This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
Closed 6 years ago.
I would appreaciate a lot if you help. I am beginner in programming. I am using Matlab. So, I have an array which is 431x1 type - double; there i have numbers 1 to 6; for ex: 1 4 5 3 2 6 6 3 3 5 4 1 ...; what I want to do is I need to make a new array where I would have each element repeat for 11 times; for ex: a(1:11)=1; a(12:22)=4; a(23:33)=5; or to illustrate differently : a=[1 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 4...];
I tried doing it in a loop but had some problems, which way could you suggest, do you know any function I could take advantage of?
First of all, it would help if you could format your code is separate blocks to make your question easier to read...
Let's say you had an array of length Nx1 as:
x = [1 2 3 4 5 ...]';
You could construct a loop and concatenate as:
for i = 1 : length(x)
for i = 1: length(x)
y(1 + (i - 1) * 11 : 1 + i * 11) = x(i); % Copy to a moving block
end
y(end) = []; % Delete the superfluous one at the end
You could also look at functions like repmat in the MATLAB help for replicating arrays.
Try this (NRepis how many times you want it repeated):
x = [1, 2, 3, 4, 5];
NRep = 5;
y = reshape(repmat(x,[NRep,1]),[1,length(x)*NRep])
Since it's a little cumbersome to write that out, I also particularly enjoy to use this "hack":
x = [1, 2, 3, 4, 5];
NRep = 5;
y = kron(x, ones(1,NRep));
Hope that helps!
P.S.: This is designed for row vectors only. Though if you need column vectors it's easy to modify.
edit: Of course, if you're post-R2015a you can just use y=repelem(x,NRep). I tend to forget about those because I work on older Matlabs (and sometimes it's not such a bad idea to be a bit backward compatible). Thanks to #rahnema1 for reminding me.
Related
This question already has answers here:
Construct a matrix from a set of coordinates
(5 answers)
Closed 6 years ago.
For example, I have a matrix like A=[1,2,3,4;5,6,7,8;9,10,11,12]. And two array x=[1,1,3,2] and y = [2,4,3,1], which represent X- and Y-coordinate.
And I want to get 4 elements in the matrix [A(1,2);A(1,4);A(3,3);A(2,1)]. I use this code: result = diag(A(x,y)); Although I get what I want, but if I deal with a large matrix, such code runs too slow for me. Dose someone have a better way?
thanks!
Probably not the faster ones, but following are some approaches to this:
A = [1 2 3 4;
5 6 7 8;
9 10 11 12];
x = [1 1 3 2];
y = [2 4 3 1];
%Approach-1 (Yours approach)
diagonal = diag(A(x,y))
%Approach-2
A1=A(x,y); LowUp=A1(tril(triu(A1))~=0)
%Approach-3
EYE= A1((eye(4,4).*A1)~=0)
%Approach-4
findeye=A1(find(eye(size(A1))))
%Approach-5
subind=A(sub2ind(size(A),x,y)).'
%Approach-6
for i=1:4
loop(i)=A(x(i),y(i));
end
loop=loop.'
You need sub2ind
A = [1,2,3,4;
5,6,7,8;
9,10,11,12];
x = [1,1,3,2];
y = [2,4,3,1];
id = sub2ind(size(A),x,y)
id =
4 10 9 2
A(id)
ans =
2 4 11 5
This question already has answers here:
Insert value at a specific spot in matlab vector or matrix
(2 answers)
Closed 7 years ago.
I think it will be better to give an example to describe my problem clearly.
Input:
a=[1 2 3 4 5 6 7 8 9 10]
b=[2 5 8]
Output:
c=[1 0 2 3 0 4 5 0 6 7]
I am trying to solve this problem using MATLAB, and looking for any efficient way or MATLAB in build function to solve this problem.
I also tried to solve it, but it requires one for loop and few if-else statements inside it.
Okay. I do not know which will work better, but I got an idea and thought to share it.
a = 1:10;
b = [2 5 8];
c = ones(1,10);
c(b) = zeros(size(b));
c(c~=0) = 1:(length(a)-length(b));
This question already has an answer here:
MATLAB: Duplicate each element of a vector? [closed]
(1 answer)
Closed 8 years ago.
It's hard to explain so I will show an example of what I would like to do:
x = [1 2 3 4 5]
I would like the outcome to be:
x = [1 1 2 2 3 3 4 4 5 5]
Preferably without the use of a for loop, but either method would be appreciative.
Thanks.
You can also use the Kronecker tensor product (kron function) which is pretty neat:
x = kron(x,ones(1,2))
x =
1 1 2 2 3 3 4 4 5 5
If you want it sorted as you have here, you could do:
y = sort([x x]);
alternatively if the order matters:
y = reshape([x;x],[1,2*length(x)])
Given A = [3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 8]
Output B = [3 4 5 6 8]
Is there a Matlab function or command to get this result? I am new to Matlab. Just now I am doing it going through for each element and keeping a counter for it. I have very big array so this is taking too much time.
Use a combination of unique and histc:
uA = unique(A); %// find unique values
B = uA(histc(A, uA)>=2); %// select those that appear at least twice
The above code gives the values that appear at least twice. If you want values that appear exactly twice, replace >= by ==.
I want to group my elements using the repeated segments in the array. The breaking is basically depend on where the repeated segments are, in my real data contains ~10000 elements and I want to know if there is a easier way to do that.
Here is a short example to clarify what I want:
Let's say I have an array,
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
What I want is to break A into [1 5 3],[6 9], and [9 5 2];
What is the easiest to code this using matlab??
Thanks.
For a vectorized solution, you can find out the places where either forward or backward differences to the neighbor are zero, and then use bwlabel (from the Image Processing Toolbox) and accumarray to gather the data.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
d = diff(A)==0;
%# combine forward and backward difference
%# and invert to identify non-repeating elments
goodIdx = ~([d,false]|[false,d]);
%# create list of group labels using bwlabel
groupIdx = bwlabel(goodIdx);
%# distribute the data into cell arrays
%# note that the first to inputs should be n-by-1
B = accumarray(groupIdx(goodIdx)',A(goodIdx)',[],#(x){x})
EDIT
Replace the last two lines of code with the following if you want the repeating elements to appear in the cell array as well
groupIdx = cumsum([1,abs(diff(goodIdx))]);
B = accumarray(groupIdx',A',[],#(x){x})
EDIT2
If you want to be able to split consecutive groups of identical numbers as well, you need to calculate groupIdx as follows:
groupIdx = cumsum([1,abs(diff(goodIdx))|~d.*~goodIdx(2:end)])
Here is a solution that works if I understand the question correctly. It can probably be optimised further.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
% //First get logical array of non consecutive numbers
x = [1 (diff(A)~=0)];
for nn=1:numel(A)
if ~x(nn)
if x(nn-1)
x(nn-1)=0;
end
end
end
% //Make a cell array using the logical array
y = 1+[0 cumsum(diff(find(x))~=1)];
x(x~=0) = y;
for kk = unique(y)
B{kk} = A(x==kk);
end
B{:}