I have a 3D image with dimensions rows x cols x deps. For every voxel in the image, I have computed a 3x3 real symmetric matrix. They are stored in the array D, which therefore has shape (rows, cols, deps, 6).
D stores the 6 unique elements of the 3x3 symmetric matrix for every voxel in my image. I need to find the Moore-Penrose pseudo inverse of all row*cols*deps matrices simultaneously/in vectorized code (looping through every image voxel and inverting is far too slow in Python).
Some of these 3x3 symmetric matrices are non-singular, and I can find their inverses, in vectorized code, using the analytical formula for the true inverse of a non-singular 3x3 symmetric matrix, and I've done that.
However, for those matrices that ARE singular (and there are sure to be some) I need the Moore-Penrose pseudo inverse. I could derive an analytical formula for the MP of a real, singular, symmetric 3x3 matrix, but it's a really nasty/lengthy formula, and would therefore involve a VERY large number of (element-wise) matrix arithmetic and quite a bit of confusing looking code.
Hence, I would like to know if there is a way to simultaneously find the MP pseudo inverse for all these matrices at once numerically. Is there a way to do this?
Gratefully,
GF
NumPy 1.8 included linear algebra gufuncs, which do exactly what you are after. While np.linalg.pinv is not gufunc-ed, np.linalg.svd is, and behind the scenes that is the function that gets called. So you can define your own gupinv function, based on the source code of the original function, as follows:
def gu_pinv(a, rcond=1e-15):
a = np.asarray(a)
swap = np.arange(a.ndim)
swap[[-2, -1]] = swap[[-1, -2]]
u, s, v = np.linalg.svd(a)
cutoff = np.maximum.reduce(s, axis=-1, keepdims=True) * rcond
mask = s > cutoff
s[mask] = 1. / s[mask]
s[~mask] = 0
return np.einsum('...uv,...vw->...uw',
np.transpose(v, swap) * s[..., None, :],
np.transpose(u, swap))
And you can now do things like:
a = np.random.rand(50, 40, 30, 6)
b = np.empty(a.shape[:-1] + (3, 3), dtype=a.dtype)
# Expand the unique items into a full symmetrical matrix
b[..., 0, :] = a[..., :3]
b[..., 1:, 0] = a[..., 1:3]
b[..., 1, 1:] = a[..., 3:5]
b[..., 2, 1:] = a[..., 4:]
# make matrix at [1, 2, 3] singular
b[1, 2, 3, 2] = b[1, 2, 3, 0] + b[1, 2, 3, 1]
# Find all the pseudo-inverses
pi = gu_pinv(b)
And of course the results are correct, both for singular and non-singular matrices:
>>> np.allclose(pi[0, 0, 0], np.linalg.pinv(b[0, 0, 0]))
True
>>> np.allclose(pi[1, 2, 3], np.linalg.pinv(b[1, 2, 3]))
True
And for this example, with 50 * 40 * 30 = 60,000 pseudo-inverses calculated:
In [2]: %timeit pi = gu_pinv(b)
1 loops, best of 3: 422 ms per loop
Which is really not that bad, although it is noticeably (4x) slower than simply calling np.linalg.inv, but this of course fails to properly handle the singular arrays:
In [8]: %timeit np.linalg.inv(b)
10 loops, best of 3: 98.8 ms per loop
EDIT: See #Jaime's answer. Only the discussion in the comments to this answer is useful now, and only for the specific problem at hand.
You can do this matrix by matrix, using scipy, that provides pinv (link) to calculate the Moore-Penrose pseudo inverse. An example follows:
from scipy.linalg import det,eig,pinv
from numpy.random import randint
#generate a random singular matrix M first
while True:
M = randint(0,10,9).reshape(3,3)
if det(M)==0:
break
M = M.astype(float)
#this is the method you need
MPpseudoinverse = pinv(M)
This does not exploit the fact that the matrix is symmetric though. You may also want to try the version of pinv exposed by numpy, that is supposedely faster, and different. See this post.
Related
I'm working on an optimization problem, but to avoid getting into the details, I'm going to provide a simple example of a bug that's been giving me headaches for a few days.
