I have a matrix with 4 rows and 400,000 columns. I need to obtain the average of 4 successive rows. That is, average of row 1 to row 4, row 5 to 8, etc.
The 4 columns should be maintained as such. I know that this may be a kindergarten level problem, but I appreciate any hints to write a program in Matlab; I have a little experience writing Matlab programs.
An example of the data can be as follows:
[1 2 3 2;
5 6 7 2;
9 6 7 6;
5 2 3 2;
9 8 7 6;
6 5 4 3;
4 3 2 1;
9 8 7 6]
I want the result as:
[5 4 5 3],[7 6 5 4]
It is not entirely clear to me how your data is layed out, so I'll give you a solution to what I think you mean.
Suppose you have
a = [
1 2 3 4 %# row 1
2 3 4 5
3 4 5 6
4 5 6 7 %# row 4
5 6 7 8 %# row 5
...
]; %# row m
and you want the average down the columns of rows 1 through 4, 5 through 8, etc.
You can do that simply by
averages = reshape(mean(reshape(a,4,4,[])),[],4)
breakdown:
A = reshape(a,4,4,[])
rearranges the data in your matrix into a 3D array. Each 3D "layer" of this array is a 4x4 matrix.
B = mean(A)
This takes the average along the columns (direction 1). Read up on help mean for more information.
C = reshape(B,[],4)
This rearranges the array of averages back to a Nx4 matrix, where N=m/4.
Related
I have this matrix:
data=[1 5402783 1
2 43359352 2
3 26118700 3
4 33091887 4
5 890931 5
6 826897 6
7 1188749 7
8 1239861 8];
I need the first column to stay as it is, sort the 2nd column (in descending order) and 'keep along' the values of the third column. If I use sort(data) it sorts all 3 columns.
I tried:
[~,idx]=sort(data(:,2),'descend');
data=data(idx,:)
but it is obviously wrong.
The output should be:
[1 43359352 2
2 33091887 4
3 26118700 3
4 5402783 1
5 1239861 8
6 1188749 7
7 890931 5
8 826897 6]
All you need to do is reassemble the data matrix in the end taking the unsorted and sorted parts:
data = [1 5402783 1
2 43359352 2
3 26118700 3
4 33091887 4
5 890931 5
6 826897 6
7 1188749 7
8 1239861 8];
[~,idx] = sort(data(:,2),'descend');
data = [data(:,1),data(idx,2:3)];
Is there any command to find mean of first 5 values then next 5 values from a total of 25 values present in a vector in MATLAB. If the dataset is X.
If anyone can help me to provide a code where I can get mean at every 5th value.
X=[4 5 6 7 2 5 7 4 2 6 7 3 2 1 5 7 8 3 4 6 8 4 2 6 8];
You can for instance reshape the vector in an array with reshape and then apply the mean function:
M = mean(reshape(X, [5, numel(X)/5]),1);
or simply
M = mean(reshape(X, 5, []),1);
But there as stated in the comments there are many other ways.
Here is one simple way to do it. Rearrange the vector into a matrix loop over the columns and take the mean of all values in each column. Store the results in a new vector.
X=[4 5 6 7 2 5 7 4 2 6 7 3 2 1 5 7 8 3 4 6 8 4 2 6 8];
Xr = reshape(X,5,5)
cols = size(Xr)(2)
avgs=zeros(1,cols)
for i= 1:cols
avgs(i) = mean(Xr(:,i))
end
Suppose that I have a matrix , let's call it A, as follows:
1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9
And I want to reshape it into a matrix like this:
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
So, basically, what I want to be done is that MATLAB first reads a block of size (2,5) and then splits the remaining matrix to the next row and then repeats this so on so forth until we get something like in my example.
I tried to do this using MATLAB's reshape command in several ways but I failed. Any help is appreciated. In case that it matters, my original data is larger. It's (2,1080). Thanks.
I don't believe you can do this in a single command, but perhaps someone will correct me. If speed isn't a huge concern a for loop should work fine.
Alternatively you can get your results by reshaping each row of A and then placing the results into every other row of a new matrix. This will also work with your larger data.
A = [1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9];
An = zeros(numel(A)/5, 5); % Set up new, empty matrix
An(1:2:end,:) = reshape(A(1,:), 5, [])'; % Write the first row of A to every other row of An
An(2:2:end,:) = reshape(A(2,:), 5, [])' % Write second row of A to remaining rows
An =
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
You may need to read more about indexing in the Matlab's documentation.
