I'm working to implement the following equation:
X =(Y.T * Y + Y.T * C * Y) ^ -1
Y is a (n x f) matrix and C is (n x n) diagonal one; n is about 300k and f will vary between 100 and 200. As part of an optimization process this equation will be used almost 100 million times so it has to be processed really fast.
Y is initialized randomly and C is a very sparse matrix with only a few numbers out of the 300k on the diagonal will be different than 0.Since Numpy's diagonal functions creates dense matrices, I created C as a sparse csr matrix. But when trying to solve the first part of the equation:
r = dot(C, Y)
The computer crashes due Memory limits. I decided then trying to convert Y to csr_matrix and make the same operation:
r = dot(C, Ysparse)
and this approach took 1.38 ms. But this solution is somewhat "tricky" since I'm using a sparse matrix to store a dense one, I wonder how efficient this really.
So my question is if is there some way of multiplying the sparse C and the dense Y without having to turn Y into sparse and improve performance? If somehow C could be represented as diagonal dense without consuming tons of memory maybe this would lead to very efficient performance but I don't know if this is possible.
I appreciate your help!
The reason the dot product runs into memory issues when computing r = dot(C,Y) is because numpy's dot function does not have native support for handling sparse matrices. What is happening is numpy thinks of the sparse matrix C as a python object, and not a numpy array. If you inspect on small scale you can see the problem first hand:
>>> from numpy import dot, array
>>> from scipy import sparse
>>> Y = array([[1,2],[3,4]])
>>> C = sparse.csr_matrix(array([[1,0], [0,2]]))
>>> dot(C,Y)
array([[ (0, 0) 1
(1, 1) 2, (0, 0) 2
(1, 1) 4],
[ (0, 0) 3
(1, 1) 6, (0, 0) 4
(1, 1) 8]], dtype=object)
Clearly the above is not the result you are interested in. Instead what you want to do is compute using scipy's sparse.csr_matrix.dot function:
r = sparse.csr_matrix.dot(C, Y)
or more compactly
r = C.dot(Y)
Try:
import numpy as np
from scipy import sparse
f = 100
n = 300000
Y = np.random.rand(n, f)
Cdiag = np.random.rand(n) # diagonal of C
Cdiag[np.random.rand(n) < 0.99] = 0
# Compute Y.T * C * Y, skipping zero elements
mask = np.flatnonzero(Cdiag)
Cskip = Cdiag[mask]
def ytcy_fast(Y):
Yskip = Y[mask,:]
CY = Cskip[:,None] * Yskip # broadcasting
return Yskip.T.dot(CY)
%timeit ytcy_fast(Y)
# For comparison: all-sparse matrices
C_sparse = sparse.spdiags([Cdiag], [0], n, n)
Y_sparse = sparse.csr_matrix(Y)
%timeit Y_sparse.T.dot(C_sparse * Y_sparse)
My timings:
In [59]: %timeit ytcy_fast(Y)
100 loops, best of 3: 16.1 ms per loop
In [18]: %timeit Y_sparse.T.dot(C_sparse * Y_sparse)
1 loops, best of 3: 282 ms per loop
First, are you really sure you need to perform a full matrix inversion in your problem ? Most of the time, one only really need to compute x = A^-1 y which is a much easier problem to solve.
If this is really so, I would consider computing an approximation of the inverse matrix instead of the full matrix inversion. Since matrix inversion is really costly. See for example the Lanczos algorithm for an efficient approximation of the inverse matrix. The approximation can be stored sparsely as a bonus. Plus, it requires only matrix-vector operations so you don't even have to store the full matrix to inverse.
As an alternative, using pyoperators, you can also use to .todense method to compute the matrix to inverse using efficient matrix vector operations. There is a special sparse container for diagonal matrices.
For an implementation of the Lanczos algorithm, you can have a look at pyoperators (disclaimer: I am one of the coauthor of this piece of software).