Say I have a 2D numpy array with observed x-y coordinates:
from scipy.optimize import distance
x = np.array([1,2], [2,3], [4,5], [5,6])
I also have a list of x-y coordinates to compare to these points (y):
y = np.array([11,13], [12, 14])
I have a function that takes the sum of manhattan differences between a value of x and all of the values in y:
def find_sum(ref_row, comp_rows):
modeled_counts = []
y = ref_row * len(comp_rows)
res = list(map(distance.cityblock, ref_row, comp_rows))
modeled_counts.append(sum(res))
return sum(modeled_counts)
Essentially, what I would like to do is find the sum of manhattan distances for every item in y with each item in x (so basically for each item in x, find the sum of the Manhattan distances between that (x,y) pair and every (x,y) pair in y).
I've tried this out with the following line of code:
z = list(map(find_sum, x, y))
However, z is of length 2 (like y), and not 4 like x. Is there a way to ensure that z is the result of consecutive one-to-all calculations? That is, I'd like to calculate the sum of all of the manhattan differences between x[0] and every set in y, and so on and so forth, so the length of z should be equal to the length of x.
Is there a simple way to do this without a for loop? My data is rather large (~ 4 million rows), so I'd really appreciate fast solutions. I'm fairly new to Python programming, so any explanations about why the solution works and is fast would be appreciated as well, but definitely isn't required!
Thanks!
This solution implements the distance in numpy, as I think it is a good example of broadcasting, which is a very useful thing to know if you need to use arrays and matrices.
By definition of Manhattan distance, you need to evaluate the sum of the absolute value of difference between each column. However, the first column of x, x[:, 0], has shape (4,) and the first column of y, y[:, 0], has shape (2,), so they are not compatible in the sense of applying subtraction: the broadcasting property says that each shape is compared starting with the trailing dimensions and two dimensions are compatible when they are equal or one of them is 1. Sadly, none of them are true for your columns.
However, you can add a new dimension of value 1 using np.newaxis, so
x[:, 0]
is array([1, 2, 4, 5]), but
x[:, 0, np.newaxis]
is
array([[1],
[2],
[4],
[5]])
and its shape is (4 ,1). Now, a matrix of shape (4, 1) subtracted by an array of shape 2 results in a matrix of shape (4, 2), by numpy's broadcasting treatment:
4 x 1
2
= 4 x 2
You can obtain the differences for each column:
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
and evaluate the sum of their absolute values:
np.abs(first_column_difference) + np.abs(second_column_difference)
which results in a (4, 2) matrix. Now, you want to sum the values for each row, so that you have 4 values:
np.sum(np.abs(first_column_difference) + np.abs(second_column_difference), axis=1)
which results in array([73, 69, 61, 57]). The rule is simple: the parameter axis will eliminate that dimension from the result, therefore using axis=1 for a (4, 2) matrix generates 4 values -- if you use axis=0, it will generate 2 values.
So, this will solve your problem:
x = np.array([[1, 2], [2, 3], [4, 5], [5, 6]])
y = np.array([[11, 13], [12, 43]])
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
z = np.abs(first_column_difference) + np.abs(second_column_difference)
print(np.sum(z, axis=1))
You can also skip the intermediate steps for each column and evaluate everything at once (it is a little bit harder to understand, so I prefer the method described above to explain what is happening):
print(np.abs(x[:, np.newaxis] - y).sum(axis=(1, 2)))
It is a general case for an n-dimensional Manhattan distance: if x is (u, n) and y is (v, n), it generates u rows by broadcasting (u, 1, n) by (v, n) = (u, v, n), then applying sum to eliminate the second and third axis.
Here is how you can do it using numpy broadcast with simplified explanation
Adjust Shape For Broadcasting
import numpy as np
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
## using np.newaxis as index add a new dimension at that position
## : give all the elements on that dimension
start_points = start_points[np.newaxis, :, :]
dest_points = dest_points[:, np.newaxis, :]
## Now lets check he shape of the point arrays
print('start_points.shape: ', start_points.shape) # (1, 4, 2)
print('dest_points.shape', dest_points.shape) # (2, 1, 2)
Lets try to understand
last element of shape represent x and y of a point, size 2
we can think of start_points as having 1 row and 4 columns of points
we can think of dest_points as having 2 rows and 1 columns of points
We can think start_points and dest_points as matrix or a table of points of size (1X4) and (2X1)
We clearly see that size are not compatible. What will happen if we perform arithmatic
operation between them? Here is where a smart part of numpy comes, called broadcast.
