I have this code
A = unidrnd(2,100,30)-1;
B = reshape(A, 100, 3, 10);
B is a multidimensional array with 10 layers of 100x3 Matrices. Now I want to perform this code,
C = length(nonzeros(all(B,2)))/100;
where the function on the right hand side of the code is suppose to generate 10 values corresponding to the result of the 10 layers, but all I get is a single value. The right hand of the code checks how many rows are all 1's. It takes the number of rows that are all 1's and divides it by 100 to obtain the fraction of the number of rows that are all 1's.
How can I obtain the result of every 100 x 3 layers of the 3D matrix using the single line of code I have shown above such that I do not have to use a loop? The result C had to be array of the results as expected.
You started out well. all(B,2) is good, it gives you the 100x1x10 matrix that's 1 where the corresponding rows are all 1's and 0 otherwise.
nonzeros, however, simply lists all of the nonzero elements of the entire matrix, in your case, a string of 1's, completely disregarding the dimensions of the array. You'd get the same results with nonzeros(A(:)) as with nonzeros(A).
[Note: nnz(A) would get you the same results as length(nonzeros(A)), but that's not what we want to do anyway.]
Since your matrix is binary (the output of all is a logical array), we can count the number of non-zero elements by summing the matrix elements. And sum gives us a dimension argument just like all, so we just sum the columns that all gave us.
C = sum(all(B,2),1)/100;
This gives you a 1x1x10 array of percentages. If you wanted that to just be a normal vector, you could use squeeze.
C = squeeze(sum(all(B,2),1)/100);
Related
for a 5 x 10 matrix, I wanted to stack the matrix in such a way that each column of the original matrix will be appended to the rows to finally end up with a 50 x 1 matrix. Basically stack the columns into 1 column. Thanks.
Here is the start of the matrix:
RR = randi(5,5,10);
For general reshaping operations, use reshape, for example:
reshape(RR,[],1)
This reshapes the array RR into an array with an indeterminate number of rows (the []) and 1 column, as you need.
Your particular case can use the shortcut mentioned by #beaker in the comments
RR(:)
This syntax is equivalent to the reshape command above, and simply lists, in one column, all elements of the array.
Both of these examples take advantage of the fact that Matlab uses column oriented storage and indexing. If, for some reason, you need to concatenate rows rather than columns, you would need to transpose the RR array first. For example:
reshape(RR', 1, []) %This concatenates the rows of RR into a single row. Note the transpose operator.
I was trying to create random magic square in Octave and tried something like rand(magic(3)), and it gave unexpected endless results something like this:
ans(:,:,1,1,2,1,1,1,1) =
0.894903 0.296415 0.143990
0.186976 0.305691 0.505485
0.224823 0.834031 0.285508
0.336706 0.318158 0.076293
On trying rand(magic(4)) and for 5,6,7... it gave a message something like this
error: out of memory or dimension too large for Octave's index type
What can be the possible reason for this vague result ?
What are you trying to do? magic(3) creates a 3-by-3 matrix in which all the rows and columns add up to the same number. rand(x) creates an n-dimensional matrix of uniformly distributed random numbers. If you call y = rand([1,2,3]) for example, you will get a 3-dimensional matrix of uniformly distributed numbers. The dimensions of y will match your input i.e. size(y) should return [1,2,3] and the number of elements will be prod(y). Thus the number of elements of rand(magic(3)) should be equal to prod(prod(magic(3))) which is 362880. If you do this for rand(magic(4)) then the number of elements would be over 20 trillion which is why you are running out of memory.
I am working on 2D rectangular packing. In order to minimize the length of the infinite sheet (Width is constant) by changing the order in which parts are placed. For example, we could place 11 parts in 11! ways.
I could label those parts and save all possible permutations using perms function and run it one by one, but I need a large amount of memory even for 11 parts. I'd like to be able to do it for around 1000 parts.
Luckily, I don't need every possible sequence. I would like to index each permutation to a number. Test a random sequence and then use GA to converge the results to find the optimal sequence.
Therefore, I need a function which gives a specific permutation value when run for any number of times unlike randperm function.
