I have the formulas below to generate a linear interpolation in q:
lsfit:{(enlist y) lsq x xexp/: til 1+z};
interp:{[xn;x;y]sum (1;xn)*flip lsfit[x;y;1]};
and the data below to interpolate:
xn:(4.7;7.5;4.9);
x:(3 5f;7.5 7.5;3 5f);
y:(1.3 1.5;2 2f;1.3 1.5);
interp'[xn;x;y]
which is generating
index value
0 enlist 1.47
1 enlist 0nf
2 enlist 1.49
why am I getting 0 in the second row?
Update: Inconsistet behaviour for other examples
xn:(6;7;8;9);
x:(6 6f;7 7f;8 8f;9 9f);
y:(1 1f;1 1f;1 1f;1 1f);
interp'[xn;x;y]
generates
index value
0 enlist 1f
1 enlist 0nf
2 enlist 0nf
3 enlist 1f
So, it looks like sometimes the formula works, rows 0 and 3, and sometimes it does not, rows 1 and 2.
How can I fix it?
Thanks!
The reason you are encountering this issue is because of the mathematical details matrix division.
Matrix division can be performed by taking the inverse of a matrix and then matrix multiplying. In q, this can be seen by performing those operations directly.
q) enlist[2 2f] lsq (1 2f;3 4f)
-1 1
q) enlist[2 2f] mmu inv (1 2f;3 4f)
-1 1
One of your input x values to lsfit is the row 7.5 7.5. With a z value of 1f, this converts that vector into a matrix (1 1;7.5 7.5) in the xexp operation. This matrix is then used in the lsq operation.
The problem then occurs because (1 1;7.5 7.5) is not invertible. A matrix is invertible if and only if the determinant is non-zero. The determinant for a 2 x 2 matrix is AD - BC. In your example, A = 1, B = 1, C = 7.5, and D = 7.5. So the determinant is zero, the matrix is not invertible, and the output from the function is Onf.
To resolve this issue, you would have to ensure that the two items in each row of x are not identical.
Hope that helps.
Related
I have a vector of simple numbers such as:
a=[1 2 3 4 5 6 7 8]
I would like to have all the numbers of the vector that fall in between [25% 75%] quartiles. However, when I use the command below:
quantile(a,[0.25 0.75])
It only gives me 2 numbers of 2 and 6 (instead of 3,4,5,6).
Do you have any solution how I can do it?
Based on the mathematical definition of a quantile, the quantile() function should not be returning {3,4,5,6} given [0.25 0.75].
A quantile of a may be thought of as the inverse of the cumulative distribution function (CDF) for a. Since the CDF Fa(x) = P(a ≤ x) is a right-continuous increasing function, its inverse Fa-1(q) will be a one-to-one function as well.
Thus quantile(0.25) can only return a single value (scalar), the smallest value x such that P(a ≤ x) = 0.25.
However, logical indexing will do the trick. See code below.
% MATLAB R2017a
a = [1 2 3 4 5 6 7 8];
Q = quantile(a,[0.25 0.75]) % returns 25th & 75th quantiles of a
aQ = a(a>=Q(1) & a<=Q(2)) % returns elements of a between 25th & 75th quantiles (inclusive)
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.
We've got a columnvector m x 1 and a matrix m x n.
For the value in row i in the columnvector we want to multiply this value with each value in the same row i of the matrix, and then sum all of these up. This is to be repeated for every row i in the vector so that we end up with a columnvector.
Want to do this with a for-loop, have this so far (where M is the matrix and v is the initial columnvector we start out with) which returns an error that says "Subscripted assignment dimension mismatch.", so I guess I messed up with my indices somehow:
for i = 1:nv
for k = 1:mM
columnvectorendresult(i,) = columnvectorendresult(i,) + v(i,:)*M(i,:);
end
end
Don't know if I'm close with what I have so far, but not fully into this just yet. Any suggestions?
In case you want to sum after multiplication, the answer of knedlsepp using the distributive property of multiplication is the logical choice. If you want to use other operations than sums or differences, than the following answer can be applied more generically
Here we go:
%// columnvector m x 1
a = randi(5,3,1)
%// matrix m x n
B = randi(5,3,2)
%// multiplication
Ba = bsxfun(#times,B,a(:))
%// sum
BaSum = sum(Ba,2)
Example:
a =
3
4
4
B =
2 5
3 1
1 1
Ba =
6 15
12 4
4 4
BaSum =
21
16
8
Instead of multiplying each entry with the same factor and then doing the summation, you should sum the rows of the matrix first and then do the multiplication. ("Use the distributive property of multiplication.")
