I have a vector for example
a = [0 1 0 3]
I want to turn a into b which equals b = [1 3].
How do I perform this in general? So I have a vector with some zero components and I want to remove the zeroes and leave just the non-zero numbers?
If you just wish to remove the zeros, leaving the non-zeros behind in a, then the very best solution is
a(a==0) = [];
This deletes the zero elements, using a logical indexing approach in MATLAB. When the index to a vector is a boolean vector of the same length as the vector, then MATLAB can use that boolean result to index it with. So this is equivalent to
a(find(a==0)) = [];
And, when you set some array elements to [] in MATLAB, the convention is to delete them.
If you want to put the zeros into a new result b, while leaving a unchanged, the best way is probably
b = a(a ~= 0);
Again, logical indexing is used here. You could have used the equivalent version (in terms of the result) of
b = a(find(a ~= 0));
but mlint will end up flagging the line as one where the purely logical index was more efficient, and thus more appropriate.
As always, beware EXACT tests for zero or for any number, if you would have accepted elements of a that were within some epsilonic tolerance of zero. Do those tests like this
b = a(abs(a) >= tol);
This retains only those elements of a that are at least as large as your tolerance.
I just came across this problem and wanted to find something about the performance, but I couldn't, so I wrote a benchmarking script on my own:
% Config:
rows = 1e6;
runs = 50;
% Start:
orig = round(rand(rows, 1));
t1 = 0;
for i = 1:runs
A = orig;
tic
A(A == 0) = [];
t1 = t1 + toc;
end
t1 = t1 / runs;
t2 = 0;
for i = 1:runs
A = orig;
tic
A = A(A ~= 0);
t2 = t2 + toc;
end
t2 = t2 / runs;
t1
t2
t1 / t2
So you see, the solution using A = A(A ~= 0) is the quicker of the two :)
I often ended up doing things like this. Therefore I tried to write a simple function that 'snips' out the unwanted elements in an easy way. This turns matlab logic a bit upside down, but looks good:
b = snip(a,'0')
you can find the function file at:
http://www.mathworks.co.uk/matlabcentral/fileexchange/41941-snip-m-snip-elements-out-of-vectorsmatrices
It also works with all other 'x', nan or whatever elements.
b = a(find(a~=0))
Data
a=[0 3 0 0 7 10 3 0 1 0 7 7 1 7 4]
Do
aa=nonzeros(a)'
Result
aa=[3 7 10 3 1 7 7 1 7 4]
Why not just, a=a(~~a) or a(~a)=[]. It's equivalent to the other approaches but certainly less key strokes.
You could use sparse(a), which would return
(1,2) 1
(1,4) 3
This allows you to keep the information about where your non-zero entries used to be.
Related
Good evening! I have a 3D vector. It has the first dimension 1. For clarity, I set it exactly the same as used in my program. "с" is like a number of experiments, in this case there are three, so I calculate the correlation function three times and then add them up.
In fact, the number of experiments is 100. I have to calculate 100 correlation functions and add them.
Tell me how you can do it automatically. And if possible, then no cycles. Thank you.
And yes, in the beginning I set the 3D vector using a loop. Is it possible to set it without a loop as well? This is certainly not my main question, but I would also like to know the answer to it.
d = [1 2 3];
c = [4 2 6];
for i = 1: length(c)
D(1,:,i) = d.*c(i);
end
D
X1 = xcorr(D(:,:,1));
X2 = xcorr(D(:,:,2));
X3 = xcorr(D(:,:,3));
X = X1+X2+X3;
With the help of a loop, my solution looks like this:
d = [1 2 3];
c = [4 2 6];
for i = 1: length(c)
D(1,:,i) = d.*c(i);
x(:,:,i) = xcorr(D(:,:,i));
end
X = sum(x,3)
It seems to be correct. Is it possible to do this without a cycle?
You can easily set your first array D without any loop, even though I don't know why you want to keep the first singleton dimension...
D(1, :, :) = d'.*c;
As for the sum of the autocorrelations, I'm not sure you can do it without a loop. The only thing that you can perhaps do is to not use an array to store the correlation for each index (if memory consumption is a problem for you) and just update the sum:
X = zeros(1, 2*length(d)-1); % initialize the sum array
for i = 1:length(c)
X = X + xcorr(D(:, :, i)); % update the sum
end
I have 2 big arrays A and b:
A: 10.000++ rows, 4 columns, not unique integers
b: vector with 500.000++ elements, unique integers
Due to the uniqueness of the values of b, I need to find the only index of b, where A(i,j) == b.
What I started with is
[rows,columns] = size(A);
B = zeros(rows,columns);
for i = 1 : rows
for j = 1 : columns
B(i,j) = find(A(i,j)==b,1);
end
end
This takes approx 5.5 seconds to compute, which is way to long, since A and b can be significantly bigger... That in mind I tried to speed up the code by using logical indexing and reducing the for-loops
[rows,columns] = size(A);
B = zeros(rows,columns);
for idx = 1 : numel(b)
B(A==b(idx)) = idx;
end
Sadly this takes even longer: 21 seconds
I even tried to do use bsxfun
for i = 1 : columns
[I,J] = find(bsxfun(#eq,A(:,i),b))
... stitch B together ...
end
but with a bigger arrays the maximum array size is quickly exceeded (102,9GB...).
