I am currently trying to create a sparse matrix that will look like this.
[[ 50. -25. 0. 0.]
[-25. 50. -25. 0.]
[ 0. -25. 50. -25.]
[ 0. 0. -25. 50.]]
But when I run it through I keep getting the value error
'data array must have rank 2' in my data array.
I am positive it is a problem with my B variable. I have tried several things but nothing is working. Any advice?
def sparse(a,b,N):
h = (b-a)/(N+1)
e = np.ones([N,1])/h**2
B = np.array([e, -2*e, e])
diags = np.array([-1,0,1])
A = spdiags(B,diags,N,N).toarray()
return A
print(sparse(0,1,4))
Just change to this:
import numpy as np
from scipy.sparse import spdiags
def sparse(a, b, N):
h = (b - a) / (N + 1)
e = np.ones(N) / h ** 2
diags = np.array([-1, 0, 1])
A = spdiags([-1 * e, 2 * e, -1 * e], diags, N, N).toarray()
return A
print(sparse(0, 1, 4))
Output
[[-50. 25. 0. 0.]
[ 25. -50. 25. 0.]
[ 0. 25. -50. 25.]
[ 0. 0. 25. -50.]]
The main change is this:
e = np.ones([N,1])/h**2
by
e = np.ones(N) / h ** 2
Note that toarray transforms the sparse matrix into a dense one, from the documentation:
Return a dense ndarray representation of this matrix.
Related
I would like to build a locally connected weight matrix that represents a locally connected neural network in pure python/numpy without deep learning frameworks like Torch or TensorFlow.
The weight matrix is a non-square 2D matrix with the dimension (number_input, number_output). (an autoencoder in my case; input>hidden)
So the function I would like to build, take the matrix dimension and the size of the receptive field (number of local connection) and give the associated weight matrix. I've already create a function like this, but for an input size of 8 and an output size of 4 (and RF = 4) my function output :
[[ 0.91822845 0. 0. 0. ]
[-0.24264655 -0.54754138 0. 0. ]
[ 0.55617366 0.12832513 -0.28733965 0. ]
[ 0.27993286 -0.33150324 0.06994107 0.61184121]
[ 0. 0.04286912 -0.20974503 -0.37633903]
[ 0. 0. -0.10386762 0.33553009]
[ 0. 0. 0. 0.09562682]
[ 0. 0. 0. 0. ]]
but I would like :
[[ 0.91822845 0. 0. 0. ]
[-0.24264655 -0.54754138 0. 0. ]
[ 0.55617366 0.12832513 0. 0. ]
[ 0 -0.33150324 0.06994107 0 ]
[ 0. 0.04286912 -0.20974503 0. ]
[ 0. 0. -0.10386762 0.33553009]
[ 0. 0. 0.11581854 0.09562682]
[ 0. 0. 0. 0.03448418]]
Here's my python code :
import numpy as np
def local_weight(input_size, output_size, RF):
input_range = 1.0 / input_size ** (1/2)
w = np.zeros((input_size, output_size))
for i in range(0, RF):
for j in range(0, output_size):
w[j+i, j] = np.random.normal(loc=0, scale=input_range, size=1)
return w
print(local_weight(8, 4, 4))
I look forward for your response!
The trick is in a small pad to work more comfortably (or control the limits).
Then you must define the step you will take with respect to the input (it is not more than the input / output). Once this is done you just have to fill in the gaps and then remove the pad.
import math
import numpy as np
def local_weight(input_size, output_size, RF):
input_range = 1.0 / input_size ** (1/2)
padding = ((RF - 1) // 2)
w = np.zeros(shape=(input_size + 2*padding, output_size))
step = float(w.shape[0] - RF) / (output_size - 1)
for i in range(output_size):
j = int(math.ceil(i * step))
j_next = j + RF
w[j:j_next, i] = np.random.normal(loc=0, scale=input_range, size=(j_next - j))
return w[padding:-padding, :]
I hope that is what you are looking for.
EDIT:
I think the implementation was misguided. I reimplement the function, we go by parts.
I calculate the radius of the receptive field (padding).
Determine the size of the W.
I calculate the step by removing the padding area so that I always stay inside.
I calculate the weights.
Remove the padding.
I am given this matrix and am trying to write a function to build this matrix for any size of n. I am told the height of the matrix is n, but not sure the width.