I don't know if it was possible when the question was asked; but nowadays, broadcasting is your friend. An n*n diagonal matrix needs only be an array of the diagonal elements to be used in a matrix product:
>>> n, f = 5, 3
>>> Y = np.random.randint(0, 10, (n, f))
>>> C = np.random.randint(0, 10, (n,))
>>> Y.shape
(5, 3)
>>> C.shape
(5,)
>>> np.all(Y.T # np.diag(C) # Y == Y.T*C # Y)
True
Do note that Y.T*C # Y is non-associative:
>>> Y.T*(C # Y)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: operands could not be broadcast together with shapes (3,5) (3,)
But Y.T # (C[:, np.newaxis]*Y) would yield the expected result:
>>> np.all(Y.T*C # Y == Y.T#(C[:, np.newaxis]*Y))
True
Related
i have an array y with shape (n,), I want to compute the inner product matrix, which is a n * n matrix
However, when I tried to do it in Python
np.dot(y , y)
I got the answer n, this is not what I am looking for
I have also tried:
np.dot(np.transpose(y),y)
np.dot(y, np.transpose(y))
I always get the same answer n
I think you are looking for:
np.multiply.outer(y,y)
or equally:
y = y[None,:]
y.T#y
example:
y = np.array([1,2,3])[None,:]
output:
#[[1 2 3]
# [2 4 6]
# [3 6 9]]
You can try to reshape y from shape (70,) to (70,1) before multiplying the 2 matrices.
# Reshape
y = y.reshape(70,1)
# Either below code would work
y*y.T
np.matmul(y,y.T)
One-liner?
np.dot(a[:, None], a[None, :])
transpose doesn't work on 1-D arrays, because you need atleast two axes to 'swap' them. This solution adds a new axis to the array; in the first argument, it looks like a column vector and has two axes; in the second argument it still looks like a row vector but has two axes.
Looks like what you need is the # matrix multiplication operator. dot method is only to compute dot product between vectors, what you want is matrix multiplication.
>>> a = np.random.rand(70, 1)
>>> (a # a.T).shape
(70, 70)
UPDATE:
Above answer is incorrect. dot does the same things if the array is 2D. See the docs here.
np.dot computes the dot product of two arrays. Specifically,
If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation).
If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a # b is preferred.
Simplest way to do what you want is to convert the vector to a matrix first using np.matrix and then using the #. Although, dot can also be used # is better because conventionally dot is used for vectors and # for matrices.
>>> a = np.random.rand(70)
(70,)
>>> a.shape
>>> a = np.matrix(a).T
>>> a.shape
(70, 1)
>>> (a # a.T).shape
(70, 70)
I am trying to construct a stack of block diagonal matrix in the form of nXMXM in numpy/scipy from a given stacks of matrices (nXmXm), where M=k*m with k the number of stacks of matrices. At the moment, I'm using the scipy.linalg.block_diag function in a for loop to perform this task:
import numpy as np
import scipy.linalg as linalg
a = np.ones((5,2,2))
b = np.ones((5,2,2))
c = np.ones((5,2,2))
result = np.zeros((5,6,6))
for k in range(0,5):
result[k,:,:] = linalg.block_diag(a[k,:,:],b[k,:,:],c[k,:,:])
However, since n is in my case getting quite large, I'm looking for a more efficient way than a for loop. I found 3D numpy array into block diagonal matrix but this does not really solve my problem. Anything I could imagine is transforming each stack of matrices into block diagonals
import numpy as np
import scipy.linalg as linalg
a = np.ones((5,2,2))
b = np.ones((5,2,2))
c = np.ones((5,2,2))
a = linalg.block_diag(*a)
b = linalg.block_diag(*b)
c = linalg.block_diag(*c)
and constructing the resulting matrix from it by reshaping
result = linalg.block_diag(a,b,c)
result = result.reshape((5,6,6))
which does not reshape. I don't even know, if this approach would be more efficient, so I'm asking if I'm on the right track or if somebody knows a better way of constructing this block diagonal 3D matrix or if I have to stick with the for loop solution.
Edit:
Since I'm new to this platform, I don't know where to leave this (Edit or Answer?), but I want to share my final solution: The highlightet solution from panadestein worked very nice and easy, but I'm now using higher dimensional arrays, where my matrices reside in the last two dimensions. Additionally my matrices are no longer of the same dimension (mostly a mixture of 1x1, 2x2, 3x3), so I adopted V. Ayrat's solution with minor changes:
def nd_block_diag(arrs):
shapes = np.array([i.shape for i in arrs])
out = np.zeros(np.append(np.amax(shapes[:,:-2],axis=0), [shapes[:,-2].sum(), shapes[:,-1].sum()]))
r, c = 0, 0
for i, (rr, cc) in enumerate(shapes[:,-2:]):
out[..., r:r + rr, c:c + cc] = arrs[i]
r += rr
c += cc
return out
which works also with array broadcasting, if the input arrays are shaped properly (i.e. the dimensions, which are to be broadcasted are not added automatically). Thanks to pandestein and V. Ayrat for your kind and fast help, I've learned a lot about the possibilites of list comprehensions and array indexing/slicing!