Can you help me find a faster solution to this? Thanks in advance!
EDIT: I extended find(A(i,j)==b,1), which speeds up the algorithm by factor 2! Thank you, but overall still too slow... ;)
The function ismember is the right tool for this:
[~,B] = ismember(A,b);
Test code:
function so
A = rand(1000,4);
b = unique([A(:);rand(2000,1)]);
B1 = op1(A,b);
B2 = op2(A,b);
isequal(B1,B2)
tic;op1(A,b);op1(A,b);op1(A,b);op1(A,b);toc
tic;op2(A,b);op2(A,b);op2(A,b);op2(A,b);toc
end
function B = op1(A,b)
B = zeros(size(A));
for i = 1:numel(A)
B(i) = find(A(i)==b,1);
end
end
function B = op2(A,b)
[~,B] = ismember(A,b);
end
I ran this on Octave, which is not as fast with loops as MATLAB. It also doesn't have the timeit function, hence the crappy timing using tic/toc (sorry for that). In Octave, op2 is more than 100 times faster than op1. Timings will be different in MATLAB, but ismember should still be the fastest option. (Note I also replaced your double loop with a single loop, this is the same but simpler and probably faster.)
If you want to repeatedly do the search in b, it is worthwhile to sort b first, and implement your own binary search. This will avoid the checks and sorting that ismember does. See this other question.
Assuming that you have positive integers you can use array indexing:
mm = max(max(A(:)),max(b(:)));
idxs = sparse(b,1,1:numel(b),mm,1);
result = full(idxs(A));
If the range of values is small you can use dense matrix instead of sparse matrix:
mm = max(max(A(:)),max(b(:)));
idx = zeros(mm,1);
idx(b)=1:numel(b);
result = idx(A);
Suppose that we have a matrix
A = [1 2 ; 3 4 ; 5 6];
and a logical matrix
B = [false true ; true true ; true false ];
I would like to obtain the row product of elements in A such that the corresponding element in B is true. In the example above, the answer is
C = [2 ; 3*4, 5] = [2 ; 12 ; 5];
One method would be to: 1) Take the power of A with respect to B; 2) Take the row product of the power matrix:
C = prod(A.^B,2);
The above command seems to perform unnecessary computation. Is there a faster way of computing C above?
Your method seems quite fast to me. If you really have a bottleneck there, you can maybe try with cheaper operators, like addition and multiplication:
C = prod(A.*B + ~B, 2);
I only tested it with octave, but it's about twice as fast.
Another less compact way, also fast in octave:
C=A; C(~B)=1; C=prod(C,2);
Here's another way, using accumarray. I doubt it's faster:
[ii, ~] = find(B); % create grouping variable
C = accumarray(ii, A(B), [], #prod); % compute product of each group
I am working with a n x 1 matrix, A, that has repeating values inside it:
A = [0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4]
which correspond to an n x 1 matrix of B values:
B = [2;4;6;8;10; 3;5;7;9;11; 4;6;8;10;12; 5;7;9;11;13]
I am attempting to produce a generalised code to place each repetition into a separate column and store it into Aa and Bb, e.g.:
Aa = [0 0 0 0 Bb = [2 3 4 5
1 1 1 1 4 5 6 7
2 2 2 2 6 7 8 9
3 3 3 3 8 9 10 11
4 4 4 4] 10 11 12 13]
Essentially, each repetition from A and B needs to be copied into the next column and then deleted from the first column
So far I have managed to identify how many repetitions there are and copy the entire column over to the next column and then the next for the amount of repetitions there are but my method doesn't shift the matrix rows to columns as such.
clc;clf;close all
A = [0;1;2;3;4;0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
B = [2;4;6;8;10;3;5;7;9;11;4;6;8;10;12;5;7;9;11;13];
desiredCol = 1; %next column to go to
destinationCol = 0; %column to start on
n = length(A);
for i = 2:1:n-1
if A == 0;
A = [ A(:, 1:destinationCol)...
A(:, desiredCol+1:destinationCol)...
A(:, desiredCol)...
A(:, destinationCol+1:end) ];
end
end
A = [...] retrieved from Move a set of N-rows to another column in MATLAB
Any hints would be much appreciated. If you need further explanation, let me know!
Thanks!
Given our discussion in the comments, all you need is to use reshape which converts a matrix of known dimensions into an output matrix with specified dimensions provided that the number of elements match. You wish to transform a vector which has a set amount of repeating patterns into a matrix where each column has one of these repeating instances. reshape creates a matrix in column-major order where values are sampled column-wise and the matrix is populated this way. This is perfect for your situation.
Assuming that you already know how many "repeats" you're expecting, we call this An, you simply need to reshape your vector so that it has T = n / An rows where n is the length of the vector. Something like this will work.
n = numel(A); T = n / An;
Aa = reshape(A, T, []);
Bb = reshape(B, T, []);
The third parameter has empty braces and this tells MATLAB to infer how many columns there will be given that there are T rows. Technically, this would simply be An columns but it's nice to show you how flexible MATLAB can be.
