Please consider my sample code:
data=[-1 0 1 2]; % data
N=[4,8,16]; % No. of desired output columns
Now Create a output matrix such that:
out=1xN % having each element of data randomly repeated exactly N/4 times
For N as a scalar it's simple:
data=[-1 0 1 2];
N= 4
R = repmat(data', 1, N); %// You actually you don't really need the '
R(randperm(numel(R)))
Related
Very new to Matlab, I usually use STATA.
I want to use the nchoosek fuction to get the sum of vectors in one matrix.
I have a 21x21 adjacency matrix, with either 0 or 1 as the inputs. I want to create a new matrix, that will give me a sum of inputs between all possible triads from the adjacency matrix.
The new matrix should look have four variables, indexes (i, j, k) - corresponding to each combination from the 21x21. And a final variable which is a sum of the inputs.
The code I have so far is:
C = nchoosek(21,3)
B = zeros(nchoosek(21,3), 4)
for i=1:C
for j=i+1:C
for k=j+1:C
B(?)=B(i, j, k, A(i)+A(j)+A(k)) #A is the 21x21 adj mat
end
end
end
I know my assignment statement is incorrect as I don't completed understand the indexing role of the ":" operator. Any help will be appreciated.
Thanks!
This might be what you want:
clear all
close all
clc
A = rand(21,21); % Replace this with actual A
rowNum = 0;
for i=1:21
for j=i+1:21
for k=j+1:21
rowNum = rowNum+1;
B(rowNum,:) = [i, j, k, sum(A(:,i)+A(:,j)+A(:,k))];
end
end
end
There are some points:
You loop for different combinations. the total number of combination is nchoosek(21,3) which you can check after 3 nested loop. Your code with for i=1:C was the first error since you're actually looping for different values of i and different values of j and k. So these just 21 values not more.
To avoid repeated combinations, it's enough to start new index after the previous one, which you've realized in your code.
There are other possible approaches such as vectorized format, but to stick to your approach, I used a counter: rowNum which is the loop counter and updated along the loop.
B(rowNum,:) means all element of rowNum'th row of the matrix B.
Below is an algorithm to find the triads in an adjacency matrix. It checks all possible triads and sums the values.
%basic adjacency matrix with two triads (1-2-5) (2-3-5)
A=[];
A(1,:) = [0 1 0 0 1];
A(2,:) = [1 0 1 0 1];
A(3,:) = [0 1 0 0 1];
A(4,:) = [0 0 0 0 1];
A(5,:) = [1 1 1 1 0];
A=A==1; %logical matrix
triads=nchoosek(1:5,3);
S=nan(size(triads,1),4);
for ct = 1:size(triads,1)
S(ct,1:3)=[A(triads(ct,1),triads(ct,2)),A(triads(ct,1),triads(ct,3)),A(triads(ct,2),triads(ct,3))];
S(ct,4)=sum(S(ct,1:3));
end
triads(find(S(:,4)==3),:)
ans =
1 2 5
2 3 5
I am trying to create a single column vector (out), which is comprised of a sequence of ones and zeros. These should occur in sets of length B and C respectively, which are repeated A number of times. For example:
out=[1
0
0
1
0
0
1
0
0]
It is currently set up as:
out=[0]; %not ideal, but used to initially define 'out'
A=3;
B=1;
C=2;
for i = 1:length(A)
for ii = 1:length(B)
out(end+1,1) = ones(ii,1);
end
for iii = 1:length(C)
out(end+1,1) = zeros(iii,1);
end
end
This is not working - current output:
out=[0
1
0]
How can I correct these loops to get the desired output? Also, is there a better way of achieving this with the given the inputs?
Many thanks.
1) You do not need to use length as this returns the length of an array type, so A,B,C will all be length of 1.
2) Just directly use the values as shown below. Also you can initialize an empty array with empty brackets []
3) If you're using the zeros and ones commands, these generate whole arrays/matrices and do not need to be in a loop. If you want to keep your loop version, just use =1 or =0
out=[]; %--> you can use this instead
A=3;
B=1;
C=2;
for i = 1:A
out(end+1:end+B,1) = ones(B,1);
out(end+1:end+C,1) = zeros(C,1);
end
... or of course to be more "Matlaby" just do what David said in the comments repmat([ones(B,1);zeros(C,1)],A,1), but the above is there to help you on your way.