Below is my code and output, is this correct? I am slightly confused by the notation of the matrix itself.
def buildMatrix(n, a):
matrix = np.zeros([n, n], dtype=float)
x_diag, y_diag = np.diag_indices_from(matrix)
for (x,y) in zip(x_diag, y_diag):
if x > (n / 2):
matrix[x][y] = -2*a
elif x == (n / 2):
matrix[x][y] = -(1 + a)
else:
matrix[x][y] = -2
if x != n - 1:
matrix[x + 1][y] = a if x >= (n / 2) else 1
matrix[x][y + 1] = a if x >= (n / 2) else 1
return matrix
Output with buildMatrix(5, 2)
[[-2. 1. 0. 0. 0.]
[ 1. -2. 1. 0. 0.]
[ 0. 1. -3. 2. 0.]
[ 0. 0. 2. -4. 2.]
[ 0. 0. 0. 2. -4.]]
Can anyone help me out?
To answer your first question, the matrix has to have a width of n in order for the matrix-vector product to be compatible.
The picture of the matrix is ambiguous on where the switch from -2 to -(1-a) to -2a occurs. In your code, you check if x==n/2 to set the switch. This is fine in python2 but will cause problems in python3 since x/2 returns 2.5. Using safer x==n//2 since n//2 return an integer in python2 as well as python3.
For generality, I'm going to assume that the switch happens at row m. The matrix can be built easier using slicing and the np.diag command.
def buildmat(n, m, a):
diag = np.zeros(n)
offdiag = np.zeros(n-1)
offdiag[0:m] = 1
offdiag[m:n-1] = a
diag[0:m] = -2
diag[m] = -(1+a)
diag[m+1:n] = -2*a
matrix = np.diag(diag) + np.diag(offdiag, 1) + np.diag(offdiag, -1)
return matrix
Running
buildmat(5, 2, 3)
produces
[[-2. 1. 0. 0. 0.]
[ 1. -2. 1. 0. 0.]
[ 0. 1. -3. 2. 0.]
[ 0. 0. 2. -4. 2.]
[ 0. 0. 0. 2. -4.]]
right now I'm working on control.matlap.tf2ss and I would like to access my array in my state space.
Here is my code
Gs = tf([P.l], [P.Jzz, 0, 0])
Cs = tf([P.Kp, P.Kd], 1)
Gcl = feedback(series(Cs, Gs), 1)
po = pole(Gcl)
num, den = tfdata(Gs)
sys = tf2ss(Gs)
print sys
Result:
A = [[ 0. 0.]
[ 1. 0.]]
B = [[-10.58350385]
[ 0. ]]
C = [[ 0. -1.]]
D = [[ 0.]]
How can I access array A, B, C, D?
For arrays of state-space models with variable numbers of states, use the syntax:
[a,b,c,d] = ssdata(sys,'cell')
I need to write a basic function that computes a 2D convolution between a matrix and a kernel.
I have recently got into Python, so I'm sorry for my mistakes.
My dissertation teacher said that I should write one by myself so I can handle it better and to be able to modify it for future improvements.
I have found an example of this function on a website, but I don't understand how the returned values are obtained.
This is the code (from http://docs.cython.org/src/tutorial/numpy.html )
from __future__ import division
import numpy as np
def naive_convolve(f, g):
# f is an image and is indexed by (v, w)
# g is a filter kernel and is indexed by (s, t),
# it needs odd dimensions
# h is the output image and is indexed by (x, y),
# it is not cropped
if g.shape[0] % 2 != 1 or g.shape[1] % 2 != 1:
raise ValueError("Only odd dimensions on filter supported")
# smid and tmid are number of pixels between the center pixel
# and the edge, ie for a 5x5 filter they will be 2.
#
# The output size is calculated by adding smid, tmid to each
# side of the dimensions of the input image.
vmax = f.shape[0]
wmax = f.shape[1]
smax = g.shape[0]
tmax = g.shape[1]
smid = smax // 2
tmid = tmax // 2
xmax = vmax + 2*smid
ymax = wmax + 2*tmid
# Allocate result image.
h = np.zeros([xmax, ymax], dtype=f.dtype)
# Do convolution
for x in range(xmax):
for y in range(ymax):
# Calculate pixel value for h at (x,y). Sum one component
# for each pixel (s, t) of the filter g.
s_from = max(smid - x, -smid)
s_to = min((xmax - x) - smid, smid + 1)
t_from = max(tmid - y, -tmid)
t_to = min((ymax - y) - tmid, tmid + 1)
value = 0
for s in range(s_from, s_to):
for t in range(t_from, t_to):
v = x - smid + s
w = y - tmid + t
value += g[smid - s, tmid - t] * f[v, w]
h[x, y] = value
return h
I don't know if this function does the weighted sum from input and filter, because I see no sum here.
I applied this with
kernel = np.array([(1, 1, -1), (1, 0, -1), (1, -1, -1)])
file = np.ones((5,5))
naive_convolve(file, kernel)
I got this matrix:
[[ 1. 2. 1. 1. 1. 0. -1.]