If you say you already know the repeated subvector, and the number of times it repeats then it is relatively straight forward:
First make your new A matrix with the repmat function.
Then remap your B vector to the same size as you new A matrix
% Given that you already have the repeated subvector Asub, and the number
% of times it repeats; An:
Asub = [0;1;2;3;4];
An = 4;
lengthAsub = length(Asub);
Anew = repmat(Asub, [1,An]);
% If you can assume that the number of elements in B is equal to the number
% of elements in A:
numberColumns = size(Anew, 2);
newB = zeros(size(Anew));
for i = 1:numberColumns
indexStart = (i-1) * lengthAsub + 1;
indexEnd = indexStart + An;
newB(:,i) = B(indexStart:indexEnd);
end
If you don't know what is in your original A vector, but you do know it is repetitive, if you assume that the pattern has no repeats you can use the find function to find when the first element is repeated:
lengthAsub = find(A(2:end) == A(1), 1);
Asub = A(1:lengthAsub);
An = length(A) / lengthAsub
Hopefully this fits in with your data: the only reason it would not is if your subvector within A is a pattern which does not have unique numbers, such as:
A = [0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0;]
It is worth noting that from the above intuitively you would have lengthAsub = find(A(2:end) == A(1), 1) - 1;, But this is not necessary because you are already effectively taking the one off by only looking in the matrix A(2:end).
Okay, so I have a script that will produce my vector of repeated integers of a certain interval, but now theres a particular instance where I need to make sure that once it is shuffled, the numbers do not repeat. So for example, I produced a vector of repeating 1-5, 36 times, shuffled. How do I ensure that there are no repeated numbers after shuffling? And to make things even more complex, I need to produce two such vectors that do not ever have the same value at the same index. For example, lets say 1:5 was repeated twice for these vectors, so then this would be what I'm looking for:
v1 v2
4 2
2 4
3 2
5 3
4 5
1 4
5 1
1 5
3 1
2 3
I made that right now by taking an example of 1 vector and just shifting it off by 1 to create another vector that will satisfy the requirements, but in my situation, that wont actually work because I can't have them be systematically dependent like that.
So I tried a recursive technique to make the script start over if the vectors did not make the cut and as expected, that did not go over so well. I hit my maximum recursive iterations and I've realized this is clearly not the way to go. Is there some other alternative?
EDIT:
So I found a way to satisfy some of the conditions I needed above in the following code:
a = nchoosek(1:5,2);
b = horzcat(a(:,2),a(:,1));
c = vertcat(a,b);
cols = repmat(c,9,1);
cols = cols(randperm(180),:);
I just need to find a way to shuffle cols that will also enforce no repeating numbers in columns, such that cols(i,1) ~= cols(i+1,1) and cols(i,2) ~= cols(i+1,2)
This works, but it probably is not very efficient for a large array:
a = nchoosek(1:5, 2);
while (any(a(1: end - 1, 1) == a(2: end, 1)) ...
|| any(a(1: end - 1, 2) == a(2: end, 2)))
random_indices = randperm(size(a, 1));
a = a(random_indices, :);
end
a
If you want something faster, the trick is to logically insert each row in a place where your conditions are satisfied, rather than randomly re-shuffling. For example:
n1 = 5;
n2 = 9;
a = nchoosek(1:n1, 2);
b = horzcat(a(:,2), a(:,1));
c = vertcat(a, b);
d = repmat(c, n2, 1);
d = d(randperm(n1 * n2), :);
% Perform an "insertion shuffle"
for k = 2: n1 * n2
% Grab row k from array d. Walk down the rows until a position is
% found where row k does not repeat with its upstairs or downstairs
% neighbors.
m = 1;
while (any(d(k,:) == d(m,:)) || any(d(k,:) == d(m+1,:)))
m = m + 1;
end
% Insert row k in the proper position.
if (m < k)
ind = [ 1: m k m+1: k-1 k+1: n1 * n2 ];
else
ind = [ 1: k-1 k+1: m k m+1: n1 * n2 ];
end
d = d(ind,:);
end
d
One way to solve this problem is to think both vectors as being created as follows:
For every row of arrays v1 and v2
Shuffle the array [1 2 3 4 5]
Set the values of v1 and v2 at the current row with the first and second value of the shuffle. Both values will always be different.
Code:
s = [1 2 3 4 5];
Nrows = 36;
solution = zeros(Nrows,2);
for k=1:Nrows
% obtain indexes j for shuffling array s
[x,j] = sort(rand(1,5));
%row k takes the first two values of shuffled array s
solution(k,1:2) = s(j(1:2));
end
v1 = solution(:,1);
v2 = solution(:,2);
Main edit: random => rand,
With this method there is no time wasted in re-rolling repeated numbers because the first and second value of shuffling [1 2 3 4 5] will always be different.
Should you need more than two arrays with different numbers the changes are simple.