I am trying to implement the QR decomposition via householder reflectors. While attempting this on a very simple array, I am getting weird numbers. Anyone who can tell me, also, why using the # vs * operator between vec and vec.T on the last line of the function definition gets major bonus points.
This has stumped two math/comp sci phds as of this morning.
import numpy as np
def householder(vec):
vec[0] += np.sign(vec[0])*np.linalg.norm(vec)
vec = vec/vec[0]
gamma = 2/(np.linalg.norm(vec)**2)
return np.identity(len(vec)) - gamma*(vec*vec.T)
array = np.array([1, 3 ,4])
Q = householder(array)
print(Q#array)
Output:
array([-4.06557377, -7.06557377, -6.06557377])
Where it should be:
array([5.09, 0, 0])
* is elementwise multiplication, # is matrix multiplication. Both have their uses, but for matrix calculations you most likely want the matrix product.
vec.T for an array returns the same array. A simple array only has one dimension, there is nothing to transpose. vec*vec.T just returns the elementwise squared array.
You might want to use vec=vec.reshape(-1,1) to get a proper column vector, a one-column matrix. Then vec*vec.T does "by accident" the correct thing. You might want to put the matrix multiplication operator there anyway.
Related
I have a numpy array of matrices which i am trying to multiply together in the form A * B * C * D where A is the first matrix, B is the second and so on. I have tried this code:
matrix = matrices[0]
for m in matrices[1:]:
matrix = np.matmul(matrix, m)
However I believe this multiplication is wrong as i get incorrect output variables, and I have triple checked the rest of my code so I believe this is the issue. How can I multiply all the matrices in this array together? Also the array length will vary depending on the input file, thus i cant use an A * B * C approach.
Your code for multiply a series of matrices together should work. Here is an example using your method with some simple matrices.
import numpy as np
matrices = []
matrices.append(np.eye(3,dtype=float))
matrices.append(np.matrix('1.0,2.0,3.0;4.0,5.0,6.0;7.0,8.0,8.0'))
matrices.append(np.eye(3,dtype=float))
matrices.append(np.linalg.inv(np.matrix('1.0,2.0,3.0;4.0,5.0,6.0;7.0,8.0,8.0')))
matrix = matrices[0]
for m in matrices[1:]:
matrix = np.matmul(matrix, m)
print(matrix)
directmul = np.matmul(matrices[1],matrices[3])
print(np.subtract(matrix,directmul))
Your problem is somewhere else: maybe how you are filling the list of matrices, or how are you filling the matrices. Have you tried unit testing on your code? Have you give a try to the python debugger?
For example, I have an equation for projection matrix which works for 1 dimensional vectors:
where P is projection matrix and T is transpose.
We know that we can't simplify this fraction more (by cancelling terms) since denominator is a dot product (thus 0 dimensional scalar, number) and numerator is a matrix (column multiplied by row is a matrix).
I'm not sure how could I define function for this equation in numpy, considering that the current function that I'm using does not differentiate between these terms, multiplication is treated as it has commutative property. I'm using numpy.multiply method:
>>> import numpy as np
>>> a = np.array(a)
>>> a*a.T
array([1, 4, 9])
>>> a.T*a
array([1, 4, 9])
As you see, both of them output vectors.
I've also tried using numpy.matmul method:
>>> np.matmul(a, a.T)
14
>>> np.matmul(a.T, a)
14
which gives dot product for both of the function calls.
I also did try numpy.dot but it obviously doesn't work for numerator terms.
From my understanding, the first function call should output matrix (since column is multiplied by row) and the second function call should output a scalar in a proper case.
Am I mistaken? Is there any method that differentiates between a multiplied by a transpose and a transpose multiplied by a?
Thank you!
Note that 1-dimensional numpy arrays are not column vectors (and operations such as transposition do not make sense). If you want to obtain a column vector you should define your array as a 2-dimensional array (with the second dimension size equal to 1).
However, you don't need to define a column vector, as numpy offers functions to do what you want by manipulating an 1D array as follows
P = np.outer(a,a)/np.inner(a,a)
Stelios' answer is the best, no doubt but for completeness you can use the # operator with 2-d arrays:
a = np.array([1,4,9])[np.newaxis]
P = (a.T # a) / (a # a.T)
I may be misunderstanding how broadcasting works in Python, but I am still running into errors.
scipy offers a number of "special functions" which take in two arguments, in particular the eval_XX(n, x[,out]) functions.
See http://docs.scipy.org/doc/scipy/reference/special.html
My program uses many orthogonal polynomials, so I must evaluate these polynomials at distinct points. Let's take the concrete example scipy.special.eval_hermite(n, x, out=None).
I would like the x argument to be a matrix shape (50, 50). Then, I would like to evaluate each entry of this matrix at a number of points. Let's define n to be an a numpy array narr = np.arange(10) (where we have imported numpy as np, i.e. import numpy as np).
