The following code generates an cell array Index [1x29], where each cell is an array [29x6]:
for i = 1 : size(P1_cell,1)
for j = 1 : size(P1_cell,2)
[Lia,Lib] = ismember(P1_cell{i,j},PATTERNS_FOR_ERANOS_cell{1},'rows');
Index1(i,j) = Lib % 29x6
end
Index{i} = Index1; % 1x29
end
How can I find the nonzero values in Index array?, i.e. generate an array with the number of non-zero values in each row of the Index1 array. I tried the following loop, but it doesn't work, it creates conflict with the previous one:
for i = 1 : length(Index)
for j = 1 : length(Index)
Non_ceros = length(find(Index{:,i}(j,:))); %% I just need the length of the find function output
end
end
I need help, Thanks in advance.
The nnz() (number of non-zeros) function can be used to evaluate the number of non-zero elements. To obtain the specific positive values you can index the array by using the indices returned by the find() function. I used some random test data but it should work for 29 by 6 sized arrays as well.
%Random test data%
Index{1} = [5 2 3 0 zeros(1,25)];
Index{2} = [9 2 3 1 zeros(1,25)];
Index{3} = [5 5 5 5 zeros(1,25)];
%Initializing and array to count the number of zeroes%
Non_Zero_Counts = zeros(length(Index),1);
for Row_Index = 1: length(Index)
%Evaluating the number of positive values%
Array = Index{Row_Index};
Non_Zero_Counts(Row_Index) = nnz(Array);
%Retrieving the positive values%
Positive_Indices = find(Array);
PositiveElements{Row_Index} = Array(Positive_Indices);
disp(Non_Zero_Counts(Row_Index) + " Non-Zero Elements ");
disp(PositiveElements{Row_Index});
end
Ran using MATLAB R2019b
for i = 1 : length(Index)
for j = 1 : length(Index)
Non_ceros(i,j) = nnz(Index{:,i}(j,:));
end
end
Related
I am trying to generate a matrix, that has all unique combinations of [0 0 1 1], I wrote this code for this:
v1 = [0 0 1 1];
M1 = unique(perms([0 0 1 1]),'rows');
• This isn't ideal, because perms() is seeing each vector element as unique and doing:
4! = 4 * 3 * 2 * 1 = 24 combinations.
• With unique() I tried to delete all the repetitive entries so I end up with the combination matrix M1 →
only [4!/ 2! * (4-2)!] = 6 combinations!
Now, when I try to do something very simple like:
n = 15;
i = 1;
v1 = [zeros(1,n-i) ones(1,i)];
M = unique(perms(vec_1),'rows');
• Instead of getting [15!/ 1! * (15-1)!] = 15 combinations, the perms() function is trying to do
15! = 1.3077e+12 combinations and it's interrupted.
• How would you go about doing in a much better way? Thanks in advance!
You can use nchoosek to return the indicies which should be 1, I think in your heart you knew this must be possible because you were using the definition of nchoosek to determine the expected final number of permutations! So we can use:
idx = nchoosek( 1:N, k );
Where N is the number of elements in your array v1, and k is the number of elements which have the value 1. Then it's simply a case of creating the zeros array and populating the ones.
v1 = [0, 0, 1, 1];
N = numel(v1); % number of elements in array
k = nnz(v1); % number of non-zero elements in array
colidx = nchoosek( 1:N, k ); % column index for ones
rowidx = repmat( 1:size(colidx,1), k, 1 ).'; % row index for ones
M = zeros( size(colidx,1), N ); % create output
M( rowidx(:) + size(M,1) * (colidx(:)-1) ) = 1;
This works for both of your examples without the need for a huge intermediate matrix.
Aside: since you'd have the indicies using this approach, you could instead create a sparse matrix, but whether that's a good idea or not would depend what you're doing after this point.
I have the following cell array A of size 1x9
A= {{'O1'} ,{'O1','O2','O3','O4'} ,{'O1','O3'} ,{'O1','O2','O3','O4'} , {},{'O1','O2','O3','O4'},{'O1','O3'},{'O1','O2','O3','O4'},{'O1','O2','O3','O4'}};
I want to sort a partition of this cell array starting the fifth element A{5:9} according to number of elements in the cell . I have tried the sort function as follows by didn't seems to work
[P,I] = sort(cellfun('length',A{5:9}));
A = A(I);
Also is there any way to keep track of the original indices of the cellarray after sorting ?
