Is it possible in matlab/octave to use the sort function to sort an array based on the relative frequency of their elements?
For example the array
m= [4,4,4,10,10,10,4,4,5]
should result in this array:
[5,10,10,10,4,4,4,4,4]
5 is the less frequent element and is on the top while 4 is the most frequent and it's on bottom.
Should one use the indices provided by histcount?
The following code first calculates how often each element occurs and then uses runLengthDecode to expand the unique elements.
m = [4,4,4,10,10,10,4,4,5];
u_m = unique(m);
elem_count = histc(m,u_m);
[elem_count, idx] = sort(elem_count);
m_sorted = runLengthDecode(elem_count, u_m(idx));
The definition of runLengthDecode is copied from this answer:
For MATLAB R2015a+:
function V = runLengthDecode(runLengths, values)
if nargin<2
values = 1:numel(runLengths);
end
V = repelem(values, runLengths);
end
For versions before R2015a:
function V = runLengthDecode(runLengths, values)
%// Actual computation using column vectors
V = cumsum(accumarray(cumsum([1; runLengths(:)]), 1));
V = V(1:end-1);
%// In case of second argument
if nargin>1
V = reshape(values(V),[],1);
end
%// If original was a row vector, transpose
if size(runLengths,2)>1
V = V.'; %'
end
end
One way would be to use accumarray to find the count of each number (I suspect you can use histcounts(m,max(m))) but then you have to clear all the 0s).
m = [4,4,4,10,10,10,4,4,5];
[~,~,subs]=unique(m);
freq = accumarray(subs,subs,[],#numel);
[~,i2] = sort(freq(subs),'descend');
m(i2)
By combinging my approach with that of m.s. you can get a simpler solution:
m = [4,4,4,10,10,10,4,4,5];
[U,~,i1]=unique(m);
freq= histc(m,U);
[~,i2] = sort(freq(i1),'descend');
m(i2)
You could count the number of repetitions with bsxfun, sort that, and apply that sorting to m:
[~, ind] = sort(sum(bsxfun(#eq,m,m.')));
result = m(ind);
Related
I have:
A matrix 3D: A = (m, n, k).
An array of choices for the third dimension corresponding to each index of the first dimension. idn = (m, 1) (wherein the value of any idn is a random integer in [1,k].
I need to capture the 2D matrix B (m,n) wherein the referred third dimension to A is taken from the corresponding choice. For example:
idn(1) = 1;
idn(2) = k;
idn(j) = k-1;
Then:
B(1,:) = A(1,:,idn(1)) = A(1,:,1);
B(2,:) = A(2,:,idn(2)) = A(2,:,k);
B(j,:) = A(j,:,idn(j)) = A(j,:,k-1);
Since idn is not constant, a simple squeeze could not help.
I have also tried the below code, but it does not work either.
B = A(:,:,idn(:));
It is very much appreciated if anyone could give me a solution.
This could be done with sub2ind and permute, but the simplest way I can think of is using linear indexing manually:
A = rand(3, 4, 5); % example data
idn = [5; 1; 2]; % example data
ind = (1:size(A,1)).' + size(A,1)*size(A,2)*(idn(:)-1); % 1st and 3rd dimensions
ind = ind + size(A,1)*(0:size(A,2)-1); % include 2nd dimension using implicit expansion
B = A(ind); % index into A to get result
I have a question regarding indexing and loops in MATLAB. I have a vector of length n (named data in the code below). I want to examine this vector 4 elements at a time inside of a for loop. How can I do this? My attempt included below does not work because it will exceed the array dimensions at the end of the loop.
for k = 1:length(data)
index = k:k+3;
cur_data = data(index);
pre_q_data1 = cur_data(1);
pre_q_data2 = cur_data(2);
% Interweaving the data
q = [pre_q_data1; pre_q_data2];
qdata = q(:)';
pre_i_data1 = cur_data(3);
pre_i_data2 = cur_data(4);
i = [pre_i_data1; pre_i_data2];
idata = i(:)';
end
You shouldn't have k go all the way to length(data) if you're planning on indexing up to k+3.
I've also taken the liberty of greatly simplifying your code, but feel free to ignore that!
for k = 1:length(data)-3
% maximum k = length(data)-3, so maximum index = length(data)-3+3=length(data)
index = k:k+3;
cur_data = data(k:k+3);
% Interweaving the data
q = cur_data(1:2); % transpose at end of line here if need be
i = cur_data(3:4); % could just use data(k+2:k+3) and not use cur_data
end
I have a vector/ or it could be array :
A = [1,2,3,4,5,1,2,3,4,5,1,2,3]
I want to extract existing different values/elements from this vector without repeating:
1,2,3,4,5
B= [1,2,3,4,5]
How can I extract it ?
