say that I have a matrix A and an integer E = 16
A = [4 3 0.987477222842622;
8 4 0.987477222842622;
3 8 0.994241214094179;
12 5 0.951729278021845;
13 4 0.972221868548579;
14 13 0.996357467281480;
16 12 0.994198825847544]
It means that
element 4 is equal to element 3 with a similarity of 0.98,
element 8 is equal to element 4 with a similarity of 0.98,
element 3 is equal to element 8 with a similarity of 0.99 etc..
please note that if element 13 is equal to element 4 (5th row) it doesn't mean that it is also equal to element 3 (since 4 is equal to 3 ,see first row). The similarity between 13 and 3 apparently is less than the threshold (0.95) to be considered equal.
In total there are E=16 elements.
I would like to find the number of distinct elements...
In this case I have 16-5=11 different elements because the first 3 couples in the matrix (4-3,8-4,3-8) should be considered as a unique element...
That because
4-3 -> 1 element
8-4 -> 1 element
3-8 -> 1 element, but since 8 is equal to 4 that is equal to 3 they should considered as a unique element..
Do you have a solution for this problem?
it's driving me crazy :)
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.
let D be a pairwise distance between two sets of observations. I used a custom distance function to represent similarity. Unlike the most of similarity measures, the way that I have chosen for calculating the distance gives large value for similar pairs.
0 1 2
0 0 4 6
1 5 9 7
2 2 1 4
In normal cases we will consider the 0 and 0 points are the most similar pair, since the distance value between them is 0. In my case the 1 and 1 are the most similar pair, since they have the largest value 9. I need to use the pairwise distance matrix to perform clustering. I need a way to convert the large values into small values that can be used by a clustering method.
If I understand correctly, you just need to reverse the sort order of the elements. The following equation converts the biggest elements into the smallest elements of a new matrix and vice-versa:
distance = max(max(D))*ones(size(D)) - D
where D is the matrix from your custom distance function. For your example above, this would result in
9 9 9 0 4 6 9 5 3
9 9 9 - 5 9 7 = 4 0 2
9 9 9 2 1 4 7 8 5
If you know that there are no negative numbers in your custom distance matrix, you could instead use the following equation, which simply offsets all elements of your matrix (to avoid division by zero) and them inverts them:
distance = 1./(D+1)
Basically the sum function calculate the sum of the columns, that is to say if we have a 4x4 matrix we would get a 1X4 vector
A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
sum(A)
ans =
34 34 34 34
But if I want to get the Summation of the rows then i have 2 methods, the first is to get the transpose of the matrix then get the summation of the transposed matrix,and finally get the transpose of the result...., The Second method is to use dimension argument for the Sum function "sum(A, 2)"
A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
sum(A,2)
ans =
34
34
34
34
The problem is here I cannot understand how this is done, If anyone could please tell me the idea/concept behind this method,
It's hard to tell exactly how sum internally works, but we can guess it does something similar to this.
Matlab stores matrices (or N-dimensional arrays) in memory using column-major order. This means the order for the elements in memory for a 3 x 4 matrix is
1 4 7 10
2 5 8 11
3 6 9 12
So it first stores element (1,1), then (1,2), then (13), then (2,1), ...
In fact, this is the order you use when you apply linear indexing (that is, index a matrix with a single number). For example, let
A = [7 8 6 2
9 0 3 5
6 3 2 1];
Then A(4) gives 8.
With this in mind, it's easy to guess that what sum(A,1) does is traverse elements consecutively: A(1)+A(2)+A(3) to obtain the sum of the first column, then A(4)+A(5)+A(6) to sum the second column, etc. In contrast, sum(A,2) proceeds in steps of size(A,1) (3 in this example): A(1)+A(4)+A(7)+A(10) to compute the sum of the first row, etc.
As a side note, this is probably related with the observed fact that sum(A,1) is faster than sum(A,2).
I'm really not sure what you are asking. sum takes two inputs, the first of which is a multidimensional array A, say.
Now let's take sA = size(A), and d between 1 and ndims(A).
To understand what B = sum(A,d) does, first we find out what the size of B is.
That's easy, sB = sA; sB(d) = 1;. So in a way, it will "reduce" the size of A along dimension d.
The rest is trivial: every element in B is the sum of elements in A along dimension d.
Basically, sum(A) = sum(A,1) which outputs the sum of the columns in the matrix. 1 indicates the columns. So, sum(A,2) outputs the sum of the rows in the matrix. 2 indicating the rows. More than that, the sum command will output the entire matrix because there is only 2 dimensions (rows and columns)
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
Here is an array i need to extract the exact values where the increasing and decreasing trend starts.
the output for the array a will be [2(first element) 2 6 9]
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
^ ^ ^ ^
| | | |
Kindly help me to get the result in MATLAB for any similar type of array..
You just have to find where the sign of the difference between consecutive numbers changes.
With some common sense and the functions diff, sign and find, you get this solution:
a = [2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2];
sda = sign(diff(a));
idx = [1 find(sda(1:end-1)~=sda(2:end))+2 ];
result = a(idx);
EDIT:
The sign function messes things up when there are two consecutive numbers which are the same, because sign(0) = 0, which is falsely identified as a trend change. You'd have to filter these out. You can do this by first removing the consecutive duplicates from the original data. Since you only want the values where the trend change starts, and not the position where it actually starts, this is easiest:
a(diff(a)==0) = [];
This is a great place to use the diff function.
Your first step will be to do the following:
B = [0 diff(a)]
The reason we add the 0 there is to keep the matrix the same length because of the way the diff function works. It will start with the first element in the matrix and then report the difference between that and the next element. There's no leading element before the first one so is just truncates the matrix by one element. We add a zero because there is no change there as it's the starting element.
If you look at the results in B now it is quite obvious where the inflection points are (where you go from positive to negative numbers).
To pull this out programatically there are a number of things you can do. I tend to use a little multiplication and the find command.
Result = find(B(1:end-1).*B(2:end)<0)
This will return the index where you are on the cusp of the inflection. In this case it will be:
ans =
4 7 13