This is how you do this in MATLAB:
columnvectorendresult = v.*sum(M,2);
Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
I'm translating some MATLAB code to Haskell using the hmatrix library. It's going well, but
I'm stumbling on the pos function, because I don't know what it does or what it's Haskell equivalent will be.
The MATLAB code looks like this:
[U,S,V] = svd(Y,0);
diagS = diag(S);
...
A = U * diag(pos(diagS-tau)) * V';
E = sign(Y) .* pos( abs(Y) - lambda*tau );
M = D - A - E;
My Haskell translation so far:
(u,s,v) = svd y
diagS = diag s
a = u `multiply` (diagS - tau) `multiply` v
This actually type checks ok, but of course, I'm missing the "pos" call, and it throws the error:
inconsistent dimensions in matrix product (3,3) x (4,4)
So I'm guessing pos does something with matrix size? Googling "matlab pos function" didn't turn up anything useful, so any pointers are very much appreciated! (Obviously I don't know much MATLAB)
Incidentally this is for the TILT algorithm to recover low rank textures from a noisy, warped image. I'm very excited about it, even if the math is way beyond me!
Looks like the pos function is defined in a different MATLAB file:
function P = pos(A)
P = A .* double( A > 0 );
I can't quite decipher what this is doing. Assuming that boolean values cast to doubles where "True" == 1.0 and "False" == 0.0
In that case it turns negative values to zero and leaves positive numbers unchanged?
It looks as though pos finds the positive part of a matrix. You could implement this directly with mapMatrix
pos :: (Storable a, Num a) => Matrix a -> Matrix a
pos = mapMatrix go where
go x | x > 0 = x
| otherwise = 0
Though Matlab makes no distinction between Matrix and Vector unlike Haskell.
But it's worth analyzing that Matlab fragment more. Per http://www.mathworks.com/help/matlab/ref/svd.html the first line computes the "economy-sized" Singular Value Decomposition of Y, i.e. three matrices such that
U * S * V = Y
where, assuming Y is m x n then U is m x n, S is n x n and diagonal, and V is n x n. Further, both U and V should be orthonormal. In linear algebraic terms this separates the linear transformation Y into two "rotation" components and the central eigenvalue scaling component.
Since S is diagonal, we extract that diagonal as a vector using diag(S) and then subtract a term tau which must also be a vector. This might produce a diagonal containing negative values which cannot be properly interpreted as eigenvalues, so pos is there to trim out the negative eigenvalues, setting them to 0. We then use diag to convert the resulting vector back into a diagonal matrix and multiply the pieces back together to get A, a modified form of Y.
Note that we can skip some steps in Haskell as svd (and its "economy-sized" partner thinSVD) return vectors of eigenvalues instead of mostly 0'd diagonal matrices.
(u, s, v) = thinSVD y
-- note the trans here, that was the ' in Matlab
a = u `multiply` diag (fmap (max 0) s) `multiply` trans v
Above fmap maps max 0 over the Vector of eigenvalues s and then diag (from Numeric.Container) reinflates the Vector into a Matrix prior to the multiplys. With a little thought it's easy to see that max 0 is just pos applied to a single element.
(A>0) returns the positions of elements of A which are larger than zero,
so forexample, if you have
A = [ -1 2 -3 4
5 6 -7 -8 ]
then B = (A > 0) returns
B = [ 0 1 0 1
1 1 0 0]
Note that we have ones corresponding to an elemnt of A which is larger than zero, and 0 otherwise.
Now if you multiply this elementwise with A using the .* notation, then you are multipling each element of A that is larger than zero with 1, and with zero otherwise. That is, A .* B means
[ -1*0 2*1 -3*0 4*1
5*1 6*1 -7*0 -8*0 ]
giving finally,
[ 0 2 0 4
5 6 0 0 ]
So you need to write your own function that will return positive values intact, and negative values set to zero.
And also, u and v does not match in dimension, for a generall SVD decomposition, so you actually would need to REDIAGONALIZE pos(diagS - Tau), so that u* diagnonalized_(diagS -tau) agrres to v