The sorting can be performed with the following code/
A= {{'O1'} ,{'O1','O2','O3','O4'} ,{'O1','O3'} ,{'O1','O2','O3','O4'} , {},
{'O1','O2','O3','O4'},{'O1','O3'},{'O1','O2','O3','O4'},{'O1','O2','O3','O4'}};
disp('Before sorting')
for ii = 1:numel(A)
fprintf('%d: %s\n',ii, cell2str(A{ii}));
end
a=A(5:end);
[P,I] = sort( cellfun(#(x) numel(x),a) );
A(5:end) = a(I);
clear a
disp('After sorting')
for ii = 1:numel(A)
fprintf('%d: %s\n',ii, cell2str(A{ii}));
end
The mapping between the initial and final order of the cell array can be found using again I:
originalOrder = 1:numel(A);
finalOrder = originalOrder;
o = originalOrder(5:end);
finalOrder(5:end) = o(I);
clear o
In your case: finalOrder = [1 2 3 4 5 7 6 8 9]
The original position of cell 6in the modified A was finalOrder(6) = 7.
Suppose I have two arrays ordered in an ascending order, i.e.:
A = [1 5 7], B = [1 2 3 6 9 10]
I would like to create from B a new vector B', which contains only the closest values to A values (one for each).
I also need the indexes. So, in my example I would like to get:
B' = [1 6 9], Idx = [1 4 5]
Note that the third value is 9. Indeed 6 is closer to 7 but it is already 'taken' since it is close to 4.
Any idea for a suitable code?
Note: my true arrays are much larger and contain real (not int) values
Also, it is given that B is longer then A
Thanks!
Assuming you want to minimize the overall discrepancies between elements of A and matched elements in B, the problem can be written as an assignment problem of assigning to every row (element of A) a column (element of B) given a cost matrix C. The Hungarian (or Munkres') algorithm solves the assignment problem.
I assume that you want to minimize cumulative squared distance between A and matched elements in B, and use the function [assignment,cost] = munkres(costMat) by Yi Cao from https://www.mathworks.com/matlabcentral/fileexchange/20652-hungarian-algorithm-for-linear-assignment-problems--v2-3-:
A = [1 5 7];
B = [1 2 3 6 9 10];
[Bprime,matches] = matching(A,B)
function [Bprime,matches] = matching(A,B)
C = (repmat(A',1,length(B)) - repmat(B,length(A),1)).^2;
[matches,~] = munkres(C);
Bprime = B(matches);
end
Assuming instead you want to find matches recursively, as suggested by your question, you could either walk through A, for each element in A find the closest remaining element in B and discard it (sortedmatching below); or you could iteratively form and discard the distance-minimizing match between remaining elements in A and B until all elements in A are matched (greedymatching):
A = [1 5 7];
B = [1 2 3 6 9 10];
[~,~,Bprime,matches] = sortedmatching(A,B,[],[])
[~,~,Bprime,matches] = greedymatching(A,B,[],[])
function [A,B,Bprime,matches] = sortedmatching(A,B,Bprime,matches)
[~,ix] = min((A(1) - B).^2);
matches = [matches ix];
Bprime = [Bprime B(ix)];
A = A(2:end);
B(ix) = Inf;
if(not(isempty(A)))
[A,B,Bprime,matches] = sortedmatching(A,B,Bprime,matches);
end
end
function [A,B,Bprime,matches] = greedymatching(A,B,Bprime,matches)
C = (repmat(A',1,length(B)) - repmat(B,length(A),1)).^2;
[minrows,ixrows] = min(C);
[~,ixcol] = min(minrows);
ixrow = ixrows(ixcol);
matches(ixrow) = ixcol;
Bprime(ixrow) = B(ixcol);
A(ixrow) = -Inf;
B(ixcol) = Inf;
if(max(A) > -Inf)
[A,B,Bprime,matches] = greedymatching(A,B,Bprime,matches);
end
end
While producing the same results in your example, all three methods potentially give different answers on the same data.