It will repeat rows of start_points to match that of dest_point making matrix of (2X4)
It will repeat columns of dest_point to match that of start_points making matrix of (2X4)
Result is arithmetic operation between every pair of elements on start_points and dest_points
Calculate the distance
diff_x_y = start_points - dest_points
print(diff_x_y.shape) # (2, 4, 2)
abs_diff_x_y = np.abs(start_points - dest_points)
man_distance = np.sum(abs_diff_x_y, axis=2)
print('man_distance:\n', man_distance)
sum_distance = np.sum(man_distance, axis=0)
print('sum_distance:\n', sum_distance)
Oneliner
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
np.sum(np.abs(start_points[np.newaxis, :, :] - dest_points[:, np.newaxis, :]), axis=(0,2))
Here is more detail explanation of broadcasting if you want to understand it more
With so many rows you can make substantial savings by using a smart algorithm. Let us for simplicity assume there is just one dimension; once we have established the algorithm, getting back to the general case is a simple matter of summing over coordinates.
The naive algorithm is O(mn) where m,n are the sizes of sets X,Y. Our algorithm is O((m+n)log(m+n)) so it scales much better.
We first have to sort the union of X and Y by coordinate and then form the cumsum over Y. Next, we find for each x in X the number YbefX of y in Y to its left and use it to look up the corresponding cumsum item YbefXval. The summed distances to all y to the left of x are YbefX times coordinate of x minus YbefXval, the distances to all y to the right are sum of all y coordinates minus YbefXval minus n - YbefX times coordinate of x.
Where does the saving come from? Sorting coordinates enables us to recycle the summations we have done before, instead of starting each time from scratch. This uses the fact that up to a sign we always sum the same y coordinates and going from left to right the signs flip one by one.
Code:
import numpy as np
from scipy.spatial.distance import cdist
from timeit import timeit
def pp(X,Y):
(m,k),(n,k) = X.shape,Y.shape
XY = np.concatenate([X.T,Y.T],1)
idx = XY.argsort(1)
Xmsk = idx<m
Ymsk = ~Xmsk
Xidx = np.arange(k)[:,None],idx[Xmsk].reshape(k,m)
Yidx = np.arange(k)[:,None],idx[Ymsk].reshape(k,n)
YbefX = Ymsk.cumsum(1)[Xmsk].reshape(k,m)
YbefXval = XY[Yidx].cumsum(1)[np.arange(k)[:,None],YbefX-1]
YbefXval[YbefX==0] = 0
XY[Xidx] = ((2*YbefX-n)*XY[Xidx]) - 2*YbefXval + Y.sum(0)[:,None]
return XY[:,:m].sum(0)
def summed_cdist(X,Y):
return cdist(X,Y,"minkowski",p=1).sum(1)
# demo
m,n,k = 1000,500,10
X,Y = np.random.randn(m,k),np.random.randn(n,k)
print("same result:",np.allclose(pp(X,Y),summed_cdist(X,Y)))
print("sort :",timeit(lambda:pp(X,Y),number=1000),"ms")
print("scipy cdist:",timeit(lambda:summed_cdist(X,Y),number=100)*10,"ms")
Sample run, comparing smart algo "sort" to naive algo implemented using cdist library function:
same result: True
sort : 1.4447695480193943 ms
scipy cdist: 36.41934019047767 ms
In python with numpy, say I have two matrices:
S, a sparse x*x matrix
M, a dense x*y matrix
Now I want to do np.dot(M, M.T) which will return a dense x*x matrix S_.
However, I only care about the cells that are nonzero in S, which means that it would not make a difference for my application if I did
S_ = S*S_
Obviously, that would be a waste of operations as I would like to leave out the irrelevant cells given in S alltogether. Remember that in matrix multiplication
S_[i,j] = np.sum(M[i,:]*M[:,j])
So I want to do this operation only for i,j such that S[i,j]=True.
Is this supported somehow by numpy implementations that run in C so that I do not need to implement it with python loops?
EDIT 1 [solved]: I still have this problem, actually M is now also sparse.
Now, given rows and cols of S, I implemented it like this:
data = np.array([ M[rows[i],:].dot(M[cols[i],:]).data[0] for i in xrange(len(rows)) ])
S_ = csr( (data, (rows,cols)) )
... but it is still slow. Any new ideas?
EDIT 2: jdehesa has given a great solution, but I would like to save more memory.
The solution was to do the following:
data = M[rows,:].multiply(M[cols,:]).sum(axis=1)
and then build a new sparse matrix from rows, cols and data.
However, when running the above line, scipy builds a (contiguous) numpy array with as many elements as nnz of the first submatrix plus nnz of the second submatrix, which can lead to MemoryError in my case.
In order to save more memory, I would like to multiply iteratively each row with its respective 'partner' column, then sum over and discard the result vector. Using simple python to implement this, basically I am back to the extremely slow version.
