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.
Related
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.
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.
This question already has answers here:
How to generate a sphere in 3D Numpy array
(5 answers)
Closed 3 months ago.
I have a ndarray of size 32x32x32. I want to create a sphere inside the array with the center at (x,y) and a radius of 4 pixels. The value of the sphere is 1 while value of the array is 0. How can this be done in python?
This is the code to generate the array:
import numpy as np
A = np.zeros((32,32,32))
print (A)
Very good question. You can try the following code. In the below mentioned code AA is the matrix that you want. =)
import numpy as np
from copy import deepcopy
''' size : size of original 3D numpy matrix A.
radius : radius of circle inside A which will be filled with ones.
'''
size, radius = 5, 2
''' A : numpy.ndarray of shape size*size*size. '''
A = np.zeros((size,size, size))
''' AA : copy of A (you don't want the original copy of A to be overwritten.) '''
AA = deepcopy(A)
''' (x0, y0, z0) : coordinates of center of circle inside A. '''
x0, y0, z0 = int(np.floor(A.shape[0]/2)), \
int(np.floor(A.shape[1]/2)), int(np.floor(A.shape[2]/2))
for x in range(x0-radius, x0+radius+1):
for y in range(y0-radius, y0+radius+1):
for z in range(z0-radius, z0+radius+1):
''' deb: measures how far a coordinate in A is far from the center.
deb>=0: inside the sphere.
deb<0: outside the sphere.'''
deb = radius - abs(x0-x) - abs(y0-y) - abs(z0-z)
if (deb)>=0: AA[x,y,z] = 1
Following is an example of the output for size=5 and radius=2 (a sphere of radius 2 pixels inside a numpy array of shape 5*5*5):
[[[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[1. 1. 1. 1. 1.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]]
I haven't printed the output for the size and radius that you had asked for (size=32 and radius=4), as the output will be very long.
Since array indexes only have a certain level of specificity (i.e. you can only subdivide down to the width, in this case 32), there's no one perfect way to represent a sphere in an array. Instead, we can treat each array index as a space of cubic area, where the [x][y][z] indices of the index represent the the cubic area's center coordinates. To create the sphere, we evaluate whether the sphere's presence in that area of space meets certain criteria.
We start with the equation for a sphere. From Wikipedia:
In analytic geometry, a sphere with center (x0, y0, z0) and radius r is the locus of all points (x, y, z) such that
(x - x0)^2 + (y - y0)^2 + (z - z0)^2 <= r^2.
For an array of dimensions N, the center will have the coordinate (N - 1) / 2 for all dimensions. (because for an even-numbered dimension, the center should be between the two center points, and for an odd-numbered dimension, the center should be an integer.) The magnitude of the radius can vary depending on where you decide the boundaries of the sphere relative to our imagined cubic array representation; re-reading the question, I notice you already gave the desired radius: 4.
There are two evaluation criteria I can think of:
Simple approach
In this approach, we will simply use a test of whether the array index's cubic area's center lies within the circle equation.
You can see Siddharth Satpathy's answer for some code using this approach.
Sophisticated approach
Ideally for me, the equation would decide whether an index lies within the sphere by assessing whether the proportion of sphere for that cubic area is greater than 50%. However, this approach unfortunately goes beyond my current working mathematical knowledge.
In regards to a discussion I had in the comments, neither approach is better than the other since they represent different perspectives: I personally imagine the array as being actually representative of the cubic area for each index, while others may imagine the indexes being the center points of these cubic areas.
Nothing above worked for me so there is my attempt:
def create_bin_sphere(arr_size, center, r):
coords = np.ogrid[:arr_size[0], :arr_size[1], :arr_size[2]]
distance = np.sqrt((coords[0] - center[0])**2 + (coords[1]-center[1])**2 + (coords[2]-center[2])**2)
return 1*(distance <= r)
where:
arr_size is a tuple with numpy array shape
center is a tuple with sphere center coords
r is a radius of the sphere
Example:
arr_size = (30,30,30)
sphere_center = (15,15,15)
r=10
sphere = create_bin_sphere(arr_size,sphere_center, r)
#Plot the result
fig =plt.figure(figsize=(6,6))
ax = fig.gca(projection='3d')
ax.voxels(sphere, facecolors=colors, edgecolor='k')
plt.show()
Vizualization rezult
Eventhough it's a little late - I recently faced the same problem and solved it somewhat like the solution supposed by Mstaino. Also, Mstainos solution doesn't work for asymetricly sized array's, because shapes will not match when calculating the distance.
