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I am trying to perform a 2d convolution in python using numpy
I have a 2d array as follows with kernel H_r for the rows and H_c for the columns
data = np.zeros((nr, nc), dtype=np.float32)
#fill array with some data here then convolve
for r in range(nr):
data[r,:] = np.convolve(data[r,:], H_r, 'same')
for c in range(nc):
data[:,c] = np.convolve(data[:,c], H_c, 'same')
data = data.astype(np.uint8);
It does not produce the output that I was expecting, does this code look OK, I think the problem is with the casting from float32 to 8bit. Whats the best way to do this
Thanks
Maybe it is not the most optimized solution, but this is an implementation I used before with numpy library for Python:
def convolution2d(image, kernel, bias):
m, n = kernel.shape
if (m == n):
y, x = image.shape
y = y - m + 1
x = x - m + 1
new_image = np.zeros((y,x))
for i in range(y):
for j in range(x):
new_image[i][j] = np.sum(image[i:i+m, j:j+m]*kernel) + bias
return new_image
I hope this code helps other guys with the same doubt.
Regards.
Edit [Jan 2019]
#Tashus comment bellow is correct, and #dudemeister's answer is thus probably more on the mark. The function he suggested is also more efficient, by avoiding a direct 2D convolution and the number of operations that would entail.
Possible Problem
I believe you are doing two 1d convolutions, the first per columns and the second per rows, and replacing the results from the first with the results of the second.
Notice that numpy.convolve with the 'same' argument returns an array of equal shape to the largest one provided, so when you make the first convolution you already populated the entire data array.
One good way to visualize your arrays during these steps is to use Hinton diagrams, so you can check which elements already have a value.
Possible Solution
You can try to add the results of the two convolutions (use data[:,c] += .. instead of data[:,c] = on the second for loop), if your convolution matrix is the result of using the one dimensional H_r and H_c matrices like so:
Another way to do that would be to use scipy.signal.convolve2d with a 2d convolution array, which is probably what you wanted to do in the first place.
Since you already have your kernel separated you should simply use the sepfir2d function from scipy:
from scipy.signal import sepfir2d
convolved = sepfir2d(data, H_r, H_c)
On the other hand, the code you have there looks all right ...
I checked out many implementations and found none for my purpose, which should be really simple. So here is a dead-simple implementation with for loop
def convolution2d(image, kernel, stride, padding):
image = np.pad(image, [(padding, padding), (padding, padding)], mode='constant', constant_values=0)
kernel_height, kernel_width = kernel.shape
padded_height, padded_width = image.shape
output_height = (padded_height - kernel_height) // stride + 1
output_width = (padded_width - kernel_width) // stride + 1
new_image = np.zeros((output_height, output_width)).astype(np.float32)
for y in range(0, output_height):
for x in range(0, output_width):
new_image[y][x] = np.sum(image[y * stride:y * stride + kernel_height, x * stride:x * stride + kernel_width] * kernel).astype(np.float32)
return new_image
It might not be the most optimized solution either, but it is approximately ten times faster than the one proposed by #omotto and it only uses basic numpy function (as reshape, expand_dims, tile...) and no 'for' loops:
def gen_idx_conv1d(in_size, ker_size):
"""
Generates a list of indices. This indices correspond to the indices
of a 1D input tensor on which we would like to apply a 1D convolution.
For instance, with a 1D input array of size 5 and a kernel of size 3, the
1D convolution product will successively looks at elements of indices [0,1,2],
[1,2,3] and [2,3,4] in the input array. In this case, the function idx_conv1d(5,3)
outputs the following array: array([0,1,2,1,2,3,2,3,4]).
args:
in_size: (type: int) size of the input 1d array.
ker_size: (type: int) kernel size.
return:
idx_list: (type: np.array) list of the successive indices of the 1D input array
access to the 1D convolution algorithm.
example:
>>> gen_idx_conv1d(in_size=5, ker_size=3)
array([0, 1, 2, 1, 2, 3, 2, 3, 4])
"""
f = lambda dim1, dim2, axis: np.reshape(np.tile(np.expand_dims(np.arange(dim1),axis),dim2),-1)
out_size = in_size-ker_size+1
return f(ker_size, out_size, 0)+f(out_size, ker_size, 1)
def repeat_idx_2d(idx_list, nbof_rep, axis):
"""
Repeats an array of indices (idx_list) a number of time (nbof_rep) "along" an axis
(axis). This function helps to browse through a 2d array of size
(len(idx_list),nbof_rep).
