y = rand(20,3);
aa= unidrnd(2,20,3) - 1;
val = ( aa & y<1.366e-04) | (~aa & y<8.298e-04);
aa(val) = ~aa(val);
I have this code.
Can any one explain to me what is happening here. I have tried to understand it step by step (debugging) but I cannot understand the purpose of using inverse '~' in line 4 and also using 'val' as indices.
y = rand(20,3);
Creates a matrix of uniformly distributed random numbers, y.
aa= unidrnd(2,20,3) - 1;
Creates a matrix of uniformly distributed random integers, that goes from 1 to 2, and then subtract one. Thus, aa is a matrix of 0s and 1s.
val = ( aa & y<1.366e-04) | (~aa & y<8.298e-04);
This line checks all the values where aa is 1AND y<1.366e-04 OR aa is 0 AND y<8.298e-04. Note that this barely happens, being y uniformly distributed numbers from 0 to 1, being them this smalls is unlikely.
aa(val) = ~aa(val);
Take all those cases computed before, and make aa change from 0 to 1 or from 1 to 0 if it happened in that index.
Related
I have the following in table in matlab
k ak
0 1
1 -0.166666667
2 0.008333333
... ...
where ak = (-1)^k/(2k+1)! but that's not really important, all that is important is that ak is a function of k and the (-1)^k.
I am trying to generate a new set of columns as below
i ai
0 1
1 0
2 -0.166666667
3 0
4 0.008333333
... ...
So far I've figured out that by stretching my indices (incrementing by 2 instead of 1) and applying the ak function it gets pretty close to what I want but it returns the absolute value.
if mod(i,2)=0
ai=(-1)^i/factorial(2*(i/2)+1);
else
ai=0;
end
How can I change this to be what I want?
You should be using this:
ai = (-1)^(ii/2)/factorial(2*(ii/2)+1);
% Notice ----^
Your if-else statements can be given a vector input and written in one-line as:
ai = ~mod(ii,2) .* (-1).^(ii/2)./factorial(2*(ii/2)+1);
%I used ii instead of i since i is for imaginary numbers in MATLAB
I have a set of independent binary random variables (say A,B,C) which take a positive value with some probability and zero otherwise, for which I have generated a matrix of 0s and 1s of all possible combinations of these variables with at least a 1 i.e.
A B C
1 0 0
0 1 0
0 0 1
1 1 0
etc.
I know the values and probabilities of A,B,C so I can calculate E(X) and E(X^2) for each. I want to treat each combination in the above matrix as a new random variable equal to the product of the random variables which are present in that combination (show a 1 in the matrix). For example, random variable Row4 = A*B.
I have created a matrix of the same size to the above, which shows the relevant E(X)s instead of the 1s, and 1s instead of the 0s. This allows me to easily calculate the vector of Expected values of the new random variables (one per combination) as the product of each row. I have also generated a similar matrix which shows E(X^2) instead of E(X), and another one which shows prob(X>0) instead of E(X).
I'm looking for a Matlab script that computes the Covariance matrix of these new variables i.e. taking each row as a random variable. I presume it will have to use the formula:
Cov(X,Y)=E(XY)-E(X)E(Y)
For example, for rows (1 1 0) and (1 0 1):
Cov(X,Y)=E[(AB)(AC)]-E(X)E(Y)
=E[(A^2)BC]-E(X)E(Y)
=E(A^2)E(B)E(C)-E(X)E(Y)
These values I already have from the matrices I've mentioned above. For each Covariance, I'm just unsure how to know which two variables appear in both rows, because for those I will have to select E(X^2) instead of E(X).
Alternatively, the above can be written as:
Cov(X,Y)=E(X)E(Y)*[1/prob(A>0)-1]
But the problem remains as the probabilities in the denominator will only be the ones of the variables which are shared between two combinations.
Any advice on how automate the computation of the Covariance matrix in Matlab would be greatly appreciated.
I'm pretty sure this is not the most efficient way to do that but that's a start:
Assume r1...n the combinations of the random variables, R is the matrix:
A B C
r1 1 0 0
r2 0 1 0
r3 0 0 1
r4 1 1 0
If you have the vector E1, E2 and ER as:
E1 = [E(A) E(B) E(C) ...]
E2 = [E(A²) E(B²) E(C²) ...]
ER = [E(r1) E(r2) E(r3) ...]
If you want to compute E(r1,r2) you can:
1) Extract the R1 and R2 columns from R
v1 = R(1,:)
v2 = R(2,:)
2) Sum both vectors in vs
vs = v1 + v2
3) Loop in vs, if you see a 2 that means the value in R2 has to be used, if you see a 1 it is the value in R1, if it is 0 do not use the value.
4) Using the loop, compute your E(r1,r2) as wanted.
This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
Closed 6 years ago.
This is a basic program but since I'm new to MATLAB, I'm not able to figure out the solution.
