly the results of a function call.
But note that this method is hard to generalize for arbitrary sizes. If the size of results is unknown in advance, it would probably be best to leave results as a vector in your downstream code.
I would like to replace the entry corresponding to the column number of an array that is part of a 3D matrix by zero. My matrix is of size IxJxJ. In each column j I can find a matrix of size IxJof which I would like to replace the jth column by zero.
You can find below an example of what I would like using a simple 3D matrix A. This example uses a loop, which is what I am trying to avoid.
A(:,:,1) = randi([1,2],5,3);
A(:,:,2) = randi([3,4],5,3);
A(:,:,3) = randi([5,6],5,3);
for i = 1:3
B = A(:,i,:);
B = squeeze(B);
B(:,i) = 0;
A(:,i,:) = B;
end
Firstly, you can replace the 4 lines of code in your for loop with just A(:,i,i) = 0;. I don't see any real need to avoid the for loop.
Using linear indexing, you can do
A((1:size(A,1)).'+size(A,1).*(size(A,2)+1).*(0:size(A,2)-1)) = 0
or for older version of Matlab without implicit expansion (pre-R2016b)
A(bsxfun(#plus,(1:size(A,1)).',size(A,1).*(size(A,2)+1).*(0:size(A,2)-1))) = 0
After some very quick testing, it actually looks like the bsxfun solution is fastest, but the differences aren't huge, your results may differ.
Use eye to create a logical mask and mutiply it by A.
A = A .* reshape(~eye(3), 1, 3, 3) ;
In MATLAB, I would like to extract a nested field for each index of a 1 x n struct (a nonscalar struct) and receive the output as a 1 x n cell array. As a simple example, suppose I start with the following struct s:
s(1).f1.fa = 'foo';
s(2).f1.fa = 'yedd';
s(1).f1.fb = 'raf';
s(2).f1.fb = 'da';
s(1).f2 = 'bok';
s(2).f2 = 'kemb';
I can produce my desired 1 x 2 cell array c using a for-loop:
n = length(s);
c = cell(1,n);
for k = 1:n
c{k} = s(k).f1.fa;
end
If I wanted to do analogously for a non-nested field, for example f2, then I could "vectorize" the operation (see this question), writing simply:
c = {s.f2};
However the same approach does not appear to work for nested fields. What then are possible ways to vectorize the above for-loop?
You cannot really vectorize it. The problem is that Matlab does not allow most forms of nested indexing, including []..
The most concise / readable option would be to concatenate s.f1 results in a structure array using [...], and then index into the new structure array with a separate call:
x = [s.f1]; c = {x.fa};
If you have a Mapping Toolbox, you could use extractfield to perform the second indexing in one expression:
c = extractfield([s.f1], 'fa');
Alternatively you could write a one-liner using arrayfun - here's a couple of options:
c = arrayfun(#(x) x.f1.fa, s, 'uni', false);
c = arrayfun(#(x) x.fa, [s.f1], 'uni', false);
Note that arrayfun and similar functions are generally slower than explicit for loops. So if the performance is critical, time / profile your code, before making a decision to get rid of the loop.
I want to compare the pixel values of two images, which I have stored in arrays.
Suppose the arrays are A and B. I want to compare the elements one by one, and if A[l] == B[k], then I want to store the match as a key value-pair in a third array, C, like so: C[l] = k.
Since the arrays are naturally quite large, the solution needs to finish within a reasonable amount of time (minutes) on a Core 2 Duo system.
This seems to work in under a second for 1024*720 matrices:
A = randi(255,737280,1);
B = randi(255,737280,1);
C = zeros(size(A));
[b_vals, b_inds] = unique(B,'first');
for l = 1:numel(b_vals)
C(A == b_vals(l)) = b_inds(l);
end
First we find the unique values of B and the indices of the first occurrences of these values.
[b_vals, b_inds] = unique(B,'first');
We know that there can be no more than 256 unique values in a uint8 array, so we've reduced our loop from 1024*720 iterations to just 256 iterations.
We also know that for each occurrence of a particular value, say 209, in A, those locations in C will all have the same value: the location of the first occurrence of 209 in B, so we can set all of them at once. First we get locations of all of the occurrences of b_vals(l) in A:
A == b_vals(l)
then use that mask as a logical index into C.
