The function setdiff(A,B,'rows') is used to return the set of rows that are in A but not B, with repetitions removed.
Is there any way to do it without removing the repetitions?
Thanks a lot.
You can use ismember instead of setdiff, to find all the rows of B that appear in A.
Because you want only those that NOT appear in A, use the ~ sign, and finally take all A rows in these rows indices:
A =
1 2 3
4 5 6
1 2 3
7 8 9
B =
4 5 6
C=A(~ismember(A,B,'rows'),:)
C =
1 2 3
1 2 3
7 8 9
Related
I have been given this problem in a MATLAB course I am doing. The instructor's solution provided there is wrong, and I have been struggling with the same problem for hours as I am a beginner who has just started coding (a science student here).
Consider a one-dimensional matrix A such as A = \[5 8 8 8 9 9 6 6 5 5 4 1 2 3 5 3 3 \]. Show the percentage frequencies of unique
elements in the matrix A in descending order.
Hint: Use the functions tabulate and sort.
How do I solve this problem using only tabulate, sort, and find functions (find is for eliminating zero frequency elements in tabulate table, which my instructor did not do)?
I tried first extracting the indices of non-zero elements in the percentage column of tabulating table using the find function, which I succeeded in doing using the following:
A = [5 8 8 8 9 9 6 6 5 5 4 1 2 3 5 3 3 ];
B = tabulate(A);
C = find(B(:,3) > 0)
But I am now struggling to return the values corresponding to the 3rd column of B using indices in C. Please help. Also please give me some alternative syntax where one can easily make a vector out of non-zero elements of a row or column easily by omitting the zeroes in that vector if it exists. Rest of the problem I'll do by myself.
With your find command, you are just finding the indices of the matrix and not the values themselves.
So you either will do something like this:
A = [5 8 8 8 9 9 6 6 5 5 4 1 2 3 5 3 3 ];
B = tabulate(A);
for i = 1:size(B,1)-1
if B(i,3) == 0
B(i,:) = [];
end
end
sortrows(B,3,'descend')
where you remove the 0 value's row.
Or since you have all the numbers with none-zero frequency you can ask for their rows. Like this:
A = [5 8 8 8 9 9 6 6 5 5 4 1 2 3 5 3 3 ];
B = tabulate(A);
C = find(B(:,3) > 0);
sortrows(B(C(:),:),3,'descend')
in a bit more elegant way. B(C(:),:) calls all the rows with first indices the indices of matrix C. Which is exactly what you are asking for. While at the same time you sort your matrix based on row 3 at a descending order.
i have data A=( 3,5,3,1,4 ) in a column and
B=[
4 6 9 1 3
2 7 2 5 7
7 3 1 8 2
4 1 6 9 1
2 5 8 3 6 ]
And i want: as in A first element is 3 and for this i want to get first element of column 3 row 1 from B which is 9. The second element of A is 5 and for this i want to get the the 2nd element of column 5 and row 2 from B which is 7 ,and do the process for all other elements . how to do this in matlab? the required elements are bold and underlined. The desired output is [9,7,1,4,3]
Read about linear indexes.
sub2ind will convert from [row col] to index.
Cols=[ 3,5,3,1,4 ];
Rows=1:length(Cols);
B=[
4 6 9 1 3
2 7 2 5 7
7 3 1 8 2
4 1 6 9 1
2 5 8 3 6 ];
Indexes=sub2ind(size(B),Rows,Cols);
Vals=B(Indexes)
if I've read well you want an elements replacement. It's quite simple
A(1)=B(1,3)
A(2)=B(2,5)
So after you have declared your two vectors you can handle them replacing singular components. In general when you have a 1-D vectors you can access to it's component position only declaring the position itself inside the parentheses, as I did for A.
Whe you have to face situation like in B, If you remember linear algebra and matrix in general B(a,b) means the element of matrix B placed in the line a and column b so you have to specify row and column to access that element.
