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.
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
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)))
Is is possible to put two for statements into one statement. Something like
A = [ 0 0 0 5
0 2 0 0
1 3 0 0
0 0 4 0];
a=size(A);
b=size(A);
ind=0;
c=0;
for ({i=1:a},{j=1:b})
end
Your question is very broad, but one thing to consider is that in MATLAB you can often take advantage of linear indexing (instead of subscripting), without actually having to reshape the array. For example,
>> A = [ 0 0 0 5
0 2 0 0
1 3 0 0
0 0 4 0];
>> A(3,2)
ans =
3
>> A(7) % A(3+(2-1)*size(A,1))
ans =
3
You can often use this to your advantage in a for loop over all the elements:
for ii=1:numel(A),
A(ii) = A(ii) + 1; % or something more useful
end
Is the same as:
for ii=1:size(A,2),
for jj=1:size(A,1),
A(jj,ii) = A(jj,ii) + 1;
end
end
But to address your specific goal in this problem, as you stated in the comments ("I am storing the non zero elements in another matrix; with elements like the index number, value, row number and column number."), of making sparse matrix representation, it comes to this:
>> [i,j,s] = find(A);
>> [m,n] = size(A);
>> S = sparse(i,j,s,m,n)
S =
(3,1) 1
(2,2) 2
(3,2) 3
(4,3) 4
(1,4) 5
But that's not really relevant to the broader question.
Actually you can combine multiple loops into one for, however it would require you to loop over a vector containing all elements rather than the individual elements.
Here is a way to do it:
iRange = 1:2;
jRange = 1:3;
[iL jL] = ndgrid(iRange,jRange);
ijRange = [iL(:) jL(:)]';
for ij = ijRange
i = ij(1); j = ij(2);
end
Note that looping over the variables may be simpler, but perhaps this method has some advantages as well.
No
read this http://www.mathworks.com/help/matlab/matlab_prog/loop-control-statements.html
i also don't see any added value even if it was possible
No I don't think you can put two for loops in one line.
Depends on your operation, you may be able to reshape it and use one for loop. If you are doing something as simple as just printing out all elements,
B = reshape(A,a*b,1);
for i=1:a*b
c = B(i);
...
end
If i read the doc of how to construct a Halton quasi-random point set and it mentions that it's possible to 'skip' the first values and then retain the 'leap' values.
Don't understand what the 'skip' and 'leap' really mean.
Have tried the following:
>> p = haltonset(1,'Skip',50,'Leap',10); d = haltonset(1,'Skip',51,'Leap',9);
>> p(2:10), d(1:9)
ans =
0.7344
0.0703
0.7891
0.4766
0.5859
0.1797
0.9922
0.3164
0.6602
ans =
0.7969
0.7344
0.8828
0.5391
0.8516
0.6484
0.9609
0.6172
0.7539
>> p(2:10) == d(1:9)
ans =
0
0
0
0
0
0
0
0
0
Thought that it might be that that this would save 10 values to p and 9 to d. Also thought that d would have the same values as p. But this was not the case.
I then tested if the 'leap' would be the same as a normal way to make a vector
- ex: (1:leap:10)
>> p = haltonset(1,'Skip',50,'Leap',1); d = haltonset(1,'Skip',50,'Leap',2);
>> p(1:2:10)==d(1:5)
ans =
1
0
0
0
0
>> p = haltonset(1,'Skip',0,'Leap',1); d = haltonset(1,'Skip',0,'Leap',2);
>> p(1:2:10)==d(1:5)
ans =
1
0
0
0
0
but this seemed not to be the case..
Can anybody give a plain English explanation of how to interpreted the 'skip' and 'leap' variables.
I find the following description to be very clear [quoting this documentation page]:
Imagine a simple 1-D sequence that produces the integers from 1 to 10.