block_diag also just iterate through shapes. Almost all time spend in copying data so you can do it whatever way your want for example with little change of source code of block_diag
arrs = a, b, c
shapes = np.array([i.shape for i in arrs])
out = np.zeros([shapes[0, 0], shapes[:, 1].sum(), shapes[:, 2].sum()])
r, c = 0, 0
for i, (_, rr, cc) in enumerate(shapes):
out[:, r:r + rr, c:c + cc] = arrs[i]
r += rr
c += cc
print(np.allclose(result, out))
# True
I don't think that you can escape all possible loops to solve your problem. One way that I find convenient and perhaps more efficient than your for loop is to use a list comprehension:
import numpy as np
from scipy.linalg import block_diag
# Define input matrices
a = np.ones((5, 2, 2))
b = np.ones((5, 2, 2))
c = np.ones((5, 2, 2))
# Generate block diagonal matrices
mats = np.array([a, b, c]).reshape(5, 3, 2, 2)
result = [block_diag(*bmats) for bmats in mats]
Maybe this can give you some ideas to improve your implementation.
I would like to vectorize a particular case of the following mathematical formula (from Table 2 and Appendix A of this paper) with numpy:
The case I would like to compute is the following, where the scaling factors under the square root can be ignored.
The term w_kij - w_ij_bar is a n x p x p matrix, where n is typically much greater than p.
I implemented 2 solutions neither of which are particularly good: one involves a double loop, while the other fills the memory with unnecessary calculations very quickly.
dummy_data = np.random.normal(size=(100, 5, 5))
# approach 1: a double loop
out_hack = np.zeros((5, 5))
for i in range(5):
for j in range(5):
out_hack[i, j] = (dummy_data.T[j, j, :]*dummy_data[:, j, i]).sum()
# approach 2: slicing a diagonal from a tensor dot product
out = np.tensordot(dummy_data.T, dummy_data, axes=1)
out = out.diagonal(0, 0, 2).diagonal(0, 0, 2)
print((out.round(6) == out_hack.round(6)).all())
>>> True
Is there a way to find middle ground between these 2 approaches?
np.einsum translates that almost literally -
np.einsum('kjj,kji->ij',dummy_data,dummy_data)
I have a large matrix A of shape (n, n, 3, 3) with n is about 5000. Now I want find the inverse and transpose of matrix A:
import numpy as np
A = np.random.rand(1000, 1000, 3, 3)
identity = np.identity(3, dtype=A.dtype)
Ainv = np.zeros_like(A)
Atrans = np.zeros_like(A)
for i in range(1000):
for j in range(1000):
Ainv[i, j] = np.linalg.solve(A[i, j], identity)
Atrans[i, j] = np.transpose(A[i, j])
Is there a faster, more efficient way to do this?
This is taken from a project of mine, where I also do vectorized linear algebra on many 3x3 matrices.
Note that there is only a loop over 3; not a loop over n, so the code is vectorized in the important dimensions. I don't want to vouch for how this compares to a C/numba extension to do the same thing though, performance wise. This is likely to be substantially faster still, but at least this blows the loops over n out of the water.