If you have a random matrix, for example a 5x5:
A(i,j) = (5 4 3 2 1
4 3 2 1 0
5 4 3 2 1
4 3 2 1 0
5 4 3 2 1)
And a second array:
B(1,j) = (4 5 6 7 8)
How can I then assign values of B to A if this only needs to be done when the value of B(1,j) is larger than any of the values from a certain colomn of A?
For example, B(1,1) = 4 and in the first colomn of A it is larger than A(1,1), A(3,1) and A(5,1), so these must be replaced by 4. In the second colomn, nothing needs to be replaced, etc.
Thanks already!
You can do this without any explicit looping using bsxfun:
A = [5 4 3 2 1
4 3 2 1 0
5 4 3 2 1
4 3 2 1 0
5 4 3 2 1];
B = [4 5 6 7 8];
A = bsxfun(#min,A,B);
Result:
A =
4 4 3 2 1
4 3 2 1 0
4 4 3 2 1
4 3 2 1 0
4 4 3 2 1
In later versions of MATLAB (2016b and later) you can even omit the bsxfun and get the same result.
A = min(A,B);
Matlab "find" may be of use to you.
https://www.mathworks.com/help/matlab/matlab_prog/find-array-elements-that-meet-a-condition.html
If you aren't concerned about speed or efficiency, you could also set up a two nested for loops with a condition (i.e. an if) statement comparing the values of A and B.
If you only interested in column wise comparison to B, you could use the increment of the outer loop in the inner loop.
for i,...
for j,...
if B(1,i) > A(j,i)
A(j,i)=B(i,j)
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 have two matrix A and B. Suppose I would like to find in each row of matrix A the smallest number, and for the same cell that this number is in Matrix A, do find the corresponding number of the same cell in matrix B. For example the number in matrix A will be in the position A(1,3), A(2,9)...and I want the corresponding number in B(1,3), B(2,9)... Is it possible to do it, or I am asking something hard for matlab. Hope someone will help me.
What you can do is use min and find the minimum across all of the rows for each column. You would actually use the second output in order to find the location of each column per row that you want to find. Once you locate these, simply use sub2ind to access the corresponding values in B. As such, try something like this:
[~,ind] = min(A,[],2);
val = B(sub2ind(size(A), (1:size(A,1)).', ind));
val would contain the output values in the matrix B which correspond to the same positions as the minimum values of each row in A. This is also assuming that A and B are the same size. As an illustration, here's an example. Let's set A and B to be a random 4 x 4 array of integers each.
rng(123);
A = randi(10, 4, 4)
B = randi(10, 4, 4)
A =
7 8 5 5
3 5 4 1
3 10 4 4
6 7 8 8
B =
2 7 8 3
2 9 4 7
6 8 4 1
6 7 3 5
By running the first line of code, we get this:
[~,ind] = min(A,[],2)
ind =
3
4
1
1
This tells us that the minimum value of the first row is the third column, the minimum value of the next row is the 4th column, and so on and so forth. Once we have these column numbers, let's access what the corresponding values are in B, so we would want row and columns (1,3), (2,4), etc. Therefore, after running the second statement, we get:
val = B(sub2ind(size(A), (1:size(A,1)).', ind))
val =
8
7
6
6
If you quickly double check the accessed positions in B in comparison to A, we have found exactly those spots in B that correspond to A.
A = randi(9,[5 5]);
B = randi(9,[5 5]);
[C,I] = min(A');
B.*(A == repmat(C',1,size(A,2)))
example,
A =
2 1 6 9 1
2 4 4 4 2
5 6 5 5 5
9 3 9 3 6
4 5 6 8 3
B =
3 5 6 8 1
9 2 9 7 1
5 6 6 5 6
4 6 1 4 5
5 3 7 1 9
ans =
0 5 0 0 1
9 0 0 0 1
5 0 6 5 6
0 6 0 4 0
0 0 0 0 9
You can use it like,
B(A == repmat(C',1,5))
ans =
9
5
5
6
6
5
4
1
1
6
9