I have one matrix and one vector, of dimensions (N, d) and (N,) respectively. For each row, I want to divide each element by the corresponding value in the vector. I was wondering if there was a vectorized implementation (to save computation time). (I'm trying to create points on the surface of a d-dimensional sphere.) Right now I'm doing this:
x = np.random.randn(N,d)
norm = np.linalg.norm(x, axis=1)
for i in range(N):
for j in range(d):
x[i][j] = x[i][j] / norm[i]
np.linalg.norm has a keepdims argument just for this:
x /= np.linalg.norm(x, axis=1, keepdims=True)
I am new to numpy and I want to solve the equation A * x = x, where A is a matrix and x is a vector.
Searching for the vector x, if it exists.
I found the np.linalg.solve()-function , but didn't got it to work as intended.
The issue here is not so much a problem with numpy as with your understanding of the linear algebra involved. The question you are asking can be rephrased as:
A # x = x
A # x = I # x
(A - I) # x = 0
This is a specific formulation of the the general eigenvector problem, with the stipulation that the eigenvalue is 1.
Numpy solves this problem with the function np.linalg.eig:
w, v = np.linalg.eig(A)
You need to check if any of the values are 1 (there could be more than one):
mask = np.isclose(w, 1)
if mask.any():
for vec in v[:, mask].T:
print(vec)
else:
print('Nope!')
The elements of v are unit vectors. Keep in mind that any scalar multiple of such a vector is also a solution.
For issues with non-invertible matrices, you may want to use scipy.linalg.svd instead:
v, w, _ = svd(A)
The rest of the procedure will be the same.
I'm trying to implement the n-mode tensor-matrix product (as defined by Kolda and Bader: https://www.sandia.gov/~tgkolda/pubs/pubfiles/SAND2007-6702.pdf) efficiently in Python using Numpy. The operation effectively gets down to (for matrix U, tensor X and axis/mode k):
Extract all vectors along axis k from X by collapsing all other axes.
Multiply these vectors on the left by U using standard matrix multiplication.
Insert the vectors again into the output tensor using the same shape, apart from X.shape[k], which is now equal to U.shape[0] (initially, X.shape[k] must be equal to U.shape[1], as a result of the matrix multiplication).
I've been using an explicit implementation for a while which performs all these steps separately:
Transpose the tensor to bring axis k to the front (in my full code I added an exception in case k == X.ndim - 1, in which case it's faster to leave it there and transpose all future operations, or at least in my application, but that's not relevant here).
Reshape the tensor to collapse all other axes.
Calculate the matrix multiplication.
Reshape the tensor to reconstruct all other axes.
Transpose the tensor back into the original order.
I would think this implementation creates a lot of unnecessary (big) arrays, so once I discovered np.einsum I thought this would speed things up considerably. However using the code below I got worse results:
import numpy as np
from time import time
def mode_k_product(U, X, mode):
transposition_order = list(range(X.ndim))
transposition_order[mode] = 0
transposition_order[0] = mode
Y = np.transpose(X, transposition_order)
transposed_ranks = list(Y.shape)
Y = np.reshape(Y, (Y.shape[0], -1))
Y = U # Y
transposed_ranks[0] = Y.shape[0]
Y = np.reshape(Y, transposed_ranks)
Y = np.transpose(Y, transposition_order)
return Y
def einsum_product(U, X, mode):
axes1 = list(range(X.ndim))
axes1[mode] = X.ndim + 1
axes2 = list(range(X.ndim))
axes2[mode] = X.ndim
return np.einsum(U, [X.ndim, X.ndim + 1], X, axes1, axes2, optimize=True)
def test_correctness():
A = np.random.rand(3, 4, 5)
for i in range(3):
B = np.random.rand(6, A.shape[i])
X = mode_k_product(B, A, i)
Y = einsum_product(B, A, i)
print(np.allclose(X, Y))
def test_time(method, amount):
U = np.random.rand(256, 512)
X = np.random.rand(512, 512, 256)
start = time()
for i in range(amount):
method(U, X, 1)
return (time() - start)/amount
def test_times():
print("Explicit:", test_time(mode_k_product, 10))
print("Einsum:", test_time(einsum_product, 10))
test_correctness()
test_times()
Timings for me:
Explicit: 3.9450525522232054
Einsum: 15.873924326896667
Is this normal or am I doing something wrong? I know there are circumstances where storing intermediate results can decrease complexity (e.g. chained matrix multiplication), however in this case I can't think of any calculations that are being repeated. Is matrix multiplication so optimized that it removes the benefits of not transposing (which technically has a lower complexity)?