C(A == b_vals(l))
All of these values will be equal to the corresponding index in B:
C(A == b_vals(l)) = b_inds(l);
Here is the updated code to consider all of the indices of a value in B (or at least as many as are necessary). If there are more occurrences of a value in A than in B, the indices wrap.
A = randi(255,737280,1);
B = randi(255,737280,1);
C = zeros(size(A));
b_vals = unique(B);
for l = 1:numel(b_vals)
b_inds = find(B==b_vals(l)); %// find the indices of each unique value in B
a_inds = find(A==b_vals(l)); %// find the indices of each unique value in A
%// in case the length of a_inds is greater than the length of b_inds
%// duplicate b_inds until it is larger (or equal)
b_inds = repmat(b_inds,[ceil(numel(a_inds)/numel(b_inds)),1]);
%// truncate b_inds to be the same length as a_inds (if necessary) and
%// put b_inds into the proper places in C
C(a_inds) = b_inds(1:numel(a_inds));
end
I haven't fully tested this code, but from my small samples it seems to work properly and on the full-size case, it only takes about twice as long as the previous code, or less than 2 seconds on my machine.
So, if I understand your question correctly, you want for each value of l=1:length(A) the (first) index k into B so that A(l) == B(k). Then:
C = arrayfun(#(val) find(B==val, 1, 'first'), A)
could give you your solution, as long as you're sure that every element will have a match. The above solution would fail otherwise, complaning that the function returned a non-scalar (because find would return [] if no match is found). You have two options:
Using a cell array to store the result instead of a numeric array. You would need to call arrayfun with 'UniformOutput', false at the end. Then, the values of A without matches in B would be those for which isempty(C{i}) is true.
Providing a default value for an index into A with no matches in B (e.g. 0 or NaN). I'm not sure about this one, but I think that you would need to add 'ErrorHandler', #(~,~) NaN to the arrayfun call. The error handler is a function that gets called when the function passed to arrayfun fails, and may either rethrow the error or compute a substitute value. Thus the #(~,~) NaN. I am not sure that it would work, however, since in this case the error is in arrayfun and not in the passed function, but you can try it.
If you have the images in arrays A & B
idx = A == B;
C = zeros(size(A));
C(idx) = A(idx);
I have three 1-d arrays where elements are some values and I want to compare every element in one array to all elements in other two.
For example:
a=[2,4,6,8,12]
b=[1,3,5,9,10]
c=[3,5,8,11,15]
I want to know if there are same values in different arrays (in this case there are 3,5,8)
The answer given by AB is correct, but it is specific for the case when you have 3 arrays that you are comparing. There is another alternative that will easily scale to any number of arrays of arbitrary size. The only assumption is that each individual array contains unique (i.e. non-repeated) values:
>> allValues = sort([a(:); b(:); c(:)]); %# Collect all of the arrays
>> repeatedValues = allValues(diff(allValues) == 0) %# Find repeated values
repeatedValues =
3
5
8
If the arrays contains repeated values, you will need to call UNIQUE on each of them before using the above solution.
Leo is almost right, should be
unique([intersect(a,[b,c]), intersect(b,c)])
c(ismember(c,a)|ismember(c,b)),
ans =
3 5 8
I think this works for all matrices.
Define what you mean by compare. If the arrays are of the same length, and you are comparing equality then you can just do foo == bar -- it's vectorized. If you need to compare in the less than/greater than sense, you can do sign(foo-bar). If the arrays are not the same length and/or you aren't comparing element-wise -- please clarify what you'd like the output of the comparison to be. For instance,
foo = 1:3;
bar = [1,2,4];
baz = 1:2;
sign(repmat(foo',1,length([bar,baz])) - repmat([bar, baz],length(foo),1))
# or, more concisely:
bsxfun(#(x,y)sign(x-y),foo',[bar,baz])
does what you ask for, but there is probably a better way depending on what you want as an output.
EDIT (OP clarified question):
To find common elements in the 3 arrays, you can simply do:
>> [intersect(a,[b,c]), intersect(b,c)]
ans =
8 3 5