I have a two column matrix of the following form:
1. 1 1
2. 1 1
3. 1 2
4. 1 2
5. 2 2
6. 2 2
7. 3 2
8. 3 2
9. 3 3
10. 4 3
11. 4 4
I would like to sample a single number from the first column using say randsample().
Let's say the results is 2.
What I would like to know is which ROW was the sample taken from? (in this case it could have been sampled both from row 5 or row 6)
Is this possible?
It's easy with find and ==:
>> A = [
1 1
1 1
1 2
1 2
2 2
2 2
3 2
3 2
3 3
4 3
4 4];
>> R = randsample(4,1)
>> find(A(:,1) == R)
R =
4
ans =
10
11
Or, as indicated by igor milla,
>> I = randi(11)
>> A(I, :)
I =
9
ans =
3 3
If you just need to sample one value, the solution as given by #igor milla is fine. But if you want to use the options given by randsample then I would recommend you to sample the column numbers rather than the sample directly.
A = rand(11,2); %suppose this is your matrix
k = 1; %This is the size of your desired sample
mysampleid = randsample(size(A,1),k)
mysample = A(mysampleid,:)
Now mysampleid contains the numbers of the columns, and mysample contains the rows that you sampled.
If you just want to sample the first column you can use A(mysampleid,1) instead.
Related
my matrix:
e =
1 2
2 3
3 3
4 3
5 2
i want to repeat value from first coloumn as much as number from the second coloumn in the same row. i want to make my matrix to be like:
e =
1 2
1 2
2 3
2 3
2 3
3 3
3 3
3 3
4 3
4 3
4 3
5 2
5 2
thank you for your help...
You can use repelem to repeat the row indices and then grab those rows from e:
new_e = e(repelem(1:size(e,1), e(:,2)), :);
If you're using a MATLAB version prior to 2015a that doesn't have repelem, here's another way to do it:
spacing = cumsum([1; e(:,2)]); % the rows of new_e where we change row values
row_indices(spacing) = 1; % make a vector with these elements = 1
row_indices = cumsum(row_indices); % convert to row indices, last index is invalid
new_e = e(row_indices(1:end-1), :); % select valid rows from e
I want to do the following:
I create a matrix with all possible permutations from 1:n, for example
n=4;
L=perms(1:n)';
I get as output as expected a 4-by-24 matrix:
L =
Columns 1 through 13
4 4 4 4 4 4 3 3 3 3 3 3
3 3 2 2 1 1 4 4 2 2 1 1
2 1 3 1 2 3 2 1 4 1 2 4
1 2 1 3 3 2 1 2 1 4 4 2
Columns 14 through 24
2 2 2 2 2 1 1 1 1 1 1
3 4 4 1 1 3 3 2 2 4 4
1 3 1 4 3 2 4 3 4 2 3
4 1 3 3 4 4 2 4 3 3 2
Now I want to use this matrix for the indexes of a for loop:
Using the first column, I want to feed the input of my loop the following indexes: i=4 j=3,2,1. Then for i=3 j=2,1. Then for i=2 j=1. i=1 is empty
This could be done just for the first column like this:
for u=4:-1:1
for v=u-1:-1:1
But will not work for other columns so I need to do the same but with the entries of matrix L, something like (it doesn't work in MATLB) for column i=1:
u=L(1:4,1)
v=L(u:L(4,1) , 1) %// where u corresponds to L(1,1) then L(2,1) then L(3,1)
(for all the columns it would look like:
for i=1:length(L)
for u=L(4*(i-1)+1:4*i)
for v=.. ?
)
This doesn't work because MATLAB takes the values of the entries and when I write L(1,1):L(4,1) it doesn't mean return the entries from line one to line four but rather all the numbers with increment 1 from the value of L(1,1) to the value of L(4,1) (here empty).
Any ideas ? thank you very much in advance
I believe something like this will solve you problem.
for col = 1:size(L,2)
rowIdx = 1;
for j = [L(:,col)]'
for k = [L(rowIdx:end,col)]'
% Do your stuff here
end
rowIdx = rowIdx + 1;
end
end
Notice how I use the values from columns from L directly as loop index variable. In a for loop statement you can basically write any row vector and the index takes those values. For example
for i = [1, 7, 11, 14, 23]
disp(i); % prints 1,7,11,14,23
end
This is true for arrays of objects, cell arrays, basically any single row matrix.
You can do it like this:
for col = 1:size(L, 2)
for I = 1:n-1
for J = I:n
i = L(I,col);
j = L(J,col);
%// As an example just print out the loop variable values
disp(sprintf('Col:%d\ti:%d\tj:%d\r\n',col,i,j))
end
end
end
Supose there is a Matrix
A =
1 3 2 4
4 2 5 8
6 1 4 9
and I have a Vector containing the "class" of each column of this matrix for example
v = [1 , 1 , 2 , 3]
How can I sum the columns of the matrix to a new matrix as column vectors each to the column of their class? In this example columns 1 and 2 of A would added to the first column of the new matrix, column 2 to the 3 to the 2nd, column 4 the the 3rd.
Like
SUM =
4 2 4
6 5 8
7 4 9
Is this possible without loops?
One of the perfect scenarios to combine the powers of accumarray and bsxfun -
%// Since we are to accumulate columns, first step would be to transpose A
At = A.' %//'
%// Create a vector of linear IDs for use with ACCUMARRAY later on
idx = bsxfun(#plus,v(:),[0:size(A,1)-1]*max(v))
%// Use ACCUMARRAY to accumulate rows from At, i.e. columns from A based on the IDs
out = reshape(accumarray(idx(:),At(:)),[],size(A,1)).'