So here's my approach in 3D which produces a sphere in the center of the array. :
# define array size and sphere radius
size = [size_x, size_y, size_z]
radius = sphere_radius
# compute center index of array
center = [int(size[0]/2), int(size[1]/2), int(size[2]/2)]
# create index grid for array
ind_0, ind_1, ind_2 = np.indices((size[0], size[1], size[2]))
# calculate "distance" of indices to center index
distance = ((ind_0 - center[0])**2 + (ind_1 - center[1])**2 + (ind_2 - center[2])**2)**.5
# create output
output = np.ones(shape = (size[0], size[1], size[2])) * (distance <= radius)
Using a combination of indexing, distance calculation and masking (all with numpy):
import numpy as np
center = (31/2, 31/2, 31/2) # if it is centered
size = (32, 32, 32)
max_dist = 4
distance = np.linalg.norm(np.subtract(np.indices(size).T, np.asarray(center)), axis=2)
#print(distance)
mask = np.ones(size) * (distance < max_dist)
print(mask)
np.indices creates an index in the form [[[(i, j, k)]]], np.substract calculates the vector difference to your center, and np.linalg.norm calculates the vector norm. The rest is just using a mask operation on the distance array.
Does that work?
EDIT: an example with (3,3,3) for clarity purposes
center = (1, 1, 1)
size = (3, 3, 3)
distance = np.linalg.norm(np.subtract(np.indices(size).T,np.asarray(center)), axis=len(center))
mask = np.ones(size) * (distance<=1)
print(mask)
>>[[[0. 0. 0.]
[0. 1. 0.]
[0. 0. 0.]]
[[0. 1. 0.]
[1. 1. 1.]
[0. 1. 0.]]
[[0. 0. 0.]
[0. 1. 0.]
[0. 0. 0.]]]
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.]]
What is the best way to fill in the lower triangle of a numpy array with zeros in place so that I don't have to do the following:
a=np.random.random((5,5))
a = np.triu(a)
since np.triu returns a copy, not a view. Preferable this would require no list indexing as well since I am working with large arrays.
Digging into the internals of triu you'll find that it just multiplies the input by the output of tri.
So you can just multiply the array in-place by the output of tri:
>>> a = np.random.random((5, 5))
>>> a *= np.tri(*a.shape)
>>> a
array([[ 0.46026582, 0. , 0. , 0. , 0. ],
[ 0.76234296, 0.5298908 , 0. , 0. , 0. ],
[ 0.08797149, 0.14881991, 0.9302515 , 0. , 0. ],
[ 0.54794779, 0.36896506, 0.92901552, 0.73747726, 0. ],
[ 0.62917827, 0.61674542, 0.44999905, 0.80970863, 0.41860336]])
Like triu, this still creates a second array (the output of tri), but at least it performs the operation itself in-place. The splat is a bit of a shortcut; consider basing your function on the full version of triu for something robust. But note that you can still specify a diagonal:
>>> a = np.random.random((5, 5))
>>> a *= np.tri(*a.shape, k=2)
>>> a
array([[ 0.25473126, 0.70156073, 0.0973933 , 0. , 0. ],
[ 0.32859487, 0.58188318, 0.95288351, 0.85735005, 0. ],
[ 0.52591784, 0.75030515, 0.82458369, 0.55184033, 0.01341398],
[ 0.90862183, 0.33983192, 0.46321589, 0.21080121, 0.31641934],
[ 0.32322392, 0.25091433, 0.03980317, 0.29448128, 0.92288577]])
I now see that the question title and body describe opposite behaviors. Just in case, here's how you can fill the lower triangle with zeros. This requires you to specify the -1 diagonal:
>>> a = np.random.random((5, 5))
>>> a *= 1 - np.tri(*a.shape, k=-1)
>>> a
array([[0.6357091 , 0.33589809, 0.744803 , 0.55254798, 0.38021111],
[0. , 0.87316263, 0.98047459, 0.00881754, 0.44115527],
[0. , 0. , 0.51317289, 0.16630385, 0.1470729 ],
[0. , 0. , 0. , 0.9239731 , 0.11928557],
[0. , 0. , 0. , 0. , 0.1840326 ]])
If speed and memory use are still a limitation and Cython is available, a short Cython function will do what you want.
Here's a working version designed for a C-contiguous array with double precision values.
cimport cython
#cython.boundscheck(False)
#cython.wraparound(False)
cpdef make_lower_triangular(double[:,:] A, int k):
""" Set all the entries of array A that lie above
diagonal k to 0. """
cdef int i, j
for i in range(min(A.shape[0], A.shape[0] - k)):
for j in range(max(0, i+k+1), A.shape[1]):
A[i,j] = 0.
This should be significantly faster than any version that involves multiplying by a large temporary array.
import numpy as np
n=3
A=np.zeros((n,n))
for p in range(n):
A[0,p] = p+1
if p >0 :
A[1,p]=p+3
if p >1 :
A[2,p]=p+4
creates a upper triangular matrix starting at 1