I'm more familiar with the subscripts style of using einsum, so worked out these equivalences:
In [194]: np.allclose(np.einsum('ij,jkl->ikl',B0,A), einsum_product(B0,A,0))
Out[194]: True
In [195]: np.allclose(np.einsum('ij,kjl->kil',B1,A), einsum_product(B1,A,1))
Out[195]: True
In [196]: np.allclose(np.einsum('ij,klj->kli',B2,A), einsum_product(B2,A,2))
Out[196]: True
With a mode parameter, your approach in einsum_product may be best. But the equivalences help me visualize the calculation better, and may help others.
Timings should basically be the same. There's an extra setup time in einsum_product that should disappear in larger dimensions.
After updating Numpy, Einsum is only slightly slower than the explicit method, with or without multi-threading (see comments to my question).
I am doing a benchmarking test in python on different ways to calculate A'*A, with A being a N x M matrix. One of the fastest ways was to use numpy.dot().
I was curious if I can obtain the same result using numpy.cov() (which gives the covariance matrix) by somehow varying the weights or by somehow pre-processing the A matrix ? But I had no success. Does anyone know if there is any relation between the product A'*A and covariance of A, where A is a matrix with N rows/observations and M columns/variables?
Have a look at the cov source. Near the end of the function it does this:
c = dot(X, X_T.conj())
Which is basically the dot product you want to perform. However, there are all kinds of other operations: checking inputs, subtracting the mean, normalization, ...
In short, np.cov will never ever be faster than np.dot(A.T, A) because internally it contains exactly that operation.
That said - the covariance matrix is computed as
Or in Python:
import numpy as np
a = np.random.rand(10, 3)
m = np.mean(a, axis=0, keepdims=True)
x = np.dot((a - m).T, a - m) / (a.shape[0] - 1)
y = np.cov(a.T)
assert np.allclose(x, y) # check they are equivalent
As you can see, the covariance matrix is equivalent to the raw dot product if you subtract the mean of each variable and divide the result by the number of samples (minus one).
Suppose I have two vectors of length 25, and I want to compute their covariance matrix. I try doing this with numpy.cov, but always end up with a 2x2 matrix.
>>> import numpy as np
>>> x=np.random.normal(size=25)
>>> y=np.random.normal(size=25)
>>> np.cov(x,y)
array([[ 0.77568388, 0.15568432],
[ 0.15568432, 0.73839014]])
Using the rowvar flag doesn't help either - I get exactly the same result.
>>> np.cov(x,y,rowvar=0)
array([[ 0.77568388, 0.15568432],
[ 0.15568432, 0.73839014]])
How can I get the 25x25 covariance matrix?
You have two vectors, not 25. The computer I'm on doesn't have python so I can't test this, but try:
z = zip(x,y)
np.cov(z)
Of course.... really what you want is probably more like:
n=100 # number of points in each vector
num_vects=25
vals=[]
for _ in range(num_vects):
vals.append(np.random.normal(size=n))
np.cov(vals)
This takes the covariance (I think/hope) of num_vects 1xn vectors
Try this:
import numpy as np
x=np.random.normal(size=25)
y=np.random.normal(size=25)
z = np.vstack((x, y))
c = np.cov(z.T)
Covariance matrix from samples vectors
To clarify the small confusion regarding what is a covariance matrix defined using two N-dimensional vectors, there are two possibilities.