Sample run -
A =
1 3 2 4 6 0
4 2 5 8 9 2
6 1 4 9 8 9
v =
1 1 2 3 3 2
out =
4 2 10
6 7 17
7 13 17
An alternative with accumarray in 2D. Generate a grid with the vector v and then apply accumarray:
A = A.';
v = [1 1 2 3];
[X, Y] = ndgrid(v,1:size(A,2));
Here X and Y look like this:
X =
1 1 1
1 1 1
2 2 2
3 3 3
Y =
1 2 3
1 2 3
1 2 3
1 2 3
Then apply accumarray:
B=accumarray([X(:) Y(:)],A(:)),
SUM = B.'
SUM =
4 2 4
6 5 8
7 4 9
As you see, using [X(:) Y(:)] create the following array:
ans =
1 1
1 1
2 1
3 1
1 2
1 2
2 2
3 2
1 3
1 3
2 3
3 3
in which the vector v containing the "class" is replicated 3 times since there are 3 unique classes that are to be summed up together.
EDIT:
As pointed out by knedlsepp you can get rid of the transpose to A and B like so:
[X2, Y2] = ndgrid(1:size(A,1),v);
B = accumarray([X2(:) Y2(:)],A(:))
which ends up doing the same. I find it a bit more easier to visualize with the transposes but that gives the same result.
How about a one-liner?
result = full(sparse(repmat(v,size(A,1),1), repmat((1:size(A,1)).',1,size(A,2)), A));
Don't optimize prematurely!
The for loop performs fine for your problem:
out = zeros(size(A,1), max(v));
for i = 1:numel(v)
out(:,v(i)) = out(:,v(i)) + A(:,i);
end
BTW: With fine, I mean: fast, fast, fast!
I am trying to flip certain parts of a matrix. I can explain better by example. Let's say that I have a matrix
M = [ 1 3 6;
1 2 4;
1 7 1;
2 9 0;
2 8 3;
2 4 2;
2 3 1;
3 6 5;
3 4 5;
3 1 9;
4 2 4;
4 8 6 ]
What I'd like to do here is take any rows with an even number in the first column, and flip the third column elements. The end result would look like this:
1 3 6
1 2 4
1 7 1
2 9 1 *
2 8 2 *
2 4 3 *
2 3 0 *
3 6 5
3 4 5
3 1 9
4 2 6 *
4 8 4 *
Note the rows marked with a star have had the elements of the third column flipped upside-down. The problem I'm having is going through each row like in a for-loop you cannot flip an entire set of rows.
Thanks in advance for any help.
Another time accumarray is the way to go:
A =[ 1 3 6 ;
1 2 4 ;
1 7 1 ;
2 9 0 ;
2 8 3 ;
2 4 2 ;
2 3 1 ;
3 6 5 ;
3 4 5 ;
3 1 9 ;
4 2 4 ;
4 8 6 ]
C = accumarray(A(:,1),A(:,3),[],#(x) {flipud(x)} ); %// get groups according to
%// first column and flip it
C = vertcat(C{:}); %// cell array returned,
%// transform to matrix
mask = ~mod(A(:,1),2); %// mask for even numbers
A(mask,3) = C(mask); %// replace masked values of 3rd column with flipped ones
returns:
A =
1 3 6
1 2 4
1 7 1
2 9 1
2 8 2
2 4 3
2 3 0
3 6 5
3 4 5
3 1 9
4 2 6
4 8 4
Certainly slower, but just for fun in two lines:
C = accumarray(A(:,1),A(:,3),[],#(x) {flipud(x)} );
A(~mod(A(:,1),2),3) = getfield( vertcat(C{:}), {~mod(A(:,1),2)});
%// well no, I won't explain it...
Edit: I assumed your first column just contains integers!
I would suggest you break the problem down into stages, something like so:
Identify blocks you wish to flip
Extract them
Flip them
Replace them
You can identify a set of even numbers using the unique and mod functions, then use a for loop over them and use logical indexing to pull/replace the blocks.
Here, try this
a = magic(5); % Some data in a 5x5 matrix
b = 1:numel(a); % Indices of <a>
Rearrange b however you want, then do a=a(b) to reassign a based on the reassigned indices of b. For example, the following code
disp(a(b));
would just return the elements of a in their original order. For your application this code should work:
a = <your matrix data>
b = 1:numel(a);
b = [b(1:27) fliplr(b(28:31)) b(32:34) fliplr(b(35:36))] % Change this part
a = reshape(a(b),size(a))
You should change b based on whatever you need it to do.
How can I multiply columns of a matrix and obtain a column vector.
Example:
A =
1 1 4
3 2 2
2 1 1
4 1 1
Expected output:
C =
4
12
2
4
Any ideas without for?
You can simply use the inbuilt prod function as prod(A,2) or prod(A')'.
For example:
>>
A = [
1 1 4
3 2 2
2 1 1
4 1 1
];
Now:
>> prod(A,2)
ans =
4
12
2
4
For more details, try help prod.
Once again, google and the Matlab documentation are your friend.
You are looking for the function prod:
prod(A,2)
A simple loop could do this for you too if you don't know the size of the matrix in hand. Just build a new vector by row-wise multiplication of elements in A.