args:
idx_list: (type: np.array or list) a 1D array of indices.
nbof_rep: (type: int) number of repetition.
axis: (type: int) axis "along" which the repetition will be applied.
return
idx_list: (type: np.array) a 1D array of indices of size len(idx_list)*nbof_rep.
example:
>>> a = np.array([0, 1, 2])
>>> repeat_idx_2d(a, 3, 0) # repeats array 'a' 3 times along 'axis' 0
array([0, 0, 0, 1, 1, 1, 2, 2, 2])
>>> repeat_idx_2d(a, 3, 1) # repeats array 'a' 3 times along 'axis' 1
array([0, 1, 2, 0, 1, 2, 0, 1, 2])
>>> b = np.reshape(np.arange(3*4), (3,4))
>>> b[repeat_idx_2d(np.arange(3), 4, 0), repeat_idx_2d(np.arange(4), 3, 1)]
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
"""
assert axis in [0,1], "Axis should be equal to 0 or 1."
tile_axis = (nbof_rep,1) if axis else (1,nbof_rep)
return np.reshape(np.tile(np.expand_dims(idx_list, 1),tile_axis),-1)
def conv2d(im, ker):
"""
Performs a 'valid' 2D convolution on an image. The input image may be
a 2D or a 3D array.
The output image first two dimensions will be reduced depending on the
convolution size.
The kernel may be a 2D or 3D array. If 2D, it will be applied on every
channel of the input image. If 3D, its last dimension must match the
image one.
args:
im: (type: np.array) image (2D or 3D).
ker: (type: np.array) convolution kernel (2D or 3D).
returns:
im: (type: np.array) convolved image.
example:
>>> im = np.reshape(np.arange(10*10*3),(10,10,3))/(10*10*3) # 3D image
>>> ker = np.array([[0,1,0],[-1,0,1],[0,-1,0]]) # 2D kernel
>>> conv2d(im, ker) # 3D array of shape (8,8,3)
"""
if len(im.shape)==2: # if the image is a 2D array, it is reshaped by expanding the last dimension
im = np.expand_dims(im,-1)
im_x, im_y, im_w = im.shape
if len(ker.shape)==2: # if the kernel is a 2D array, it is reshaped so it will be applied to all of the image channels
ker = np.tile(np.expand_dims(ker,-1),[1,1,im_w]) # the same kernel will be applied to all of the channels
assert ker.shape[-1]==im.shape[-1], "Kernel and image last dimension must match."
ker_x = ker.shape[0]
ker_y = ker.shape[1]
# shape of the output image
out_x = im_x - ker_x + 1
out_y = im_y - ker_y + 1
# reshapes the image to (out_x, ker_x, out_y, ker_y, im_w)
idx_list_x = gen_idx_conv1d(im_x, ker_x) # computes the indices of a 1D conv (cf. idx_conv1d doc)
idx_list_y = gen_idx_conv1d(im_y, ker_y)
idx_reshaped_x = repeat_idx_2d(idx_list_x, len(idx_list_y), 0) # repeats the previous indices to be used in 2D (cf. repeat_idx_2d doc)
idx_reshaped_y = repeat_idx_2d(idx_list_y, len(idx_list_x), 1)
im_reshaped = np.reshape(im[idx_reshaped_x, idx_reshaped_y, :], [out_x, ker_x, out_y, ker_y, im_w]) # reshapes
# reshapes the 2D kernel
ker = np.reshape(ker,[1, ker_x, 1, ker_y, im_w])
# applies the kernel to the image and reduces the dimension back to the one of original input image
return np.squeeze(np.sum(im_reshaped*ker, axis=(1,3)))
I tried to add a lot of comments to explain the method but the global idea is to reshape the 3D input image to a 5D one of shape (output_image_height, kernel_height, output_image_width, kernel_width, output_image_channel) and then to apply the kernel directly using the basic array multiplication. Of course, this methods is then using more memory (during the execution the size of the image is thus multiply by kernel_height*kernel_width) but it is faster.
To do this reshape step, I 'over-used' the indexing methods of numpy arrays, especially, the possibility of giving a numpy array as indices into a numpy array.
This methods could also be used to re-code the 2D convolution product in Pytorch or Tensorflow using the base math functions but I have no doubt in saying that it will be slower than the existing nn.conv2d operator...
I really enjoyed coding this method by only using the numpy basic tools.
