I have a matrix with 1 column of data in it. The column has 1556480 points of data in it. Call the matrix Vmatrix. I have another matrix with 1520 values. Call this Vmean_matrix. Is it possible to that a for loop can be created to subtract the first value in Vmean_matrix from the first 1024 values in Vmatrix and the second value in Vmean_matrix from the values 1025 - 2048 in matrix Vmatrix and so on?
Reshape Vmatrix into a 1024-row matrix, reshape Vmean_matrix into a single row, and subtract with bsxfun:
result = bsxfun(#minus, reshape(Vmatrix, 1024, []), Vmean_matrix(:).'); %'// 1024 rows
result = result(:); %// linearize if needed
This may be a way:
% // Vmatrix = ...
% // Vmean_matrix = ...
len = length(Vmean_matrix);
sub = [];
for ii = 0 : len - 1
sub = [sub; Vmatrix( ii*1024+1 : (ii+1)*1024 ) - Vmean_matrix(ii+1)];
end
Or to make it faster, you can write it like this way:
% // Vmatrix = ...
% // Vmean_matrix = ...
len = length(Vmean_matrix);
sub = zeros(length(Vmatrix), 1);
for ii = 0 : len - 1
sub( ii*1024+1 : (ii+1)*1024 ) = Vmatrix( ii*1024+1 : (ii+1)*1024 ) - Vmean_matrix(ii+1);
end
Related
I have two arrays, der_pos and der_neg, which contain indices of an array interpolated. I want to get all the indices for the maximum values of interpolated in the intervals:
der_pos(1):der_neg(1)
der_pos(2):der_neg(2)
etc...
E.g.:
interpolated = [1,5,3,2,7,10,8,14,4]
der_pos = [1,6]
der_neg = [4,9]
So, I would like to obtain the indices:
[2,8]
Because:
in the interval der_pos(1):der_neg(1) → 1:4 → interpolated(1:4) = [1,5,3,2] the max is 5 which is at index 2.
in the interval der_pos(2):der_neg(2) → 6:9 → interpolated(6:9) = [10,8,14,4] the max is 14 which is at index 8.
I managed to do it using a for loop:
interpolated = [1,5,3,2,7,10,8,14,4];
der_pos = [1,6];
der_neg = [4,9];
indices = zeros(1,length(der_pos));
for i = [1:length(der_pos)]
[peak, index] = max(interpolated(der_pos(i):der_neg(i)));
indices(i) = index + der_pos(i) - 1;
endfor
indices % gives [2,8]
But is there a more concise way of doing this?
Here's a sample code. The function findpeaks returns all the peak. The loop then keeps indices of peaks that are in the wanted range.
I added a test to avoid errors in case of no found peak (index will be -1), and to keep the first peak if two peaks are found. You can use a cell if you want to keep all peaks in an interval.
interpolated = [1,5,3,2,7,10,8,14,4];
der_pos = [1 6 7 ];
der_neg = [4 9 8];
[~,i]=findpeaks(interpolated);
indices= -1+zeros(size(der_pos,2),1);
for loopi = 1:length(i)
val = i(i>=der_pos(loopi)&i<=der_neg(loopi));
if ~isempty(val)
indices(loopi) = val(1);
end
end
Here's a way:
interpolated = [1,5,3,2,7,10,8,14,4]; % data
der_pos = [1,6]; % data
der_neg = [4,9]; % data
m = bsxfun(#ge, 1:numel(interpolated), der_pos(:)) .* ...
bsxfun(#le, 1:numel(interpolated), der_neg(:)); % each row is a mask that contains
% 1 for values in the interval and 0 for values outside the interval
m(~m) = NaN; % replace 0 by NaN
[val, result] = max(bsxfun(#times, m, interpolated(:).'), [], 2); % val is the maximum
% of each row, and result is its column index. The maximum is 1 for rows that
% contain at least a 1, and NaN for rows that only contain NaN
result(isnan(val)) = 0; % If the maximum value was NaN the result is set to 0
% (or maybe use NaN), to indicate that the interval was empty
This gives 0 for empty intervals. For example, der_pos = [1,6,8]; der_neg = [4,9,6]; produce result = [2;8;0].