[ 2. 3. 1. 1. 1. -1. -2.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 2. 1. -1. -1. -1. -3. -2.]
[ 1. 0. -1. -1. -1. -2. -1.]]
I tried to do a manual calculation (on paper) for the first full iteration of the function and I got 'h[0,0] = 0', because of the matrix product: 'filter[0, 0] * matrix[0, 0]', but the function returns 1. I am very confused with this.
If anyone can help me understand what is going on here, I would be very grateful. Thanks! :)
Yes, that function computes the convolution correctly. You can check this using scipy.signal.convolve2d
import numpy as np
from scipy.signal import convolve2d
kernel = np.array([(1, 1, -1), (1, 0, -1), (1, -1, -1)])
file = np.ones((5,5))
x = convolve2d(file, kernel)
print x
Which gives:
[[ 1. 2. 1. 1. 1. 0. -1.]
[ 2. 3. 1. 1. 1. -1. -2.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 2. 1. -1. -1. -1. -3. -2.]
[ 1. 0. -1. -1. -1. -2. -1.]]
It's impossible to know how to explain all this to you since I don't know where to start, and I don't know how all the other explanations aren't working for you. I think, though, that you are doing all of this as a learning exercise so you can figure this out for yourself. From what I've seen on SO, asking big questions on SO is not a substitute for working it through yourself.
Your specific question of why does
h[0,0] = 0
in your calculation not match this matrix is a good one. In fact, both are correct. The reason for mismatch is that the output of the convolution doesn't have the mathematical indices specified, but instead they are implied. The center, which is mathematically indicated by the indices [0,0] corresponds to x[3,3] in the matrix above.
import scipy.sparse.linalg as scial
import scipy.sparse as scisp
import numpy
def buildB(A,x,col_size_A):
d = numpy.zeros(col_size_A)
for index in xrange(col_size_A):
d[index] = 2*x[index]-1
tmp = scisp.spdiags(d,0,col_size_A,col_size_A)
return scisp.bmat([[A],[tmp]])
def buildQ(l,row_size_A):
q = numpy.zeros(row_size_A)
for index in xrange(row_size_A):
q[index] = 2*l[index]
return scisp.spdiags(q,0,row_size_A,row_size_A)
def buildh(A,x,b,col_size_A):
p = A.dot(x)
p = numpy.subtract(p, b)
quad = numpy.zeros(col_size_A)
for index in xrange(col_size_A):
quad[index] = x[index]*x[index]-x[index]
return numpy.concatenate((p, quad))
def ini():
A = numpy.array([[1,1],[1,-1]])
b = [1, 0]
c = [1, 1]
col_size_A = 2
row_size_A = 2
main(A,b,c,col_size_A,row_size_A)
def main(A,b,c, col_size_A, row_size_A):
x = numpy.zeros(col_size_A)
l = numpy.zeros(row_size_A*2)
eps = 10e-6
k = 0
while True:
B = buildB(A,x,col_size_A)
Q = buildQ(l[row_size_A/2:row_size_A+1], col_size_A)
Bt = B.transpose()
h = buildh(A,x,b,col_size_A)
g = numpy.add(c,Bt.dot(l))
F = numpy.concatenate((g, h))
print "Iteration " + str(k),
tol = numpy.amax(F)
print "- Tol "+ str(tol)
if tol < eps:
print "Done"
break
tF = -numpy.concatenate((c, h))
FGrad2 = scisp.csc_matrix(scisp.bmat([[Q,Bt],[B, None]]))
print FGrad2
print FGrad2.todense()
print " "
print tF
xdelta = scial.spsolve(FGrad2,tF)
print xdelta
x = x + xdelta[0:col_size_A]
l = x[col_size_A:]
k = k + 1
if __name__ == "__main__":
ini()
The output is:
(2, 0) 1.0
(3, 0) 1.0
(4, 0) -1.0
(2, 1) 1.0
(3, 1) -1.0
(5, 1) -1.0
(0, 2) 1.0
(1, 2) 1.0
(0, 3) 1.0
(1, 3) -1.0
(0, 4) -1.0
(1, 5) -1.0
[[ 0. 0. 1. 1. -1. 0.]
[ 0. 0. 1. -1. 0. -1.]
[ 1. 1. 0. 0. 0. 0.]
[ 1. -1. 0. 0. 0. 0.]
[-1. 0. 0. 0. 0. 0.]
[ 0. -1. 0. 0. 0. 0.]]
lda must be >= MAX(N,1): lda=2 N=3BLAS error: Parameter number 7 passed to cblas_dtrsv had an invalid value
[-1. -1. 1. -0. -0. -0.]