Is there a fast way of solving this problem?
Here is how you can do it with NumPy/SciPy, both for dense and sparse M matrices:
import numpy as np
import scipy.sparse as sp
# Coordinates where S is True
S = np.array([[0, 1],
[3, 6],
[3, 4],
[9, 1],
[4, 7]])
# Dense M matrix
# Random big matrix
M = np.random.random(size=(1000, 2000))
# Take relevant rows and compute values
values = np.sum(M[S[:, 0]] * M[S[:, 1]], axis=1)
# Make result matrix from values
result = np.zeros((len(M), len(M)), dtype=values.dtype)
result[S[:, 0], S[:, 1]] = values
# Sparse M matrix
# Construct sparse M as COO matrix or any other way
M = sp.coo_matrix(([10, 20, 30, 40, 50], # Data
([0, 1, 3, 4, 6], # Rows
[4, 4, 5, 5, 8])), # Columns
shape=(1000, 2000))
# Convert to CSR for fast row slicing
M_csr = M.tocsr()
# Take relevant rows and compute values
values = M_csr[S[:, 0]].multiply(M_csr[S[:, 1]]).sum(axis=1)
values = np.squeeze(np.asarray(values))
# Construct COO sparse matrix from values
result = sp.coo_matrix((values, (S[:, 0], S[:, 1])), shape=(M.shape[0], M.shape[0]))
Hi I'm relatively new here and trying to do some calculations with numpy. I'm experiencing a long elapse time from one particular calculation and can't work out any faster way to achieve the same thing.
Basically its part of a ray triangle intersection algorithm and I need to calculate all the vector cros products from two matrices of different sizes.
The code I was using was :
allhvals1 = numpy.cross( dirvectors[:,None,:], trivectors2[None,:,:] )
where dirvectors is an array of n* vectors (xyz) and trivectors2 is an array of m*vectors(xyz). allhvals1 is an array of the cross products of size n*M*vector (xyz).
This works but is very slow. It's essentially the n*m matrix of each vector from each array. Hope that you understand. The sizes of each varies from approx 1-4000 depending on parameters (I basically chunk the dirvectors dependent on size).
Any advice appreciated. Unfortunately my matrix math is somewhat flakey.
If you look at the source code of np.cross, it basically moves the xyz dimension to the front of the shape tuple for all arrays, and then has the calculation of each of the components spelled out like this:
x = a[1]*b[2] - a[2]*b[1]
y = a[2]*b[0] - a[0]*b[2]
z = a[0]*b[1] - a[1]*b[0]
In your case, each of those products requires allocating huge arrays, so the overall behavior is not very efficient.
Lets set up some test data:
u = np.random.rand(1000, 3)
v = np.random.rand(2000, 3)
In [13]: %timeit s1 = np.cross(u[:, None, :], v[None, :, :])
1 loops, best of 3: 591 ms per loop
We can try to compute it using Levi-Civita symbols and np.einsum as follows:
eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1
In [14]: %timeit s2 = np.einsum('ijk,uj,vk->uvi', eijk, u, v)
1 loops, best of 3: 706 ms per loop
In [15]: np.allclose(s1, s2)
Out[15]: True
So while it works, it has worse performance. The thing is that np.einsum has trouble when there are more than two operands, but has optimized pathways for two or less. So we can try to rewrite it in two steps, to see if it helps:
In [16]: %timeit s3 = np.einsum('iuk,vk->uvi', np.einsum('ijk,uj->iuk', eijk, u), v)
10 loops, best of 3: 63.4 ms per loop
In [17]: np.allclose(s1, s3)
Out[17]: True
Bingo! Close to an order of magnitude improvement...
Some performance figures for NumPy 1.11.0 with a=numpy.random.rand(n,3), b=numpy.random.rand(n,3):
The nested einsum is about twice as fast as cross for the largest n tested.
While writing dynamic simulations for underwater vehicles I have found this method for fast cross product:
https://github.com/simena86/Simulink-Underwater-Robotics-Simulator/blob/master/3rdparty/gnc_mfiles/Smtrx.m
Which works well, it is written in Matlab but the code is very simple. Just read the comments at the top.
I would like a numpy-sh way of vectorizing the calculation of eigenvalues, such that I can feed it a matrix of matrices and it would return a matrix of the respective eigenvalues.
For example, in the code below, B is the block 6x6 matrix composed of 4 copies of the 3x3 matrix A.