Normally I would run screaming from for and while loops in Matlab, but in this case I cannot see how the solution could be vectorized. At least it is O(N) (or near enough, depending on how many equally-close matches to each A(i) there are in B). It would be pretty simple to code the following in C and compile it into a mex file, to make it run at optimal speed, but here's a pure-Matlab solution:
function [out, ind] = greedy_nearest(A, B)
if nargin < 1, A = [1 5 7]; end
if nargin < 2, B = [1 2 3 6 9 10]; end
ind = A * 0;
walk = 1;
for i = 1:numel(A)
match = 0;
lastDelta = inf;
while walk < numel(B)
delta = abs(B(walk) - A(i));
if delta < lastDelta, match = walk; end
if delta > lastDelta, break, end
lastDelta = delta;
walk = walk + 1;
end
ind(i) = match;
walk = match + 1;
end
out = B(ind);
You could first get the absolute distance from each value in A to each value in B, sort them and then get the first unique value to a sequence when looking down in each column.
% Get distance from each value in A to each value in B
[~, minIdx] = sort(abs(bsxfun(#minus, A,B.')));
% Get first unique sequence looking down each column
idx = zeros(size(A));
for iCol = 1:numel(A)
for iRow = 1:iCol
if ~ismember(idx, minIdx(iRow,iCol))
idx(iCol) = minIdx(iRow,iCol);
break
end
end
end
The result when applying idx to B
>> idx
1 4 5
>> B(idx)
1 6 9
I have multiple array, number can be arbitrary. but the size of all array is same. How do i add each element of with respective element of all the arrays and maybe save it in another array
A1 = [1 2 3 4 5 6]
A2 = [1 2 3 4 5 6]
.
.
.
.
final = [1+1+1+... 2+2+2+.... 3+3+3+3.... 4+4+4.... 5+5+5+5... 6+6+6+6...]
As your arrays are all the same length you can just add the arrays forming a new array.
final = A1+A2
This function searches in your workspace looking for all variables containing capital 'A'. The for loop adds all found variables. If there are other variables containing 'A', other restrictions has to be made.
variables = who %# all variable names from workspace
index = strmatch('A',variables) %# indices matching "A"
newarray = 0
for j = 1:numel(index)
tmp = eval(char(variables(index(j)))); %# store variable in tmp
newarray = newarray + tmp; %# sum
end
If you have an unknown number of A's, you can try something like this:
final = 0
i = 1
while exist(['A' num2str(i)]) == 1 % ['A' num2str(i)] constructs the variable name, eval calls it
final = final + eval(['A' num2str(i)]);
i = i + 1;
end
This should work as long as the variables are stored in the workspace, are of the same length and are named A1, A2, A3, ... A9, A10, ...
Let's say you have this structure (as you write in the comments):
main = struct('err',{1:6,5:10,1:6,1:6},'seg_err',{1:6,5:10,1:6,5:10});
you can convert it to matrix:
m = vertcat(main.seg_err);;
And than take the sum in a simple command:
final = sum(m)
which results:
final =
12 16 20 24 28 32
and thanks to #beaker :)
I have created a logical array of 1's and 0's using the following code:
nWindow = 10;
LowerTotInitial = std(LowerTot(1:nWindow));
UpperTotInitial = std(UpperTot(1:nWindow));
flag = 0;
flagArray = zeros(length(LowerTot), 1);
for n = 1 : nData0 - nWindow
for k = 0 : nWindow - 1
if LowerTot(n + k) < 0.1*LowerTotInitial || UpperTot(n + k) < 0.1*UpperTotInitial
flag = 1;
flagArray(n) = 1;
else
flag = 0;
end
end
end
This returns flagArray, an array of 0's and 1's. I am trying to find the index of the first 1 in the array. ie. 1 = flagArray(index). I am confused as to what is the best way to accomplish this!
What you call an entry number is referred to as an index in MATLAB-speak. To find the index of the first matching element in an array you can use the FIND function:
>> x = [0 0 1 0 1 0];
>> find(x, 1, 'first')
ans =
3
Try this ind = find(flagArray, k, 'first')
with k =1
Read this Matlab Docs - find