The intervals may overlap. For example, der_pos = [1,6,3]; der_neg = [4,9,7]; produce result = [2;8;6].
We are trying to write a function that takes arr and counts how many 0's and 1's appear in sequence. The output should be a 2D array where column 1 is how many appear in sequence and column 2 is which token it is (0 or 1). Our function below
function [token] = tokenizeSignal(arr)
matA = diff(find(diff([log_vector;-1])));
addA = zeros(size(matA, 1),1);
matA = [matA, addA];
matB = diff(find(diff([log_vector;0])));
addB = ones(size(matB, 1), 1);
matB = [matB, addB];
[nRowsA, nCols] = size(matA);
nRowsB = size(matB, 1);
AB = zeros(nRowsA + nRowsB, nCols);
AB(1:2:end, :) = matA;
AB(2:2:end, :) = matB;
token = AB;
works with
arr = [0; 0; 0; 1; 1; 1; 0];
but nothing else because it adds random integers into the matrix. Why does it do this and how can I fix it?
Here is code that takes any array arr and produces what you want:
% input checking/processing
% ... convert the input into a column vector
arr = arr(:);
% ... check that the input is nonempty and numeric
if ~isnumeric(arr), error('Bad input'); end
if isempty(arr), error('Bad input'); end
% determine the starting indices of each sequence in arr
I = [1 ; find(diff(arr)) + 1];
% determine the values of each of these sequences
values = arr(I);
% determine the length of each of these sequences
value_counts = [diff(I) ; length(arr) - max(I) + 1];
% produce the output
token = [value_counts, values];
I have an array A (I have written so as to make it similar to the matrix that I am using) :
%%%%%%%%%%%%% This is Matrix %%%%%%%%%%%%%%%%%%%%
a = 3; b = 240; c = 10; d = 30; e = 1;
mtx1 = a.*rand(30,1) + a;
mtx2 = round((b-c).*rand(30,1));
mtx3 = round((d-e).*rand(30,1));
mtx4 = -9999.*ones(30,1);
A = [mtx1 mtx2 mtx3 mtx4];
for i = 10:12
for ii = 17 :19
A(i,:)= -9999;
A(ii,:)= 999;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
I would calculate some statistical values, excluding from the calculation the values **-9999 and 999.
the statistical values must be calculated with respect to each column.
the columns represent respectively: the wind speed, direction, and
other parameters
I wrote a code but it is not correct
[nr,ncc]=size(A);
for i=1:ncc
B = A(:,i); %// Temp Vector
Oup=1; Odw=1; %// for Vector Control
while Oup>0 %// || Odw>0 % Oup>0 OR Odw>0 , Oup>0 && (AND) Odw>0
B=sort(B,'descend');
U = find(B<999 & B>-9999); % find for each column of the temp
%vector
Oup = length(U); % Calculates the length
B(U)=[]; % Delete values -9999 and 9999
end
% calculates parameters with the vector temp
count(i)=length(B);
med(i)=mean(B);
devst(i)=std(B);
mediana(i)=median(B);
vari(i)=var(B);
kurt(i)=kurtosis(B);
Asimm(i)=skewness(B);
Interv(i)=range(B);
Mass(i)=max(B);
Mini(i)=min(B);
if length(B)<nr
B(length(B)+1:nr)=nan;
end
C(:,i)=B(:); %//reconstruction of the original matrix
end
would you have any suggestions?
If your data set is in A, and you want to operate on it with a function f, just use logical indexing, i.e.:
f(A( ~(A==999 & A==-9999) )) =...
Alternatively, use find and linear indexing:
ind = find( ~(A==999 & A==-9999) );
f(A(ind)) = ....
I have a MATLAB program like this
for m = 1:2
%# Some code to calculate a matrix (Ytotale)
%# Size of Ytotale is (1200 * 144) %%
%#...
