I am trying to create two vectors
first:
[ 10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10 ]
second:
[ 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 ]
I had almost no problems with the first one:
i = 1:10;
for t = 1: 2*length(i)-1
y1(t) = abs(length(x)-t)+1;
end
But there are some problems with the second...
Does anyone have any idea how I can create it using the same for loop?Thanks in advance
If you want to do it with a loop:
N = 10;
for t = 1:2*N-1
y2(t) = -abs(t-N)+N;
end
But its probably easier to use the following, first make an array 1:N, and then concatenate the array N-1:-1:1:
y2 = [1:N, N-1:-1:1]
I have an IxS matrix called numerateur and I would like to set one element per row to zero. However, the column to which this element belongs changes for each row. I can do it with a loop but would like to use a more efficient method.
You can find my loop below:
choice is an Ix1 vector containing, for each row, the column number of the element I want to set to zero in numerateur.
for i = 1:I
rank1 = choice(i);
numerateur(i,rank1) = 0;
end
You need to convert the column indices to linear indices:
ind = (1:size(numerateur,1)) + (choice(:).'-1)*size(numerateur,1);
numerateur(ind) = 0;
Example:
>> numerateur = randi(9,5,7)
numerateur =
8 7 8 3 2 8 4
7 8 6 3 6 1 3
8 9 5 5 5 9 5
4 4 2 7 4 9 9
1 6 7 3 8 7 9
>> choice = [2; 3; 6; 5; 1];
>> ind = (1:size(numerateur,1)) + (choice(:).'-1)*size(numerateur,1);
>> numerateur(ind) = 0;
>> numerateur
numerateur =
8 0 8 3 2 8 4
7 8 0 3 6 1 3
8 9 5 5 5 0 5
4 4 2 7 0 9 9
0 6 7 3 8 7 9
Use implicit expansion to find logical indices:
numerateur(choice == 1:size(numerator,2)) = 0;
Alternatively you can use sparse matrix to generate logical indices:
numerateur(sparse(1:I,choice,true,I,S)) = 0;
This question already has an answer here:
Multiply 2D Matrix with vector to span third dimension - MATLAB
(1 answer)
Closed 7 years ago.
Here what I would like to achieve:
I have a matrix C
C=[1 2 3; 4 5 6; 7 8 9];
And a vector a
a=[1 2];
I would like to make such an operation, that each element of the a vector is multiplied with C ( scalar multiplication ) and out comes a 3-dimensional array D:
(:,:,1) =
1 2 3
4 5 6
7 8 9
(:,:,2) =
2 4 6
8 10 12
14 16 18
It would certainly work with a loop, but, since I'll need this operation on multiple occasions, a oneliner would be a great saver.
This is a beautiful example for the use of bsxfun and reshape. While #thewaywewalks proposes first calling bsxfun and reshaping the result, I'd suggest the opposite. This makes one of the key concepts of bsxfun - the singleton dimension expansion - more clear:
out = bsxfun(#times,C,reshape(a,1,1,[]))
ans(:,:,1) =
1 2 3
4 5 6
7 8 9
ans(:,:,2) =
2 4 6
8 10 12
14 16 18
With reshape(a,1,1,[]), you make a be in the third dimension. If you now apply bsxfun, it will multiply the matrix C with each element of a.
Some reshape'ing and some bsxfun will do:
out = reshape(bsxfun(#mtimes, C(:), a(:).'), [size(C),numel(a)] )
As suggested in hbaderts answer one could also use bsxfun's capability of dimension expansion, and provide a permuted vector of factors:
out = bsxfun(#mtimes,C,permute(a,[3,1,2]))
out(:,:,1) =
1 2 3
4 5 6
7 8 9
out(:,:,2) =
2 4 6
8 10 12
14 16 18
I have another method for the benchmark compare...
IMO it's the neatest way, at least for the syntax/readability term:
out = reshape(kron(a,C),[size(C),numel(a)]);
out(:,:,1) =
1 2 3
4 5 6
7 8 9
out(:,:,2) =
2 4 6
8 10 12
14 16 18
Another possibility is to use matrix multiplication of C as a column vector times a as a row vector (this gives all element-wise products), and then reshape the result:
out = reshape(C(:)*a, size(C,1), size(C,2), numel(a));
EDIT (BENCHMARKING): Since several solutions (including mine below) have been suggested, here is some rough benchmarking to compare the different solutions, using larger arrays:
a=1:10;
N=1000; timers=zeros(N,6);
for ii=1:N; C=rand(400);
tic; out = repmat(C,[1,1,numel(a)]).*reshape(repelem(a,size(C,1),size(C,2)),[size(C),numel(a)]); timers(ii,1)=toc;
tic; out = bsxfun(#times,C,reshape(a,1,1,[])); timers(ii,2)=toc;
tic; out = reshape(C(:)*a, size(C,1), size(C,2), numel(a)); timers(ii,3)=toc;
tic; out = bsxfun(#mtimes,C,permute(a,[3,1,2])); timers(ii,4)=toc;
tic; out = reshape(bsxfun(#mtimes, C(:), a(:).'), [size(C),numel(a)] ); timers(ii,5)=toc;
tic; out = reshape(kron(a,C),[size(C),numel(a)]); timers(ii,6)=toc;
end;
mean(timers)
ans =
0.0080863 0.0032406 0.0041718 0.015166 0.0074462 0.0033051
... suggesting that #hbaderts solution is fastest, then #Adiel's, then #Luis Mendo's, then #thewaywewalk's (1), then mine, then #thewaywewalk's (2).