I have a column vector "Time" in which I want to print value "1" in first 147 cells, followed by "2" in 148 to 2*147 cells and so on. For that, I have written the following script:
Trial>> c=1;
Trial>> k=0;
Trial>> for i = c:146+c
Time(i,1)=1+k;
c=i;
k=k+1;
end
I know I need to iterate the loop over "Time(i,1)=1+k;" before it executes the next statement. I tried using break but that's not supposed to work. Can anyone suggest me the solution to get the desired results?(It was quite simple in C with just the use of curly braces.)
I am sure you don't want to run c=i; in every iteration.
My code should work for you:
x = 10; % Replace 10 by the max number you need in your array.
k = 1;
for i = 1 : x * 147
Time(i, 1) = k;
if rem(i, 147) == 0
k = k + 1;
end
end
This is the prime example of a piece of code that should be vectorized can help you understand vectorization. Your code can be written like this:
n = 147;
reps = 10; %% Replace this by the maximum number you want your matrix to have
Time = reshape(bsxfun(#plus, zeros(n,1), 0:reps), 1, []);
Explanation:
Let A be a column vector (1 column, n rows), and B be a row vector (1 row, m columns.
What bsxfun(#plus, A, B) will do here is to add all elements in A with all elements in B, like this:
A(1)+B(1) A(1)+B(2) A(1)+B(3) ... A(1)+B(m)
A(2)+B(1) A(2)+B(2) ............. A(2)+B(m)
............................................
A(n)+B(1) A(n)+B(2) .............. A(n)+B(m)
Now, for the two vectors we have: zeros(n,1), and 0:reps, this will give us;
0+0 0+1 0+2 0+reps
0+0 0+1 0+2 0+reps
% n rows of this
So, what we need to do now is place each column underneath each other, so that you will have the column with zeros first, then the row with ones, ... and finally the one with reps (147 in your case).
This can be achieved by reshaping the matrix:
reshape(bsxfun(#plus, zeros(n,1), 0:reps), [], 1);
^ ^ ^ ^
| | | Number of rows in the new matrix. When [] is used, the appropriate value will be chosen by Matlab
| | Number of rows in the new matrix
| matrix to reshape
reshape command
Another approach is using kron:
kron(ones(reps+1, 1) * 0:(n-1)
For the record, a review of your code:
You should always preallocate memory for matrices that are created inside loops. In this case you know it will become a matrix of dimensions ((reps+1)*n-by-1). This means you should do Time = zeros((reps+1)*n, 1);. This will speed up your code a lot.
You shouldn't use i and j as variable names in Matlab, as they denote the imaginary unit (sqrt(-1)). You can for instance do: for ii = 1:(n*147) instead.
You don't want c=i inside the loop, when the loop is supposed to go from c to c + 146. That doesn't make much sense.
You can use repmat,
x = 10; % Sequence length (or what ever it can be called)
M = repmat(1:x,147,1); % Replicate array 1:x for 147 columns
M = M(:); % Reshape the matrix so that is becomes a column vector.
I can assume that this is a task to practice for loops, but this will work.
An alternative solution may be to do
n = 147;
reps = 10;
a = ceil( (1:(n*reps)) / n);
You first construct an array with the length you want. Then you divide, and round of upwards. 1 to 147 will then become 1.
How can I construct a scrambled matrix with 128 rows and 32 columns in vb.net or Matlab?
Entries of the matrix are numbers between 1 and 32 with the condition that each row mustn't contain duplicate elements and rows mustn't be duplicates.
This is similar to #thewaywewalk's answer, but makes sure that the matrix has no repeated rows by testing if it does and in that case generating a new matrix:
done = 0;
while ~done
[~, matrix] = sort(rand(128,32),2);
%// generate each row as a random permutation, independently of other rows.
%// This line was inspired by randperm code
done = size(unique(matrix,'rows'),1) == 128;
%// in the event that there are repeated rows: generate matrix again
end
If my computations are correct, the probability that the matrix has repteated rows (and thus has to be generated again) is less than
>> 128*127/factorial(32)
ans =
6.1779e-032
Hey, it's more likely that a cosmic ray will spoil a given run of the program! So I guess you can safely remove the while loop :-)
With randperm you can generate one row:
row = randperm(32)
if this vector wouldn't be that long you could just use perms to find all permutations:
B = perms(randperm(32))
but it's memory-wise too much! ( 32! = 2.6313e+35 rows )
so you can use a little loop:
N = 200;
A = zeros(N,32);
for ii = 1:N
A(ii,:) = randperm(32);
end
B = unique(A, 'rows');
B = B(1:128,:);
For my tests it was sufficient to use N = 128 directly and skip the last two lines, because with 2.6313e+35 possibly permutations the probability that you get a correct matrix with the first try is very high. But to be sure that there are no row-duplicates choose a higher number and select the first 128 rows finally. In case the input vector is relatively short and the number of desired rows close to the total number of possible permutations use the proposed perms(randperm( n )).
small example for intergers from 1 to 4 and a selection of 10 out of 24 possible permutations:
N = 20;
A = zeros(N,4);
for ii = 1:N
A(ii,:) = randperm(4);
end
B = unique(A, 'rows');
B = B(1:10,:);
returns:
B =
1 2 3 4
1 2 4 3
1 3 4 2
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
some additional remarks for the choice of N:
I made some test runs, where I used the loop above to find all permutations like perms does. For vector lengths of n=4 to n=7 and in each case N = factorial(n): 60-80% of the rows are unique.