For your example, it is easy to do the following
A=[1 2 3 4 5 1 2 3 4 5; 0 2 4 6 8 1 3 5 7 9]
a1=A(:,1:5); % extract all rows, and columns from 1 to 5
a2=A(:,6:end); % extract all rows, and columns from 6 to end
B=[a1;a2] % construct a new matrix.
It is not difficult to build some sort of loops to extract the rest.
Here's a way you can do it in one line using the reshape and permute commands:
B = reshape(permute(reshape(A,2,5,[]), [1,3,2]), [], 5);
The reshape(A,2,5,[]) command reshapes your A matrix into a three-dimensional tensor of dimension 2 x 5 x nblocks, where nblocks is the number of blocks in A in the horizontal direction. The permute command then swaps the 2nd and 3rd dimensions of this 3D tensor, so that it becomes a 2 x nblocks x 5 tensor. The final reshape command then transforms the 3D tensor into a matrix of dimension (2*nblocks) x 5.
Looking at the results at each stage may give you a better idea of what's happening.
I'm very new to scilab syntax and can't seem to find a way to extract the even and odd elements of a matrix into two separate matrix, suppose there's a matrix a:
a=[1,2,3,4,5,6,7,8,9]
How do I make two other matrix b and c which will be like
b=[2 4 6 8] and c=[1 3 5 7 9]
You can separate the matrix by calling row and column indices:
a=[1,2,3,4,5,6,7,8,9];
b=a(2:2:end);
c=a(1:2:end);
[2:2:end] means [2,4,6,...length(a)] and [1:2:end]=[1,3,5,...length(a)]. So you can use this tip for every matrix for example if you have a matrix a=[5,4,3,2,1] and you want to obtain the first three elements:
a=[5,4,3,2,1];
b=a(1:1:3)
b=
1 2 3
% OR YOU CAN USE
b=a(1:3)
If you need elements 3 to 5:
a=[5,4,3,2,1];
b=a(3:5)
b=
3 2 1
if you want to elements 5 to 1, i.e. in reverse:
a=[5,4,3,2,1];
b=a(5:-1:1);
b=
1 2 3 4 5
a=[1,2,3,4,5,6,7,8,9];
b = a(mod(a,2)==0);
c = a(mod(a,2)==1);
b =
2 4 6 8
c =
1 3 5 7 9
Use mod to check whether the number is divisible by 2 or not (i.e. is it even) and use that as a logical index into a.
The title is about selecting rows of a matrix, while the body of the question is about elements of a vector ...
With Scilab, for rows just do
a = [1,2,3 ; 4,5,6 ; 7,8,9];
odd = a(1:2:$, :);
even = a(2:2:$, :);
Example:
--> a = [
5 4 6
3 6 5
3 5 4
7 0 7
8 7 2 ];
--> a(1:2:$, :)
ans =
5 4 6
3 5 4
8 7 2
--> a(2:2:$, :)
ans =
3 6 5
7 0 7
I was given the following Question:
Write a function call zigzag that takes in a 2-dimensional array A and return a 1- dimensional array created by traverse through A in zigzag way starting at position (1,1).
Example:
A =[1 2 3 4 5 6
7 8 9 1 3 4
3 4 5 6 3 1
3 4 5 6 7 8]
zigzag(A) should return:
[1 2 3 4 5 6 4 3 1 9 8 7 3 4 5 6 3 1 8 7 6 5 4 3]
The way I solved it, I am not sure if this is a correct method to do it. I would be glad to know if this is perfect and how I could improve my answer:
function B=zigzag(A)
[r,c]=size(A);
B= reshape(A’,1,:);
m=0
n=0
For r>m+2
m=m+2;
n=n+1;
For i=1:c
B(nc+i)=B(2cn-i+1);
End
End
disp(B)
If it gives you the right output, then you're certainly doing something right. However, what I would have done was access the even rows of your matrix, reverse the directions so that they're displayed in reverse order, transpose your matrix then unravel it.
The reason why we transpose it is because when we unravel a matrix in MATLAB, this means that the columns of the matrix are stacked on top of each other so that one single vector is produced. We want the rows to be stacked on top of each other and making the even rows in reverse order will allow you to do the zigzag that you expect. If you want the rows to be stacked on top of each other, you need to transpose the matrix first so that rows become columns, and when you unravel this matrix, you'll stack the rows on top of each other instead to create a single vector.
Something like this:
B = A; %// Make a copy
B(2:2:end,:) = fliplr(B(2:2:end,:)); %// Flip even rows
B = reshape(B.', 1, []); %// Unravel
With your example, I get:
B =
Columns 1 through 13
1 2 3 4 5 6 4 3 1 9 8 7 3
Columns 14 through 24
4 5 6 3 1 8 7 6 5 4 3