I would to use cumsum to find the sum up to 5 for example, but I want the end not all the intermediate values.
For example:
A = [1:5];
B = cumsum(A)
returns
B =
1 3 6 10 15
I just want 15 not the rest.
What I have tried:
B = cumsum(A)(end) but it gives me this error:
??? Error: ()-indexing must appear last in an index expression.
The last element of cumsum is per definition equal to the sum. Hence you just need:
A = [1:5];
B = sum(A)
Of course you can also do what you tried to do, with a detour:
A = [1:5];
B = cumsum(A);
B(end)
You can use
C = B(length(A))
or
C = B(end)
or all in one line
C = sum(A)
All these will give you
C =
15
Related
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
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
What I'm trying to do
I have an array of numbers:
>> A = [2 2 2 2 1 3 4 4];
And I want to find the array indices where each number can be found:
>> B = arrayfun(#(x) {find(A==x)}, 1:4);
In other words, this B should tell me:
>> for ii=1:4, fprintf('Item %d in location %s\n',ii,num2str(B{ii})); end
Item 1 in location 5
Item 2 in location 1 2 3 4
Item 3 in location 6
Item 4 in location 7 8
It's like the 2nd output argument of unique, but instead of the first (or last) occurrence, I want all the occurrences. I think this is called a reverse lookup (where the original key is the array index), but please correct me if I'm wrong.
How can I do it faster?
What I have above gives the correct answer, but it scales terribly with the number of unique values. For a real problem (where A has 10M elements with 100k unique values), even this stupid for loop is 100x faster:
>> B = cell(max(A),1);
>> for ii=1:numel(A), B{A(ii)}(end+1)=ii; end
But I feel like this can't possibly be the best way to do it.
We can assume that A contains only integers from 1 to the max (because if it doesn't, I can always pass it through unique to make it so).
That's a simple task for accumarray:
out = accumarray(A(:),(1:numel(A)).',[],#(x) {x}) %'
out{1} = 5
out{2} = 3 4 2 1
out{3} = 6
out{4} = 8 7
However accumarray suffers from not being stable (in the sense of unique's feature), so you might want to have a look here for a stable version of accumarray, if that's a problem.
Above solution also assumes A to be filled with integers, preferably with no gaps in between. If that is not the case, there is no way around a call of unique in advance:
A = [2.1 2.1 2.1 2.1 1.1 3.1 4.1 4.1];
[~,~,subs] = unique(A)
out = accumarray(subs(:),(1:numel(A)).',[],#(x) {x})
To sum up, the most generic solution, working with floats and returning a sorted output could be:
[~,~,subs] = unique(A)
[subs(:,end:-1:1), I] = sortrows(subs(:,end:-1:1)); %// optional
vals = 1:numel(A);
vals = vals(I); %// optional
out = accumarray(subs, vals , [],#(x) {x});
out{1} = 5
out{2} = 1 2 3 4
out{3} = 6
out{4} = 7 8
Benchmark
function [t] = bench()
%// data
a = rand(100);
b = repmat(a,100);
A = b(randperm(10000));
%// functions to compare
fcns = {
#() thewaywewalk(A(:).');
#() cst(A(:).');
};
% timeit
t = zeros(2,1);
for ii = 1:100;
t = t + cellfun(#timeit, fcns);
end
format long
end
function out = thewaywewalk(A)
[~,~,subs] = unique(A);
[subs(:,end:-1:1), I] = sortrows(subs(:,end:-1:1));
idx = 1:numel(A);
out = accumarray(subs, idx(I), [],#(x) {x});
end
function out = cst(A)
[B, IX] = sort(A);
out = mat2cell(IX, 1, diff(find(diff([-Inf,B,Inf])~=0)));
end
0.444075509687511 %// thewaywewalk
0.221888202987325 %// CST-Link
Surprisingly the version with stable accumarray is faster than the unstable one, due to the fact that Matlab prefers sorted arrays to work on.
This solution should work in O(N*log(N)) due sorting, but is quite memory intensive (requires 3x the amount of input memory):
[U, X] = sort(A);
B = mat2cell(X, 1, diff(find(diff([Inf,U,-Inf])~=0)));
I am curious about the performance though.
I have two vectors P and D. I have written the following code which looks at each element of D and then finds the same element in P (both have same elements just in different order) and returns the index of that element in P. The following code is able to do that.
for i=1:17
ind = find(ismember(P,D(i)));
ind
msgbox(sprintf('\n i is: %d\n',ind));
end
The problem is that I want each 'ind' value to be stored in an array.
for i=1:17
ind(i) = find(ismember(P,D(i)));
msgbox(sprintf('\n i is: %d\n',ind));
end
returns an error.
Is it possible assign find(ismember) to different entries of an array?
Thanks!
Edited:
My vectors P and D look as follows:
P = {'Fz' 'Fp1' 'Cz' 'T3' 'T4'}
D = {'T4' 'Cz' 'T3' 'Fp1' 'Fz'}
The error I'm getting at the moment is:
In an assignment A(:) = B, the number of elements in A and B must be the same.
You don't have to implement this yourself, ismember can already do it:
P = {'Fz' 'Fp1' 'Cz' 'T3' 'T4'};
D = {'T4' 'Cz' 'T3' 'Fp1' 'Fz'};
[~, ind] = ismember(D, P)
results in
ind =
5 3 4 2 1
But if you really want to do it yourself – your code basically works already
ind = zeros(size(D));
for i = 1 : numel(D)
ind(i) = find(ismember(P, D(i)));
end
and gives the same result.
I have a vector a = [1 2 3 4 5] how would I make it so b = '12345'?
I have tried b = num2str(a) but it outputs 1 2 3 4 5.
You can specify the format in num2str, much as you would in C's function sprintf:
b = num2str(a,'%i');
Or use sprintf:
b = sprintf('%i',a);
If a contains only single-digit numbers, you can also convert to char directly:
b = char(a+'0');
You need to convert yor vector to a single number first (assuming all elements are in range 0..9):
a = 1:5;
num = ( 10.^((numel(a)-1):-1:0) ) * a'; %'
b = num2str( num )
You can try this code here.