I have two matrix M x N, For simplicity, we take 4x4:
Matrix A:
1 4 2 5
4 5 8 2
3 4 5 6
2 3 5 8
Matrix B:
10 11 12 13
56 11 23 45
34 44 33 25
25 63 35 78
If an element of matrix A is greater then 5, then we change it from matrix B.
At the end we must get a matrix C:
1 4 2 5
4 5 23 2
3 4 5 25
2 3 5 78
How can I make it, should I use something like logical indexing..
Yes, you should use logical indexing:
C = A;
C(C>5) = B(C>5);
This means that every element in C that is >5 is set to the corresponding value in B.
or
C = A.*(A<=5) + B.*(A>5);
The comparisons in the parentheses create arrays with 0 and 1, so the first multiplication sets all elements of A to zero that should be taken from B, and second multiplication sets all elements of B to zero that should be taken from A.
Related
Suppose I have a matrix of dimension [4x4], and a vector of [16x1], I need to multiply every 4 element in the vector in one element in the matrix, (instead of multiplying element in row by element in vector), how can I do that using loop ?
For example here below, the results of the first four elements in the resulted vector as shown in the below example, then the same thing for the second, third and fourth rows in the matrix. :
So the results in that example is supposed to be [16x1]
Thank you
Using kron you can use this one-liner:
%A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
%v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
sum(kron(A,ones(4,4)).'.*v).'/4
I use the kronecker tensor product to "replicate" 4x4 time the A matrice. After that it's pure algebra.
This is just matrix multiplication in disguise... If your tall vector was a matrix of the same size as the matrix shown, where each highlighted block is a row, it's matrix multiplication. We can set this up, then reshape back into a vector.
You can use indexing to turn this into simple matrix multiplication. A question I answered already today details how the below indexing works using bsxfun, then we just reshape at the end:
% Setup
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
% Matrix mutliplication
r = numel(v)/size(A,1);
b = A * v( bsxfun( #plus, (1:r:numel(v)).', 0:r-1 ) );
% Make result a column vector
b = reshape( b.', [], 1 );
See if this is what you want:
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
r = reshape(sum(bsxfun(#times, permute(A, [3 2 1]), permute(reshape(v, 1, [], size(A,2)), [2 3 1])), 2), [], 1);
which gives
r =
17
17
17
17
41
41
41
41
65
65
65
65
89
89
89
89
There are details that I assumed, but this shoudl do the trick:
A=reshape(1:16,4,4).';
b=repelem([2,0,1,3],1,4).';
c=[];
for row=1:size(A,1)
c=[ c; sum(reshape(repelem(A(row,:),4).*b.',4,[]),2)];
end
I am assuming here that your demo for the vector is just a bad example and that you wont have repeated values, otherwise an easier version can be achieved by just not doing 3/4ths of the multiplications.
If you do not have access to repelem, have a look at alterative codes that do the same thing:Element-wise array replication in Matlab
I need to find the the size of bin with maximum and minimum element. I am using histc function in MATLAB.
Here is what I am doing,
A=[1 2 3 11 22 3 4 55 6 7 2 33 44 5 22]
edges = [10 inf];
N = histc(A,edges)
it gives N=[6,0]; means there are 6 elements having values greater than 10. Now I want to count what is the maximum count in a bin for my condition.
here it should be 2 as there are two instances where we have two integers satisfying my condition 11 22 and 33 44
How to count it in MATLAB.
Here you go;
A=[1 2 3 11 22 3 4 55 6 7 2 33 44 5 22]
arr=diff([0 (find(~(A>10))) numel(A)+1]) -1;
arr(find(arr(1,:)==0))=[];
largest=max(arr); % longest sequence of occurences of numbers > 10
smallest=min(arr); % smallest sequence of occurences of numbers > 10
Cheers!!
I need to loop through coloumn 1 of a matrix and return (i) when I have come across ALL of the elements of another vector which i can predefine.
check_vector = [1:43] %% I dont actually need to predefine this - i know I am looking for the numbers 1 to 43.
matrix_a coloumn 1 (which is the only coloumn i am interested in looks like this for example
1
4
3
5
6
7
8
9
10
11
12
13
14
16
15
18
17
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
1
3
4
2
6
7
8
We want to loop through matrix_a and return the value of (i) when we have hit all of the numbers in the range 1 to 43.
In the above example we are looking for all the numbers from 1 to 43 and the iteration will end round about position 47 in matrix_a because it is at this point that we hit number '2' which is the last number to complete all numbers in the sequence 1 to 43.
It doesnt matter if we hit several of one number on the way, we count all those - we just want to know when we have reached all the numbers from the check vector or in this example in the sequence 1 to 43.
Ive tried something like:
completed = []
for i = 1:43
complete(i) = find(matrix_a(:,1) == i,1,'first')
end
but not working.
Assuming A as the input column vector, two approaches could be suggested here.
Approach #1
With arrayfun -
check_vector = [1:43]
idx = find(arrayfun(#(n) all(ismember(check_vector,A(1:n))),1:numel(A)),1)+1
gives -
idx =
47
Approach #2
With customary bsxfun -
check_vector = [1:43]
idx = find(all(cumsum(bsxfun(#eq,A(:),check_vector),1)~=0,2),1)+1
To find the first entry at which all unique values of matrix_a have already appeared (that is, if check_vector consists of all unique values of matrix_a): the unique function almost gives the answer:
[~, ind] = unique(matrix_a, 'first');
result = max(ind);
Someone might have a more compact answer but is this what your after?
maxIndex = 0;
for ii=1:length(a)
[f,index] = ismember(ii,a);
maxIndex=max(maxIndex,max(index));
end
maxIndex
Here is one solution without a loop and without any conditions on the vectors to be compared. Given two vectors a and b, this code will find the smallest index idx where a(1:idx) contains all elements of b. idx will be 0 when b is not contained in a.
a = [ 1 4 3 5 6 7 8 9 10 11 12 13 14 16 15 18 17 19 20 21 22 23 24 25 26 ...
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 1 3 4 2 6 7 8 50];
b = 1:43;
[~, Loca] = ismember(b,a);
idx = max(Loca) * all(Loca);
Some details:
ismember(b,a) checks if all elements of b can be found in a and the output Loca lists the indices of these elements within a. The index will be 0, if the element cannot be found in a.
idx = max(Loca) then is the highest index in this list of indices, so the smallest one where all elements of b are found within a(1:idx).
all(Loca) finally checks if all indices in Loca are nonzero, i.e. if all elements of b have been found in a.
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
everyone
I have problem when make iteration for two variable but, it's combine just in one vector or array
At first i write my input for iteration w(0) as w
w=[1 50];
for number 1, I use array
e=0:1:(n-1);
f=0:2:(2*n-2); %for 50 in column 2.
I try to use this code
w=[1 50];
ww=kron(ones((n),1),w)
e=0:1:(n-1);
f=0:2:(2*n-2);
r=[e',f']
x=ww+r
and the output is
ww =
1 50
1 50
1 50
1 50
1 50
1 50
r =
0 0
1 2
2 4
3 6
4 8
5 10
x =
1 50
2 52
3 54
4 56
5 58
6 60
I want x is output just in one array, in example
x =
1
50
2
52
3
54
4
56
5
58
6
60
where w=[1 50] can be use difference addition for iteration
Apply this to your x matrix:
x = reshape(x.',[],1);
See reshape doc for details.
Here is a simple method to create your vector from scratch:
x = [1:6;50:2:60];
x(:)
Or with your variables:
x = [e; f];
x(:)