So FGrad2 seems to be a valid csc matrix and tF a valid numpy.array.
What is wrong with this code? I don't even know why the error is before the print of tF even so the error is behind at spsolve
Edit
Ok i fixed that, it is because the first guess for the parameters was wrong leading to a singular matrix, but suppling a valid guess for l, leads to wrong calculation of spsolve
as mentioned i labeled all output as you can see spsolve returns the wrong calculation.
$FGrad2 * xdelta != tF$
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import scipy.sparse.linalg as scial
import scipy.sparse as scisp
import numpy
def buildB(A,x,col_size_A):
d = numpy.zeros(col_size_A)
for index in xrange(col_size_A):
d[index] = 2*x[index]-1
tmp = scisp.spdiags(d,0,col_size_A,col_size_A)
return scisp.bmat([[A],[tmp]])
def buildQ(l,row_size_A):
q = numpy.zeros(row_size_A)
for index in xrange(row_size_A):
q[index] = 2*l[index]
return scisp.spdiags(q,0,row_size_A,row_size_A)
def buildh(A,x,b,col_size_A):
p = A.dot(x)
p = numpy.subtract(p, b)
quad = numpy.zeros(col_size_A)
for index in xrange(col_size_A):
quad[index] = x[index]*x[index]-x[index]
return numpy.concatenate((p, quad))
def ini():
A = numpy.array([[1,1],[1,0]])
b = [1, 0]
c = [1, 1]
col_size_A = 2
row_size_A = 2
main(A,b,c,col_size_A,row_size_A)
def main(A,b,c, col_size_A, row_size_A):
x = numpy.zeros(col_size_A)
x[0] = 0
x[1] = 1
l = numpy.ones(row_size_A*2)
eps = 10e-6
k = 0
while True:
B = buildB(A,x,col_size_A)
Q = buildQ(l[row_size_A:], col_size_A)
Bt = B.transpose()
h = buildh(A,x,b,col_size_A)
g = numpy.add(c,Bt.dot(l))
F = numpy.concatenate((g, h))
print "Iteration " + str(k),
tol = numpy.amax(numpy.absolute(F))
print "- Tol "+ str(tol)
if tol < eps:
print "Done"
print x
break
tF = -numpy.concatenate((c, h))
FGrad2 = scisp.csc_matrix(scisp.bmat([[Q,Bt],[B, None]]))
print "FGrad2"
print FGrad2.todense()
print "tF"
print tF
xdelta = scial.spsolve(FGrad2,tF)
print "spsolution"
print xdelta
print ""
x = x + xdelta[0:col_size_A]
l = xdelta[col_size_A:]
k = k + 1
if __name__ == "__main__":
ini()
Output:
Iteration 0 - Tol 3.0
FGrad2
[[ 2. 0. 1. 1. -1. 0.]
[ 0. 2. 1. 0. 0. 1.]
[ 1. 1. 0. 0. 0. 0.]
[ 1. 0. 0. 0. 0. 0.]
[-1. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 0. 0.]]
tF
[-1. -1. -0. -0. -0. -0.]
spsolution
[-1. -1. -0. -0. -0. -0.]
I think this is failing for you because your matrix is singular. E.g. convert to dense and use the regular numpy.linalg.solve:
>>> xdelta = numpy.linalg.solve(FGrad2.todense(), tF)
...
raise LinAlgError('Singular matrix')
numpy.linalg.linalg.LinAlgError: Singular matrix
The error I get is:
File "stack27538259.py", line 62, in main
xdelta = scial.spsolve(FGrad2,tF)
File "/usr/lib/python2.7/dist-packages/scipy/sparse/linalg/dsolve/linsolve.py", line 143, in spsolve
b, flag, options=options)
RuntimeError: superlu failure (singular matrix?) at line 100 in file scipy/sparse/linalg/dsolve/SuperLU/SRC/dsnode_bmod.c
As xnx wrote, FGrad2 is singular.
np.linalg.det(FGrad2.todense()) # 0.0
(scipy version 0.14.0)
after the change I get:
/usr/lib/python2.7/dist-packages/scipy/sparse/linalg/dsolve/linsolve.py:145: MatrixRankWarning: Matrix is exactly singular
and
spsolution
[ nan nan nan nan nan nan]
and an infinite loop unless I add k counter and break.
Documentation for cblas_dtrsv may be found (here)
Accordingly,
the routine solves a triangular system A*X = B (presumably)
lda is the leading dimension of matrix B
N is the order of the matrix A
the error message says lda = 2 and N = 3 but lda must be >= MAX(N,1)
Perhaps this helps track down the problem.