I would appreciate for any help please
Try this,
A = [1,2,3,4,5,1,2,3,4,5,1,2,3]
y = unique(A)
B = unique(A) returns the same values as in a but with no repetitions. The resulting vector is sorted in ascending order. A can be a cell array of strings.
B = unique(A,'stable') does the same as above, but without sorting.
B = unique(A,'rows') returns the unique rows ofA`.
[B,i,j] = unique(...) also returns index vectors i and j such that B = A(i) and A = B(j) (or B = A(i,:) and A = B(j,:)).
Reference: http://cens.ioc.ee/local/man/matlab/techdoc/ref/unique.html
Documentation: https://uk.mathworks.com/help/matlab/ref/unique.html
The answers below are correct but if the user does not want to sort the data, you can use unique with the parameter stable
A = [1,2,3,4,5,1,2,3,4,5,1,2,3]
B = unique(A,'stable')
pvec = 1:3;
for i = 1:3
p=pvec(i);
for m = 1:p
erfun=erfc(5/(2*sqrt(p-m)));
suma(m) = sum(erfun)
end
end
I want to save sum of all values of erfun for every p i.e I want to have 3 values in final array but every value in the array should be sum of all the values of erfun for one p.
Similar questions have been addressed but I could not apply them in my case.
Minimal fix method (to your own code)
The following modification to your code will yield your requested results
suma = zeros(3,1);
pvec = 1:3;
for i = 1:3
p=pvec(i);
for m = 1:p
erfun=erfc(5/(2*sqrt(p-m)));
suma(i) = suma(i) + erfun; %// <-- modified here
end
end
Where I've also included suma = zeros(3,1), which I assume that you also have in your code (however not shown in your question); pre-allocating suma with sufficient entries.
Alternative method (arrayfun)
Another solution, you can make use of the arrayfun command to get rid of the inner for loop:
suma = zeros(1,3);
pvec = 1:3;
for i = 1:3
p=pvec(i);
suma(i) = sum(arrayfun(#(x) erfc(5/(2*sqrt(p-x))), 1:p));
end
Alternative method #2 (arrayfun)
An even more condensed solution, including also the purpose of the outer for loop in your arrayfun call:
suma = arrayfun(#(x) ...
sum(erfc(5./(2*sqrt(kron(x, ones(1,x-1)) - 1:(x-1))))), pvec)
Here we've made use of the kron command, which will be implicitly used in the arrayfun command above as follows
kron(1, []) = [] %// empty array
kron(2, [1]) = 2
kron(3, [1 1]) = [3 3]
and used the fact that erfc addition from 1/sqrt(0) is always 0 (i.e., erfc(Inf) = 0, and hence we needn't evaluate the case m=p as it yields no addition to our sum).
Result
All of the above methods yield the result
suma =
0
0.0004
0.0128
I would like to compute the maximum and, more importantly, its coordinates of an N-by-N...by-N array, without specifying its dimensions.
For example, let's take:
A = [2 3];
B = [2 3; 3 4];
The function (lets call it MAXI) should return the following values for matrix A:
[fmax, coor] = MAXI(A)
fmax =
3
coor =
2
and for matrix B:
[fmax, coor] = MAXI(B)
fmax =
4
coor=
2 2
The main problem is not to develop a code that works for one class in particular, but to develop a code that as quickly as possible works for any input (with higher dimensions).
To find the absolute maximum, you'll have to convert your input matrix into a column vector first and find the linear index of the greatest element, and then convert it to the coordinates with ind2sub. This can be a little bit tricky though, because ind2sub requires specifying a known number of output variables. For that purpose we can employ cell arrays and comma-separated lists, like so:
[fmax, coor] = max(A(:));
if ismatrix(A)
C = cell(1:ndims(A));
[C{:}] = ind2sub(size(A), coor);
coor = cell2mat(C);
end
EDIT: I've added an additional if statement that checks if the input is a matrix or a vector, and in case of the latter it returns the linear index itself as is.
In a function, it looks like so:
function [fmax, coor] = maxi(x)
[fmax, coor] = max(A(:));
if ismatrix(A)
C = cell(1:ndims(A));
[C{:}] = ind2sub(size(A), coor);
coor = cell2mat(C);
end
Example
A = [2 3; 3 4];
[fmax, coor] = maxi(A)
fmax =
4
coor =
2 2