How about some modulo arithmetic?
result = double(mod(0:(B+C)*A-1, B+C)<B).';
Example:
>> B = 2; %// number of ones in each period
>> C = 4; %// number of zeros in each period
>> A = 3; %// number of periods
>> result = double(mod(0:(B+C)*A-1, B+C)<B).'
result =
1
1
0
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
I can suggest 2 ways:
a)Using for loop-
A=3;
B=2;
C=3;
OneVector=ones(1,B); % B is the length of ones.
zeroVector=zeros(1,C); % C is the length of zeros.
combinedVector=cat(2,OneVector,zeroVector);
Warehouse=[]; % to save data
for(i=1:A)
Warehouse=cat(2,Warehouse,combinedVector);
end
b)using repmat:
OneVector=ones(1,B); % B is the length of ones.
zeroVector=zeros(1,C); % C is the length of zeros.
combinedVector=cat(2,OneVector,zeroVector);
Warehouse=repmat(combinedVector, [A,1]);
I hope, this will solve your problem.
I have a matrix with constant consecutive values randomly distributed throughout the matrix. I want the indices of the consecutive values, and further, I want a matrix of the same size as the original matrix, where the number of consecutive values are stored in the indices of the consecutive values. For Example
original_matrix = [1 1 1;2 2 3; 1 2 3];
output_matrix = [3 3 3;2 2 0;0 0 0];
I have struggled mightily to find a solution to this problem. It has relevance for meteorological data quality control. For example, if I have a matrix of temperature data from a number of sensors, and I want to know what days had constant consecutive values, and how many days were constant, so I can then flag the data as possibly faulty.
temperature matrix is number of days x number of stations and I want an output matrix that is also number of days x number of stations, where the consecutive values are flagged as described above.
If you have a solution to that, please provide! Thank you.
For this kind of problems, I made my own utility function runlength:
function RL = runlength(M)
% calculates length of runs of consecutive equal items along columns of M
% work along columns, so that you can use linear indexing
% find locations where items change along column
jumps = diff(M) ~= 0;
% add implicit jumps at start and end
ncol = size(jumps, 2);
jumps = [true(1, ncol); jumps; true(1, ncol)];
% find linear indices of starts and stops of runs
ijump = find(jumps);
nrow = size(jumps, 1);
istart = ijump(rem(ijump, nrow) ~= 0); % remove fake starts in last row
istop = ijump(rem(ijump, nrow) ~= 1); % remove fake stops in first row
rl = istop - istart;
assert(sum(rl) == numel(M))
% make matrix of 'derivative' of runlength
% don't need last row, but needs same size as jumps for indices to be valid
dRL = zeros(size(jumps));
dRL(istart) = rl;
dRL(istop) = dRL(istop) - rl;
% remove last row and 'integrate' to get runlength
RL = cumsum(dRL(1:end-1,:));
It only works along columns since it uses linear indexing. Since you want do something similar along rows, you need to transpose back and forth, so you could use it for your case like so:
>> original = [1 1 1;2 2 3; 1 2 3];
>> original = original.'; % transpose, since runlength works along columns
>> output = runlength(original);
>> output = output.'; % transpose back
>> output(output == 1) = 0; % see hitzg's comment
>> output
output =
3 3 3
2 2 0
0 0 0
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
I have written a for loop code and I want to write in more succinct way without using a for loop, but instead use matrix conditional.
I am teaching myself matlab and I would appreciate any feedback.
I want to create a new matrix, the first column is y, and the second column is filled with zero except for the y's whose indices are contained in the indices matrix. And in the latter case, add 1 instead of 0.
Thanks.
y=[1;2;3;4;5;6;7];
indices=[1;3;5];
[m,n]=size(y);
tem=zeros(m,1);
data=[y,tem];
[r,c]=size(indices);
for i=1:r
a=indices(i);
data(a,2 )=1;
end
Output:
data =
1 1
2 0
3 1
4 0
5 1
6 0
7 0
A shorter alternative:
data = [y(:), full(sparse(indices, 1, 1, numel(y), 1))];
The resulting matrix data is composed of two column vectors: y(:) and a sparse array, with "1"s at the positions corresponding to indices.
Using proper initialization and sparse matrices can be really useful in MATLAB.
How about
data = zeros( m, 2 );
data(:,1) = y;
data( indices, 2 ) = 1;