So, calling
scipy.special.eval_hermite(narr, matrix)
should return Hermitian polynomials H_0(matrix), H_1(matrix), H_2(matrix), etc. Each H_X(matrix) is of the shape (50,50), the shape of the original input matrix.
Then, I would like to sum these values. So, I call
matrix1 = np.sum( [scipy.eval_hermite(narr, matrix)], axis=0 )
but I get a broadcasting error!
ValueError: operands could not be broadcast together with shapes (10,) (50,50)
I can solve this with a for loop, i.e.
matrix2 = np.sum( [scipy.eval_hermite(i, matrix) for i in narr], axis=0)
This gives me the correct answer, and the output matrix2.shape = (50,50). But using this for loop slows down my code, big time. Remember, we are working with entries of matrices.
Is there a way to do this without a for loop?
eval_hermite broadcasts n with x, then evaluates Hn(x) at each point. Thus, the output shape will be the result of broadcasting n with x. So, if you want to make this work, you'll have to make n and x have compatible shapes:
import scipy.special as ss
import numpy as np
matrix = np.ones([100,100]) # example
narr = np.arange(10) # example
ss.eval_hermite(narr[:,None,None], matrix).shape # => (10, 100, 100)
But note that this might actually be faster:
out = np.zeros_like(matrix)
for n in narr:
out += ss.eval_hermite(n, matrix)
In testing, it appears to be between 5-10% faster than np.sum(...) of above.
The documentation for these functions is skimpy, and a lot of the code is compiled, so this is just based on experimentation:
special.eval_hermite(n, x, out=None)
n apparently is a scalar or array of integers. x can be an array of floats.
special.eval_hermite(np.ones(5,int)[:,None],np.ones(6)) gives me a (5,6) result. This is the same shape as what I'd get from np.ones(5,int)[:,None] * np.ones(6).
The np.ones(5,int)[:,None] is a (5,1) array, np.ones(6) a (6,), which for this purpose is equivalent of (1,6). Both can be expanded to (5,6).
So as best I can tell, broadcasting rules in these special functions is the same as for operators like *.
Since special.eval_hermite(nar[:,None,None], x) produces a (10,50,50), you just apply sum to axis 0 of that to produce the (50,50).
special.eval_hermite(nar[:,Nar,Nar], x).sum(axis=0)
Like I wrote before, the same broadcasting (and summing) rules apply for this hermite as they do for a basic operation like *.
I'm wondering if there is a simple way to multiply a numpy matrix by a scalar. Essentially I want all values to be multiplied by the constant 40. This would be an nxn matrix with 40's on the diagonal, but I'm wondering if there is a simpler function to use to scale this matrix. Or how would I go about making a matrix with the same shape as my other matrix and fill in its diagonal?
Sorry if this seems a bit basic, but for some reason I couldn't find this in the doc.
If you want a matrix with 40 on the diagonal and zeros everywhere else, you can use NumPy's function fill_diagonal() on a matrix of zeros. You can thus directly do:
N = 100; value = 40
b = np.zeros((N, N))
np.fill_diagonal(b, value)
This involves only setting elements to a certain value, and is therefore likely to be faster than code involving multiplying all the elements of a matrix by a constant. This approach also has the advantage of showing explicitly that you fill the diagonal with a specific value.
If you want the diagonal matrix b to be of the same size as another matrix a, you can use the following shortcut (no need for an explicit size N):
b = np.zeros_like(a)
np.fill_diagonal(b, value)
Easy:
N = 100
a = np.eye(N) # Diagonal Identity 100x100 array
b = 40*a # Multiply by a scalar
If you actually want a numpy matrix vs an array, you can do a = np.asmatrix(np.eye(N)) instead. But in general * is element-wise multiplication in numpy.
I have done a series of transformations using glMultMatrix. How do I multiply a vector (nX, nY, nZ, 1) to the matrix I have from the transformation? How do I get that matrix to multiply with a vector?
pyglet.gl.lib.GLException: invalid operation
I am getting above error if I use glMultMatrix. I need to call this multiplication between glBegin and glEnd.
If I'm reading your question right, you want something that returns the result of a Matrix-vector product between the current ModelView/Projection matrix and a vector you specify.
In that case, OpenGL can't do the multiplication for you. Instead, you need to extract the current matrix and do the multiplication yourself:
import numpy as np
someVector = np.array([2,3,4,5])
glMatrixMode(GL_MODELVIEW)
glMultMatrix(...)
modelViewMatrix = glGetDoublev(GL_MODELVIEW_MATRIX)
result = np.dot(modelViewMatrix, someVector)
Depending on what you're trying to do, you might need to get both the ModelView and Projection matrices and multiply them first.