For example, function(5,6) should always return say [1 4 3 2 5 6] for 6 parts. I don't need the sequences in a specific order, but the function should give the same sequence for same index. and also for some other index, the sequence should not be same as this one.
So far, I have used randperm function to generate random sequence for around 2000 iterations and finding a best sequence out of it by comparing length, but this works only for few number of parts. Also using randperm may result in repeated sequence instead of unique sequence.
Here's a picture of what I have done.
I can't save the outputs of randperm because I won't have a searchable function space. I don't want to find the length of the sheet for all sequences. I only need do it for certain sequence identified by certain index determined by genetic algorithm. If I use randperm, I won't have the sequence for all indexes (even though I only need some of them).
For example, take some function, 'y = f(x)', in the range [0,10] say. For each value of x, I get a y. Here y is my sheet length. x is the index of permutation. For any x, I find its sequence (the specific permutation) and then its corresponding sheet length. Based on the results of some random values of x, GA will generate me a new list of x to find a more optimal y.
I need a function that duplicates perms, (I guess perms are following the same order of permutations each time it is run because perms(1:4) will yield same results when run any number of times) without actually storing the values.
Is there a way to write the function? If not, then how do i solve my problem?
Edit (how i approached the problem):
In Genetic Algorithm, you need to crossover parents(permutations), But if you crossover permutations, you will get the numbers repeated. for eg:- crossing over 1 2 3 4 with 3 2 1 4 may result something like 3 2 3 4. Therefore, to avoid repetition, i thought of indexing each parent to a number and then convert the number to binary form and then crossover the binary indices to get a new binary number then convert it back to decimal and find its specific permutation. But then later on, i discovered i could just use ordered crossover of the permutations itself instead of crossing over their indices.
More details on Ordered Crossover could be found here
Below are two functions that together will generate permutations in lexographical order and return the nth permutation
For example, I can call
nth_permutation(5, [1 2 3 4])
And the output will be [1 4 2 3]
Intuitively, how long this method takes is linear in n. The size of the set doesn't matter. I benchmarked nth_permutations(n, 1:1000) averaged over 100 iterations and got the following graph
So timewise it seems okay.
function [permutation] = nth_permutation(n, set)
%%NTH_PERMUTATION Generates n permutations of set in lexographical order and
%%outputs the last one
%% set is a 1 by m matrix
set = sort(set);
permutation = set; %First permutation
for ii=2:n
permutation = next_permute(permutation);
end
end
function [p] = next_permute(p)
%Following algorithm from https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
%Find the largest index k such that p[k] < p[k+1]
larger = p(1:end-1) < p(2:end);
k = max(find(larger));
%If no such index exists, the permutation is the last permutation.
if isempty(k)
display('Last permutation reached');
return
end
%Find the largest index l greater than k such that p[k] < p[l].
larger = [false(1, k) p(k+1:end) > p(k)];
l = max(find(larger));
%Swap the value of p[k] with that of p[l].
p([k, l]) = p([l, k]);
%Reverse the sequence from p[k + 1] up to and including the final element p[n].
p(k+1:end) = p(end:-1:k+1);
end
I'm struggling with one of my matlab assignments. I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est. Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them. The cycle repeats 10 times chosing different values each time and adding different random number. Here is a part of my code:
steps=10;
for s=1:steps
for i=1:1:5
rl(s,i)=m_est(randi(numel(m_est)));
rl_nr(s,i)=find(rl(s,i)==m_est);
a=-1;
b=1;
r(s)=(b-a)*rand(1,1)+a;
end
pert_layers(s,:)=rl(s,:)+r(s);
M=repmat(m_est',s,1);
end
for k=steps
for m=1:1:5
M_pert=M;
M_pert(1:k,rl_nr(k,1:m))=pert_layers(1:k,1:m);
end
end
In matrix M I am storing 10 initial models and want to replace the random numbers with indices from rl_nr matrix into those stored in pert_layers matrix. However, the last loop responsible for assigning values from pert_layers to rl_nr indices does not work properly.
Does anyone know how to solve this?