This is the basic sequence and the first three points are [1,2,3]:
Now look at how Scramble, Leap, and Skip work together:
Scramble: Scrambling shuffles the points in one of several
different ways. In this example, assume a scramble turns the sequence
into 1,3,5,7,9,2,4,6,8,10. The first three points are now [1,3,5]:
Skip: A Skip value specifies the number of initial points to
ignore. In this example, set the Skip value to 2. The sequence is now
5,7,9,2,4,6,8,10 and the first three points are [5,7,9]:
Leap: A Leap value specifies the number of points to ignore for
each one you take. Continuing the example with the Skip set to 2, if
you set the Leap to 1, the sequence uses every other point. In this
example, the sequence is now 5,9,4,8 and the first three points are
[5,9,4]:
EDIT:
Let me show with an example:
%# create 1D sequences (x: picked, .: ignored)
p00 = haltonset(1,'Skip',0,'Leap',0); %# xxxxxxxxxxxxxxx
p50 = haltonset(1,'Skip',5,'Leap',0); %# .....xxxxxxxxxx
p02 = haltonset(1,'Skip',0,'Leap',2); %# x..x..x..x..x..
p52 = haltonset(1,'Skip',5,'Leap',2); %# .....x..x..x..x
%# each pair of these are equal
[p50(1:10) p00(6:15)] %# skip vs. noskip
[p02(1:5) p00(1:3:13)] %# leap vs. noleap
[p52(1:4) p00(6:3:15)] %# skip+leap vs. noskip+noleap
In general:
skip = 50;
leap = 10;
p00 = haltonset(1,'Skip',0,'Leap',0);
p = haltonset(1,'Skip',skip,'Leap',leap);
num = 9;
[p(1:num) p00(skip+1:leap+1:num*leap+num-leap+skip)]
new_img is ==>
new_img = zeros(height, width, 3);
curMean is like this: [double, double, double]
new_img(rows,cols,:) = curMean;
so what is wrong here?
The line:
new_img(rows,cols,:) = curMean;
will only work if rows and cols are scalar values. If they are vectors, there are a few options you have to perform the assignment correctly depending on exactly what sort of assignment you are doing. As Jonas mentions in his answer, you can either assign values for every pairwise combination of indices in rows and cols, or you can assign values for each pair [rows(i),cols(i)]. For the case where you are assigning values for every pairwise combination, here are a couple of the ways you can do it:
Break up the assignment into 3 steps, one for each plane in the third dimension:
new_img(rows,cols,1) = curMean(1); %# Assignment for the first plane
new_img(rows,cols,2) = curMean(2); %# Assignment for the second plane
new_img(rows,cols,3) = curMean(3); %# Assignment for the third plane
You could also do this in a for loop as Jonas suggested, but for such a small number of iterations I kinda like to use an "unrolled" version like above.
Use the functions RESHAPE and REPMAT on curMean to reshape and replicate the vector so that it matches the dimensions of the sub-indexed section of new_img:
nRows = numel(rows); %# The number of indices in rows
nCols = numel(cols); %# The number of indices in cols
new_img(rows,cols,:) = repmat(reshape(curMean,[1 1 3]),[nRows nCols]);
For an example of how the above works, let's say I have the following:
new_img = zeros(3,3,3);
rows = [1 2];
cols = [1 2];
curMean = [1 2 3];
Either of the above solutions will give you this result:
>> new_img
new_img(:,:,1) =
1 1 0
1 1 0
0 0 0
new_img(:,:,2) =
2 2 0
2 2 0
0 0 0
new_img(:,:,3) =
3 3 0
3 3 0
0 0 0
Be careful with such assignments!
a=zeros(3);
a([1 3],[1 3]) = 1
a =
1 0 1
0 0 0
1 0 1
In other words, you assign all combinations of row and column indices. If that's what you want, writing
for z = 1:3
newImg(rows,cols,z) = curMean(z);
end
should get what you want (as #gnovice suggested).
However, if rows and cols are matched pairs (i.e. you'd only want to assign 1 to elements (1,1) and (3,3) in the above example), you may be better off writing
for i=1:length(rows)
newImg(rows(i),cols(i),:) = curMean;
end