def adjoint(A):
"""compute inverse without division by det; ...xv3xc3 input, or array of matrices assumed"""
AI = np.empty_like(A)
for i in xrange(3):
AI[...,i,:] = np.cross(A[...,i-2,:], A[...,i-1,:])
return AI
def inverse_transpose(A):
"""
efficiently compute the inverse-transpose for stack of 3x3 matrices
"""
I = adjoint(A)
det = dot(I, A).mean(axis=-1)
return I / det[...,None,None]
def inverse(A):
"""inverse of a stack of 3x3 matrices"""
return np.swapaxes( inverse_transpose(A), -1,-2)
def dot(A, B):
"""dot arrays of vecs; contract over last indices"""
return np.einsum('...i,...i->...', A, B)
A = np.random.rand(2,2,3,3)
I = inverse(A)
print np.einsum('...ij,...jk',A,I)
for the transpose:
testing a bit in ipython showed:
In [1]: import numpy
In [2]: x = numpy.ones((5,6,3,4))
In [3]: numpy.transpose(x,(0,1,3,2)).shape
Out[3]: (5, 6, 4, 3)
so you can just do
Atrans = numpy.transpose(A,(0,1,3,2))
to transpose the second and third dimensions (while leaving dimension 0 and 1 the same)
for the inversion:
the last example of http://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.inv.html#numpy.linalg.inv
Inverses of several matrices can be computed at once:
from numpy.linalg import inv
a = np.array([[[1., 2.], [3., 4.]], [[1, 3], [3, 5]]])
>>> inv(a)
array([[[-2. , 1. ],
[ 1.5, -0.5]],
[[-5. , 2. ],
[ 3. , -1. ]]])
So i guess in your case, the inversion can be done with just
Ainv = inv(A)
and it will know that the last two dimensions are the ones it is supposed to invert over, and that the first dimensions are just how you stacked your data. This should be much faster
speed difference
for the transpose: your method needs 3.77557015419 sec, and mine needs 2.86102294922e-06 sec (which is a speedup of over 1 million times)
for the inversion: i guess my numpy version is not high enough to try that numpy.linalg.inv trick with (n,n,3,3) shape, to see the speedup there (my version is 1.6.2, and the docs i based my solution on are for 1.8, but it should work on 1.8, if someone else can test that?)
Numpy has the array.T properties which is a shortcut for transpose.
For inversions, you use np.linalg.inv(A).
As posted by wim A.I also works on matrix. e.g.
print (A.I)
for numpy-matrix object, use matrix.getI.
e.g.
A=numpy.matrix('1 3;5 6')
print (A.getI())
Assume an n-dimensional array of observations that are reshaped to be a 2d-array with each row being one observation set. Using this reshape approach, np.polyfit can compute 2nd order fit coefficients for the entire ndarray (vectorized):
fit = np.polynomial.polynomialpolyfit(X, Y, 2)
where Y is shape (304000, 21) and X is a vector. This results in a (304000,3) array of coefficients, fit.
Using an iterator it is possible to call np.polyval(fit, X) for each row. This is inefficient when a vectorized approach may exist. Could the fit result be applied to the entire observation array without iterating? If so, how?
This is along the lines of this SO question.
np.polynomial.polynomial.polyval takes multidimensional coefficient arrays:
>>> x = np.random.rand(100)
>>> y = np.random.rand(100, 25)
>>> fit = np.polynomial.polynomial.polyfit(x, y, 2)
>>> fit.shape # 25 columns of 3 polynomial coefficients
(3L, 25L)
>>> xx = np.random.rand(50)
>>> interpol = np.polynomial.polynomial.polyval(xx, fit)
>>> interpol.shape # 25 rows, each with 50 evaluations of the polynomial
(25L, 50L)
And of course:
>>> np.all([np.allclose(np.polynomial.polynomial.polyval(xx, fit[:, j]),
... interpol[j]) for j in range(25)])
True
np.polynomial.polynomial.polyval is a perfectly fine (and convenient) approach to efficient evaluation of polynomial fittings.
However, if 'speediest' is what you are looking for, simply constructing the polynomial inputs and using the rudimentary numpy matrix multiplication functions results in slightly faster ( roughly 4x faster) computational speeds.
Setup
Using the same setup as above, we'll create 25 different line fittings.
>>> num_samples = 100000
>>> num_lines = 100
>>> x = np.random.randint(0,100,num_samples)
>>> y = np.random.randint(0,100,(num_samples, num_lines))
>>> fit = np.polyfit(x,y,deg=2)
>>> xx = np.random.randint(0,100,num_samples*10)
Numpy's polyval Function
res1 = np.polynomial.polynomial.polyval(xx, fit)
Basic Matrix Multiplication
inputs = np.array([np.power(xx,d) for d in range(len(fit))])
res2 = fit.T.dot(inputs)
Timing the functions
Using the same parameters above...
%timeit _ = np.polynomial.polynomial.polyval(xx, fit)
1 loop, best of 3: 247 ms per loop
%timeit inputs = np.array([np.power(xx, d) for d in range(len(fit))]);_ = fit.T.dot(inputs)
10 loops, best of 3: 72.8 ms per loop
To beat a dead horse...
mean Efficiency bump of ~3.61x faster. Speed fluctuations probably come from random computer processes in background.