Best regards
Your code uses a lot of loops and in this particular circumstance, it's quite inefficient. It's better if you actually vectorize your code. As such, let me go through your problem description one point at a time and let's code up each part (if applicable):
I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est.
I'm interpreting this as you having an array m_est of 100 elements, and with this array, you wish to create 10 different "models", where each model is 5 elements sampled from m_est. rl will store these values from m_est while rl_nr will store the indices / locations of where these values originated from. Also, for each model, you wish to add a random value to every element that is part of this model.
Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them.
Instead of doing this with a for loop, generate all of your random indices in one go. Since you have 10 steps, and we wish to sample 5 points per step, you have 10*5 = 50 points in total. As such, why don't you use randperm instead? randperm is exactly what you're looking for, and we can use this to generate unique random indices so that we can ultimately use this to sample from m_est. randperm generates a vector from 1 to N but returns a random permutation of these elements. This way, you only get numbers enumerated from 1 to N exactly once and we will ensure no repeats. As such, simply use randperm to generate 50 elements, then reshape this array into a matrix of size 10 x 5, where the number of rows tells you the number of steps you want, while the number of columns is the total number of points per model. Therefore, do something like this:
num_steps = 10;
num_points_model = 5;
ind = randperm(numel(m_est));
ind = ind(1:num_steps*num_points_model);
rl_nr = reshape(ind, num_steps, num_points_model);
rl = m_est(rl_nr);
The first two lines are pretty straight forward. We are just declaring the total number of steps you want to take, as well as the total number of points per model. Next, what we will do is generate a random permutation of length 100, where elements are enumerated from 1 to 100, but they are in random order. You'll notice that this random vector uses only a value within the range of 1 to 100 exactly once. Because you only want to get 50 points in total, simply subset this vector so that we only get the first 50 random indices generated from randperm. These random indices get stored in ind.
Next, we simply reshape ind into a 10 x 5 matrix to get rl_nr. rl_nr will contain those indices that will be used to select those entries from m_est which is of size 10 x 5. Finally, rl will be a matrix of the same size as rl_nr, but it will contain the actual random values sampled from m_est. These random values correspond to those indices generated from rl_nr.
Now, the final step would be to add the same random number to each model. You can certainly use repmat to replicate a random column vector of 10 elements long, and duplicate them 5 times so that we have 5 columns then add this matrix together with rl.... so something like:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
r = repmat(r, 1, num_points_model);
M_pert = rl + r;
Now M_pert is the final result you want, where we take each model that is stored in rl and add the same random value to each corresponding model in the matrix. However, if I can suggest something more efficient, I would suggest you use bsxfun instead, which does this replication under the hood. Essentially, the above code would be replaced with:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
M_pert = bsxfun(#plus, rl, r);
Much easier to read, and less code. M_pert will contain your models in each row, with the same random value added to each particular model.
The cycle repeats 10 times chosing different values each time and adding different random number.
Already done in the above steps.
I hope you didn't find it an imposition to completely rewrite your code so that it's more vectorized, but I think this was a great opportunity to show you some of the more advanced functions that MATLAB has to offer, as well as more efficient ways to generate your random values, rather than looping and generating the values one at a time.
Hopefully this will get you started. Good luck!
i have two matrices
r=10,000x2
q=10,000x2
i have to find out those rows of q which are one value or both values(as it is a two column matrix) different then r and allocate them in another matrix, right now i am trying this.i cannot use isequal because i want to know those rows
which are not equal this code gives me the individual elements not the complete rows different
can anyone help please
if r(:,:)~=q(:,:)
IN= find(registeredPts(:,:)~=q(:,:))
end
You can probably do this using ismember. Is this what you want? Here you get the values from q in rows that are different from r.
q=[1,2;3,4;5,6]
r=[1,2;3,5;5,6]
x = q(sum(ismember(q,r),2) < 2,:)
x =
3 4
What this do:
ismember creates an array with 1's in the positions where q == r, and 0 in the remaining positions. sum(.., 2) takes the column sum of each of these rows. If the sum is less than 2, that row is included in the new array.
Update
If the values might differ some due to floating point arithmetic, check out ismemberf from the file exchange. I haven't tested it myself, but it looks good.