So for small n I would recommend to choose N as follows to be absolutely on the safe side:
N = min( [Q factorial(n)] )*2;
where Q is the number of permutations you want. For bigger n you either run out of memory while searching for all permutations, or the desired subset is so small compared to the number of all possible permutations that repetition is very unlikely! (Cosmic Ray theory linked by Luis Mendo)
Your requirements are very loose and allow many different possibilities. The most efficient solution I can think off that meets these requirements is as follows:
p = perms(1:6);
[p(1:128,:) repmat(7:32,128,1)]
Say I have matrices A and B. I want to create a third matrix C where
A = [1,0,0,1]
B = [1,0,1,0]
C = [11, 00, 01, 10]
Is there such a function in Matlab? If not how would I go about creating a function that does this?
Edit: C is not literal numbers. They are concatenated values of A,B element-wise.
Edit2: The actual problem I am dealing with is I have 10 large matrices of the size [x,y] where x,y > 1000. The elements in these matrices all have 0s and 1s. No other number. What I need to accomplish is have the element [x1,y1] in matrix 1 to be appended to the element in [x1,y1] of matrix 2. and then that value to be appended to [x1,y1] of matrix 3.
Another example:
A = [1,1,1,1;
0,0,0,0]
B = [0,0,0,0;
1,1,1,1]
C = [1,0,1,0;
0,1,0,1]
And I need a matrix D where
D = [101, 100, 101, 101; 010, 011, 010, 011]
I recommend that you avoid manipulating binary numbers as strings where possible. It seems tempting, and there are cases where matlab provides a more elegant solution if you treat a binary number as a string, but you cannot perform binary arithmetic on strings. You can always work with integers as if they were binary (they are stored as bits in your machine after all), and then just display them as binary numbers using dec2bin when necessary.
A = [1,0,0,1]
B = [1,0,1,0]
C = bitshift(A,1)+B;
display(dec2bin(C));
In the other case you show in your question you could use:
A = [1,1,1,1; 0,0,0,0];
B = [0,0,0,0; 1,1,1,1];
C = [1,0,1,0; 0,1,0,1];
D = bitshift(A,2) + bitshift(B,1) + C;
You can also convert an arbitrary length row vector of zeros and ones into its decimal equivalent by defining this simple function:
mat2dec = #(x) x*2.^(size(x,2)-1:-1:0)';
This will also work for matrices too. For example
>> M = [0 0 1; 0 1 1; 0 1 0; 1 1 0; 1 1 1; 1 1 0; 1 0 0];
>> dec2bin(mat2dec(M))
ans =
001
011
010
110
111
110
100
In my experience, treating binary numbers as strings obfuscates your code and is not very flexible. For example, try adding two binary "strings" together. You have to use bin2dec every time, so why not just leave the numbers as numbers until you want to display them? You have already run into some of the issues caused by strings of different lengths too. You will be amazed how one simple change can break everything when treating numbers as strings. The worst part is that an algorithm may work great for one set of data and not for another. If all I test with is two-bit binary numbers and a three-bit number somehow sneaks its way in, I may not see an error, but my results will be inexplicably incorrect. I realize that this is a very subjective issue, and I think that I definitely stand in the minority on StackOverflow, so take it for what it's worth.
It depends how you want the output formatted. You could apply bitshift to the numerical values and convert to binary:
>> b = dec2bin(bitshift(A,1)+B)
b =
11
00
01
10
For a general matrix Digits:
>> Digits
Digits =
1 0 0 0
0 1 0 0
1 0 1 1
1 1 0 0
1 0 0 0
1 0 1 1
1 1 0 0
1 0 1 0
0 1 1 1
0 1 1 1
>> Ddec = D*(2.^(size(D,2)-1:-1:0))'; % dot product
>> dec2bin(Ddec)
ans =
1000
0100
1011
1100
1000
1011
1100
1010
0111
0111
Another way to write that is dec2bin(sum(D .* repmat(2.^(size(D,2)-1:-1:0),size(D,1),1),2)).
For your larger problem, with 10 large matrixes (say M1, M2, ..., M10), you can build the starting Digits matrix by:
Digits = [M1(:) M2(:) M3(:) M4(:) M5(:) M6(:) M7(:) M8(:) M9(:) M10(:)];
If that is reverse the order of the digits, just do Digits = fliplr(Digits);.
If you would rather not reshape anything you can compute the matrix decimal values from the matrices of digits as follows:
M = cat(3,A,B,C);
Ddec = sum(bsxfun(#times,M,permute(2.^(size(M,3)-1:-1:0),[1 3 2])),3)
I see some quite extensive answers so perhaps this is thinking too simple, but how about just this assuming you have vectors of ones and zeros representing your binary numbers:
A = [1,0,0,1];
B = [1,0,1,0];
C = 10*A + B
This should give you the numbers you want, you may want to add leading zeros. Of course this method can easily be expanded to append multiple matrices, you just need to make sure there is a factor (base) 10 between them before adding.