C is what I would like to see as output, i.e. an array of dimension (2,2,3) (because A has 3 eigenvalues).
This is of course a very simplified example, in the general case the matrices A can have any size (although they are still square), and the matrix B is not necessarily formed of copies of A, but different A1, A2, etc (all of same size but containing different elements).
import numpy as np
A = np.array([[0, 1, 0],
[0, 2, 0],
[0, 0, 3]])
B = np.bmat([[A, A], [A,A]])
C = np.array([[np.linalg.eigvals(B[0:3,0:3]),np.linalg.eigvals(B[0:3,3:6])],
[np.linalg.eigvals(B[3:6,0:3]),np.linalg.eigvals(B[3:6,3:6])]])
Edit: if you're using a version of numpy >= 1.8.0, then np.linalg.eigvals operates over the last two dimensions of whatever array you hand it, so if you reshape your input to an (n_subarrays, nrows, ncols) array you'll only have to call eigvals once:
import numpy as np
A = np.array([[0, 1, 0],
[0, 2, 0],
[0, 0, 3]])
# the input needs to be an array, since matrices can only be 2D.
B = np.repeat(A[np.newaxis,...], 4, 0)
# for arbitrary input arrays you could do something like:
# B = np.vstack(a[np.newaxis,...] for a in input_arrays)
# but for this to work it will be necessary for each element in
# 'input_arrays' to have the same shape
# eigvals will operate over the last two dimensions of the array and return
# a (4, 3) array of eigenvalues
C = np.linalg.eigvals(B)
# reshape this output so that it matches your original example
C.shape = (2, 2, 3)
If your input arrays don't all have the same dimensions, e.g. input_arrays[0].shape == (2, 2), input_arrays[1].shape == (3, 3) etc. then you could only vectorize this calculation across subsets with matching dimensions.
If you're using an older version of numpy then unfortunately I don't think there's any way to vectorize the calculation of the eigenvalues over multiple input arrays - you'll just have to loop over your inputs in Python instead.
You could just do something like this
C = np.array([[np.linalg.eigvals(B[i:i+3, j:j+3])
for i in xrange(0, B.shape[0], 3)]
for j in xrange(0, B.shape[1], 3)])
Perhaps a nicer approach is to use the block_view function from https://stackoverflow.com/a/5078155/1352250:
B_blocks = block_view(B)
C = np.array([[np.linalg.eigvals(m) for m in v] for v in B_blocks])
Update
As ali_m points out, this method is a form of syntactic sugar that will not reduce the overhead incurred from calling eigvals a large number of times. While this overhead should be small if each matrix it is applied to is large-ish, for the 6x6 matrices that the OP is interested in, it is not trivial (see the comments below; according to ali_m, there might be a factor of three difference between the version I give above, and the version he posted that uses Numpy >= 1.8.0).
Using Python & Numpy, I would like to:
Consider each row of an (n columns x
m rows) matrix as a vector
Weight each row (scalar
multiplication on each component of
the vector)
Add each row to create a final vector
(vector addition).
The weights are given in a regular numpy array, n x 1, so that each vector m in the matrix should be multiplied by weight n.
Here's what I've got (with test data; the actual matrix is huge), which is perhaps very un-Numpy and un-Pythonic. Can anyone do better? Thanks!
import numpy
# test data
mvec1 = numpy.array([1,2,3])
mvec2 = numpy.array([4,5,6])
start_matrix = numpy.matrix([mvec1,mvec2])
weights = numpy.array([0.5,-1])
#computation
wmatrix = [ weights[n]*start_matrix[n] for n in range(len(weights)) ]
vector_answer = [0,0,0]
for x in wmatrix: vector_answer+=x
Even a 'technically' correct answer has been all ready given, I'll give my straightforward answer:
from numpy import array, dot
dot(array([0.5, -1]), array([[1, 2, 3], [4, 5, 6]]))
# array([-3.5 -4. -4.5])
This one is much more on with the spirit of linear algebra (and as well those three dotted requirements on top of the question).
Update:
And this solution is really fast, not marginally, but easily some (10- 15)x faster than all ready proposed one!
It will be more convenient to use a two-dimensional numpy.array than a numpy.matrix in this case.
start_matrix = numpy.array([[1,2,3],[4,5,6]])
weights = numpy.array([0.5,-1])
final_vector = (start_matrix.T * weights).sum(axis=1)
# array([-3.5, -4. , -4.5])
The multiplication operator * does the right thing here due to NumPy's broadcasting rules.