Yfinal = Ytotale;
for l = 1:1200
i = l;
j = retard(l,1);
if Yfinal(i,j) == 0
Yfinal(i,j:end) = circshift(Yfinal(i,j:end),[retard(l,2) retard(l,2)]);
for j = retard(l,1):retard(l,1)+retard(l,2)-1
Yfinal(i,j) = 1;
end
else
Yfinal(i,j:end) = circshift(Yfinal(i,j:end),[retard(l,2) retard(l,2)]);
for j = retard(l,1):retard(l,1)+retard(l,2)-1
Yfinal(i,j) = 0;
end
end
end
%# ( Here i , j are index of matrix Ytotale , and l is the index
%# of matrix retard of size (1200 * 2)
for i =1:1200
not_char(i,1) = sum(Yfinal(i,1:144));
req(i,1) = sum(Ytotale(i,1:144));
end
final = req - not_char;
ve_delay = sum(Yfinal(:,1:144))';
end
The total process will iterate from m = 1 to 2 and two Ytotale matrix will form, hence I want to store the value of ve_delay and final in a row matrix for each Ytotale , but my code overwrites the matrix values .
please help...
This answer is adapted from the comment by macduf
Try final{m} = req - not_char; and ve_delay{m} = sum(Yfinal(:,1:144)); . These values are now stored in a cell matrix (the curly bracket notation). You can convert the cell array into a regular array afterward.
I have two matrices of big sizes, which are something similar to the following matrices.
m; with size 1000 by 10
n; with size 1 by 10.
I would like to subtract each element of n from all elements of m to get ten different matrices, each has size of 1000 by 10.
I started as follows
clc;clear;
nrow = 10000;
ncol = 10;
t = length(n)
for i = 1:nrow;
for j = 1:ncol;
for t = 1:length(n);
m1(i,j) = m(i,j)-n(1);
m2(i,j) = m(i,j)-n(2);
m3(i,j) = m(i,j)-n(3);
m4(i,j) = m(i,j)-n(4);
m5(i,j) = m(i,j)-n(5);
m6(i,j) = m(i,j)-n(6);
m7(i,j) = m(i,j)-n(7);
m8(i,j) = m(i,j)-n(8);
m9(i,j) = m(i,j)-n(9);
m10(i,j) = m(i,j)-n(10);
end
end
end
can any one help me how can I do it without writing the ten equations inside the loop? Or can suggest me any convenient way especially when the two matrices has many columns.
Why can't you just do this:
m01 = m - n(1);
...
m10 = m - n(10);
What do you need the loop for?
Even better:
N = length(n);
m2 = cell(N, 1);
for k = 1:N
m2{k} = m - n(k);
end
Here we go loopless:
nrow = 10000;
ncol = 10;
%example data
m = ones(nrow,ncol);
n = 1:ncol;
M = repmat(m,1,1,ncol);
N = permute( repmat(n,nrow,1,ncol) , [1 3 2] );
result = bsxfun(#minus, M, N );
%or just
result = M-N;
Elapsed time is 0.018499 seconds.
or as recommended by Luis Mendo:
M = repmat(m,1,1,ncol);
result = bsxfun(#minus, m, permute(n, [1 3 2]) );
Elapsed time is 0.000094 seconds.
please make sure that your input vectors have the same orientation like in my example, otherwise you could get in trouble. You should be able to obtain that by transposements or you have to modify this line:
permute( repmat(n,nrow,1,ncol) , [1 3 2] )
according to your needs.
You mentioned in a comment that you want to count the negative elements in each of the obtained columns:
A = result; %backup results
A(A > 0) = 0; %set non-negative elements to zero
D = sum( logical(A),3 );
which will return the desired 10000x10 matrix with quantities of negative elements. (Please verify it, I may got a little confused with the dimensions ;))
Create the three dimensional result matrix. Store your results, for example, in third dimension.
clc;clear;
nrow = 10000;
ncol = 10;
N = length(n);
resultMatrix = zeros(nrow, ncol, N);
neg = zeros(ncol, N); % amount of negative values
for j = 1:ncol
for i = 1:nrow
for t = 1:N
resultMatrix(i,j,t) = m(i,j) - n(t);
end
end
for t = 1:N
neg(j,t) = length( find(resultMatrix(:,j,t) < 0) );
end
end