MATLAB: Locating entered vector values

locaterowvector

I have a vector x=(0:1:9) and the user has to input a vector (single integers). How can i locate the values the user input in the x variable? Can it be done without using loops?

x=[0 1 2 3 4 5 6 7 8 9];

% User input vector

a=[1 4 0 0];

% I was trying this but it fails when i run the code

>> [r,c]=find(ColorNum(rst(1:end)));

Best Answer

ind = find(ismember(x, a))

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MATLAB: Locating Peaks Naively

In case you don't have the Signal Processing Toolbox, or its findpeaks function is not general enough, you could try the following function - though I don't claim it's the the fastest way possible!

function peaks = peakfind(v, x)
%PEAKFIND finds local peaks
%   PEAKS = PEAKFIND(V, X) returns the indices of elements of vector V such
%   that V(P) is greater than V(P + K) for all integers K such that K >=
%   -X, K <= X and K ~= 0. Only peaks for which P > X and P < length(V)+1-X
%   are returned.
%
%   V must be a row or a column vector. X must be a positive integer. The
%   result is a vector containing the indices P.
% Argument checking omitted for speed.
pks = true;
va = v(x+1 : end);
for k = 1:x
   vb = v(x+1-k : end-k);
   pksa = va > vb;
   pksb = vb > va;
   pks = pks & pksa(1 : end-x) & pksb(1+k : end-x+k);
end
peaks = find(pks) + x;
end

One thing to note: if you define a peak to be an element that is greater than all its neighbours, repeated values in the input vector will never be seen as a peak. So [ 1 100 100 1] (with X=1) has no peaks using your definition. If you want to find such plateaux, you can do it by breaking the symmetry and finding one of the values. In the code above, you could replace "pksb = vb > va" with "pksb = ~pksa" (and amending the description) to achieve this.

Best Answer

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MATLAB: Locating Peaks Naively

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