My solution:
Another option, using repmat and reshape (no bsxfun):
out = repmat(C,[1,1,numel(a)]).*reshape(repelem(a,size(C,1),size(C,2)),[size(C),numel(a)])
out(:,:,1) =
1 2 3
4 5 6
7 8 9
out(:,:,2) =
2 4 6
8 10 12
14 16 18
This is the element-wise multiplication of two arrays. The first is your original matrix C repeated numel(a) times in the third dimension:
repmat(C,[1,1,numel(a)])
ans(:,:,1) =
1 2 3
4 5 6
7 8 9
ans(:,:,2) =
1 2 3
4 5 6
7 8 9
The second is the same size as the first, with each slice containing the corresponding element of a:
reshape(repelem(a,size(C,1),size(C,2)),[size(C),numel(a)])
ans(:,:,1) =
1 1 1
1 1 1
1 1 1
ans(:,:,2) =
2 2 2
2 2 2
2 2 2
I have a 1 x 15 array of values:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
I need to rearrange them into a 3 x 5 matrix using a for loop:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
How would I do that?
I'm going to show you three methods. One where you need to have a for loop, and two others when you don't:
Method #1 - for loop
First, create a matrix that is 3 x 5, then keep track of an index that will go through your array. After, create a double for loop that will help you populate the array.
index = 1;
array = 1 : 15; %// Array we wish to access
matrix = zeros(3,5); %// Initialize
for m = 1 : 3
for n = 1 : 5
matrix(m,n) = array(index);
index = index + 1;
end
end
matrix =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Method #2 - Without a for loop
Simply put, use reshape:
matrix = reshape(1:15, 5, 3).';
matrix =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
reshape will take a vector and restructure it into a matrix so that you populate the matrix by columns first. As such, we want to put 1 to 5 in the first column, 6 to 10 in the second and 11 to 15 in the third column. Therefore, our output matrix is in fact 5 x 3. When you see this, this is actually the transposed version of the matrix we want, which is why you do .' to transpose the matrix back.
Method #3 - Another method without a for loop (tip of the hat goes to Luis Mendo)
You can use vec2mat, and specify that you need to have 5 columns worth for your matrix:
matrix = vec2mat(1:15, 5);
matrix =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
vec2mat takes a vector and reshapes it into a matrix of as many columns as you specify in the second parameter. In this case, we need 5 columns.
For the sake of (bsx)fun, here is another option...
bsxfun(#plus,1:5,[0:5:10]')
ans =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
less readable, maybe faster, but who cares if it is such a small of an array...
A = [ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ] ;
A = reshape( A' , 3 , 5 ) ;
A' = 1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
I have an array A, and want to reshape the last four elements of each column into a 2x2 matrix. I would like the results to be stored in a cell array B.
For example, given:
A = [1:6; 3:8; 5:10]';
I would like B to contain three 2x2 arrays, such that:
B{1} = [3, 5; 4, 6];
B{2} = [5, 7; 6, 8];
B{3} = [7, 9; 8, 10];
I can obviously do this in a for loop using something like reshape(A(end-3:end, ii), 2, 2) and looping over ii. Can anyone propose a vectorized method, perhaps using something similar to cellfun that can apply an operation repeatedly to columns of an array?
The way I do this is to look at the desired indices and then figure out a way to generate them, usually using some form of repmat. For example, if you want the last 4 items in each column, the (absolute) indices into A are going to be 3,4,5,6, then add the number of rows to that to move to the next column to get 9,10,11,12 and so on. So the problem becomes generating that matrix in terms of your number of rows, number of columns, and the number of elements you want from each column (I'll call it n, in your case n=4).
octave:1> A = [1:6; 3:8; 5:10]'
A =
1 3 5
2 4 6
3 5 7
4 6 8
5 7 9
6 8 10
octave:2> dim=size(A)
dim =
6 3
octave:3> n=4
n = 4
octave:4> x=repmat((dim(1)-n+1):dim(1),[dim(2),1])'
x =
3 3 3
4 4 4
5 5 5
6 6 6
octave:5> y=repmat((0:(dim(2)-1)),[n,1])
y =
0 1 2
0 1 2
0 1 2
0 1 2
octave:6> ii=x+dim(1)*y
ii =
3 9 15
4 10 16
5 11 17
6 12 18
octave:7> A(ii)
ans =
3 5 7
4 6 8
5 7 9
6 8 10
octave:8> B=reshape(A(ii),sqrt(n),sqrt(n),dim(2))
B =
ans(:,:,1) =
3 5
4 6
ans(:,:,2) =
5 7
6 8
ans(:,:,3) =
7 9
8 10
Depending on how you generate x and y, you can even do away with the multiplication, but I'll leave that to you. :D
IMO you don't need a cell array to store them either, a 3D matrix works just as well and you index into it the same way (but don't forget to squeeze it before you use it).
I gave a similar answer in this question.