If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
Related
I would like to sum specific columns of each row in a matrix using a for loop. Below I have included a simplified version of my problem. As of right now, I am calculating the column sums individually, but this is not effective as my actual problem has multiple matrices (data sets).
a = [1 2 3 4 5 6; 4 5 6 7 8 9];
b = [2 2 3 4 4 6; 3 3 3 4 5 5];
% Repeat the 3 lines of code below for row 2 of matrix a
% Repeat the entire process for matrix b
c = sum(a(1,1:3)); % Sum columns 1:3 of row 1
d = sum(a(1,4:6)); % Sum columns 4:6 of row 1
e = sum(a(1,:)); % Sum all columns of row 1
I would like to know how to create a for loop that automatically loops through and sums the specific columns of each row for each matrix that I have.
Thank you.
Here is a solution that you don't need to use for loop.
Assuming that you have a matrix a of size 2x12, and you want to do the row sums every 4 columns, then you can use reshape() and squeeze() to get the final result:
k = 4;
a = [1:12
13:24];
% a =
% 1 2 3 4 5 6 7 8 9 10 11 12
% 13 14 15 16 17 18 19 20 21 22 23 24
s = squeeze(sum(reshape(a,size(a,1),k,[]),2));
and you will get
s =
10 26 42
58 74 90
Consider a row vector A and row vector B. For example:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
A has previously been checked to be a subset of B. By subset, I specifically mean that all elements in A can be found in B. I know that elements in A will not ever repeat. However, the elements in B are free to repeat as many or as few times as they like. I checked this condition using:
is_subset = all(ismember(A,B));
With all that out of the way, I need to know the indices of the elements of A within B including the times when these elements repeat within B. For the example A and B above, the output would be:
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
Use ismember to find the relevant logical indices. Then convert them to linear indices using find.
C = find(ismember(B,A));
You can find the difference of each element of A with B, and get the indices you want. Something like below:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
tol = 10^-3 ;
N = length(A) ;
iwant = cell(N,1) ;
for i = 1:N
idx = abs(A(i)-B)<=tol ;
iwant{i} = find(idx) ;
end
iwant = [iwant{:}] ;
I have a 4x9 matrix, and I need to calculate the sum of all numbers in every other column of c starting with the first. Can anyone point me in the right direction? I know we have to use the function sum() but that's about it.
I used Octave rather than MATLAB, but this works for me:
A = randi(10,4,9)
B = A(:, 1:2:9)
C = sum(B)
Generate a 4x9 matrix with random numbers between 1 and 10, then create a sub-matrix with each row, and given columns 1:2:9 means starting from the first column and ending on the 9th, choose every second column, then sum up each column. Example output:
>> A = randi(10,4,9)
A =
1 3 6 8 2 8 4 8 10
3 6 10 4 6 4 6 2 8
4 3 9 2 7 10 6 9 6
8 5 3 9 3 8 4 6 10
>> B = A(:, 1:2:9)
B =
1 6 2 4 10
3 10 6 6 8
4 9 7 6 6
8 3 3 4 10
>> C = sum(B)
C =
16 28 18 20 34
You could also take the sum of matrix C using the sum() first and then select every other element from the result starting from the 1st element.
tmpC = sum(C);
result = tmpC(1:2:end)
I have a matrix A in Matlab of dimension Nx(N-1), e.g.
N=5;
A=[1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16;
17 18 19 20];
I want to rearrange the elements of A in a certain way. Specifically I want to create a matrix B of dimension (N-1)xN such that:
for i=1,...,N,
B(:,i) collects
1) the first i-1 elements of the i-1th column of A and
2) the last N-i elements of the ith column of A.
Notice that for i=1 the i-1th column of A does not exist and therefore 1) is skipped; similarly, for i=N theith column of A does not exist and therefore 2) is skipped.
In the example above
B=[5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16];
This code does what I want. I am asking your help to vectorise it in an efficient way.
B=zeros(N-1,N);
for i=1:N
if i>1 && i<N
step1=A(1:i-1,i-1);
step2=A(i+1:N,i);
B(:,i)=[step1;step2];
elseif i==1
B(:,i)=A(i+1:N,i);
elseif i==N
B(:,i)=A(1:i-1,i-1);
end
end
Extract the lower and upper triangular matrices of A. Then reassemble them with a "diagonal shift":
u = triu(A);
l = tril(A,-1);
B = padarray(u(1:end-1,:),[0 1],'pre') + padarray(l(2:end,:),[0 1],'post');
Another valid approach using logical indexing combined with tril and triu:
B = zeros(size(A'));
B(tril(true(size(B)))) = A(tril(true(size(A)), -1));
B(triu(true(size(B)), 1)) = A(triu(true(size(A))));
Result:
>> B
B =
5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16
Say I have a matrix A whose first column contains item IDs with repetition and second column contains their weights.
A= [1 40
3 33
2 12
4 22
2 10
3 6
1 15
6 29
4 10
1 2
5 18
5 11
2 8
6 25
1 14
2 11
4 28
3 38
5 35
3 9];
I now want to find the difference of each instance of A and its associated minimum weight. For that, I make a matrix B with its first column containing the unique IDs from column 1 of A, and its column 2 containing the associated minimum weight found from column 2 of A.
B=[1 2
2 8
3 6
4 10
5 11
6 25];
Then, I want to store in column 3 of A the difference of each entry and its associated minimum weight.
A= [1 40 38
3 33 27
2 12 4
4 22 12
2 10 2
3 6 0
1 15 13
6 29 4
4 10 0
1 2 0
5 18 7
5 11 0
2 8 0
6 25 0
1 14 12
2 11 3
4 28 18
3 38 32
5 35 24
3 9 3];
This is the code I wrote to do this:
for i=1:size(A,1)
A(i,3) = A(i,1) - B(B(:,1)==A(i,2),2);
end
But this code takes a long time to execute as it needs to loop through B every time it loops through A. That is, it has a complexity of size(A) x size(B). Is there a better way to do this without using loops, that would execute faster?
You can use accumarray to first compute the minimum value in the second column of A for each unique value in the first column of A. We can then index into the result using the first column of A and compare to the second column of A to create the third column.
% Compute the mins
min_per_group = accumarray(A(:,1), A(:,2), [], #min);
% Compute the difference between the second column and the minima
A(:,3) = A(:,2) - min_per_group(A(:,1));