The question you have to ask yourself is whether you consider:
each vector as N realizations/samples of one single variable (for example two 3-dimensional vectors [X1,X2,X3] and [Y1,Y2,Y3], where you have 3 realizations for the variables X and Y respectively)
each vector as 1 realization for N variables (for example two 3-dimensional vectors [X1,Y1,Z1] and [X2,Y2,Z2], where you have 1 realization for the variables X,Y and Z per vector)
Since a covariance matrix is intuitively defined as a variance based on two different variables:
in the first case, you have 2 variables, N example values for each, so you end up with a 2x2 matrix where the covariances are computed thanks to N samples per variable
in the second case, you have N variables, 2 samples for each, so you end up with a NxN matrix
About the actual question, using numpy
if you consider that you have 25 variables per vector (took 3 instead of 25 to simplify example code), so one realization for several variables in one vector, use rowvar=0
# [X1,Y1,Z1]
X_realization1 = [1,2,3]
# [X2,Y2,Z2]
X_realization2 = [2,1,8]
numpy.cov([X,Y],rowvar=0) # rowvar false, each column is a variable
Code returns, considering 3 variables:
array([[ 0.5, -0.5, 2.5],
[-0.5, 0.5, -2.5],
[ 2.5, -2.5, 12.5]])
otherwise, if you consider that one vector is 25 samples for one variable, use rowvar=1 (numpy's default parameter)
# [X1,X2,X3]
X = [1,2,3]
# [Y1,Y2,Y3]
Y = [2,1,8]
numpy.cov([X,Y],rowvar=1) # rowvar true (default), each row is a variable
Code returns, considering 2 variables:
array([[ 1. , 3. ],
[ 3. , 14.33333333]])
Reading the documentation as,
>> np.cov.__doc__
or looking at Numpy Covariance, Numpy treats each row of array as a separate variable, so you have two variables and hence you get a 2 x 2 covariance matrix.
I think the previous post has right solution. I have the explanation :-)
I suppose what youre looking for is actually a covariance function which is a timelag function. I'm doing autocovariance like that:
def autocovariance(Xi, N, k):
Xs=np.average(Xi)
aCov = 0.0
for i in np.arange(0, N-k):
aCov = (Xi[(i+k)]-Xs)*(Xi[i]-Xs)+aCov
return (1./(N))*aCov
autocov[i]=(autocovariance(My_wector, N, h))
You should change
np.cov(x,y, rowvar=0)
onto
np.cov((x,y), rowvar=0)
What you got (2 by 2) is more useful than 25*25. Covariance of X and Y is an off-diagonal entry in the symmetric cov_matrix.
If you insist on (25 by 25) which I think useless, then why don't you write out the definition?
x=np.random.normal(size=25).reshape(25,1) # to make it 2d array.
y=np.random.normal(size=25).reshape(25,1)
cov = np.matmul(x-np.mean(x), (y-np.mean(y)).T) / len(x)
As pointed out above, you only have two vectors so you'll only get a 2x2 cov matrix.
IIRC the 2 main diagonal terms will be sum( (x-mean(x))**2) / (n-1) and similarly for y.
The 2 off-diagonal terms will be sum( (x-mean(x))(y-mean(y)) ) / (n-1). n=25 in this case.
according the document, you should expect variable vector in column:
If we examine N-dimensional samples, X = [x1, x2, ..., xn]^T
though later it says each row is a variable
Each row of m represents a variable.
so you need input your matrix as transpose
x=np.random.normal(size=25)
y=np.random.normal(size=25)
X = np.array([x,y])
np.cov(X.T)
and according to wikipedia: https://en.wikipedia.org/wiki/Covariance_matrix
X is column vector variable
X = [X1,X2, ..., Xn]^T
COV = E[X * X^T] - μx * μx^T // μx = E[X]
you can implement it yourself:
# X each row is variable
X = X - X.mean(axis=0)
h,w = X.shape
COV = X.T # X / (h-1)
i don't think you understand the definition of covariance matrix.
If you need 25 x 25 covariance matrix, you need 25 vectors each with n data points.