One of the most obvious is to hard code the kernel.
img = img.convert('L')
a = np.array(img)
out = np.zeros([a.shape[0]-2, a.shape[1]-2], dtype='float')
out += a[:-2, :-2]
out += a[1:-1, :-2]
out += a[2:, :-2]
out += a[:-2, 1:-1]
out += a[1:-1,1:-1]
out += a[2:, 1:-1]
out += a[:-2, 2:]
out += a[1:-1, 2:]
out += a[2:, 2:]
out /= 9.0
out = out.astype('uint8')
img = Image.fromarray(out)
This example does a box blur 3x3 completely unrolled. You can multiply the values where you have a different value and divide them by a different amount. But, if you honestly want the quickest and dirtiest method this is it. I think it beats Guillaume Mougeot's method by a factor of like 5. His method beating the others by a factor of 10.
It may lose a few steps if you're doing something like a gaussian blur. and need to multiply some stuff.
Try to first round and then cast to uint8:
data = data.round().astype(np.uint8);
I wrote this convolve_stride which uses numpy.lib.stride_tricks.as_strided. Moreover it supports both strides and dilation. It is also compatible to tensor with order > 2.
import numpy as np
from numpy.lib.stride_tricks import as_strided
from im2col import im2col
def conv_view(X, F_s, dr, std):
X_s = np.array(X.shape)
F_s = np.array(F_s)
dr = np.array(dr)
Fd_s = (F_s - 1) * dr + 1
if np.any(Fd_s > X_s):
raise ValueError('(Dilated) filter size must be smaller than X')
std = np.array(std)
X_ss = np.array(X.strides)
Xn_s = (X_s - Fd_s) // std + 1
Xv_s = np.append(Xn_s, F_s)
Xv_ss = np.tile(X_ss, 2) * np.append(std, dr)
return as_strided(X, Xv_s, Xv_ss, writeable=False)
def convolve_stride(X, F, dr=None, std=None):
if dr is None:
dr = np.ones(X.ndim, dtype=int)
if std is None:
std = np.ones(X.ndim, dtype=int)
if not (X.ndim == F.ndim == len(dr) == len(std)):
raise ValueError('X.ndim, F.ndim, len(dr), len(std) must be the same')
Xv = conv_view(X, F.shape, dr, std)
return np.tensordot(Xv, F, axes=X.ndim)
%timeit -n 100 -r 10 convolve_stride(A, F)
#31.2 ms ± 1.31 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)
Super simple and fast convolution using only basic numpy:
import numpy as np
def conv2d(image, kernel):
# apply kernel to image, return image of the same shape
# assume both image and kernel are 2D arrays
# kernel = np.flipud(np.fliplr(kernel)) # optionally flip the kernel
k = kernel.shape[0]
width = k//2
# place the image inside a frame to compensate for the kernel overlap
a = framed(image, width)
b = np.zeros(image.shape) # fill the output array with zeros; do not use np.empty()
# shift the image around each pixel, multiply by the corresponding kernel value and accumulate the results
for p, dp, r, dr in [(i, i + image.shape[0], j, j + image.shape[1]) for i in range(k) for j in range(k)]:
b += a[p:dp, r:dr] * kernel[p, r]
# or just write two nested for loops if you prefer
# np.clip(b, 0, 255, out=b) # optionally clip values exceeding the limits
return b
def framed(image, width):
a = np.zeros((image.shape[0]+2*width, image.shape[1]+2*width))
a[width:-width, width:-width] = image
# alternatively fill the frame with ones or copy border pixels
return a
Run it:
Image.fromarray(conv2d(image, kernel).astype('uint8'))
Instead of sliding the kernel along the image and computing the transformation pixel by pixel, create a series of shifted versions of the image corresponding to each element in the kernel and apply the corresponding kernel value to each of the shifted image versions.
This is probably the fastest you can get using just basic numpy; the speed is already comparable to C implementation of scipy convolve2d and better than fftconvolve. The idea is similar to #Tatarize. This example works only for one color component; for RGB just repeat for each (or modify the algorithm accordingly).
Typically, Convolution 2D is a misnomer. Ideally, under the hood,
whats being done is a correlation of 2 matrices.
pad == same
returns the output as the same as input dimension
It can also take asymmetric images. In order to perform correlation(convolution in deep learning lingo) on a batch of 2d matrices, one can iterate over all the channels, calculate the correlation for each of the channel slices with the respective filter slice.
For example: If image is (28,28,3) and filter size is (5,5,3) then take each of the 3 slices from the image channel and perform the cross correlation using the custom function above and stack the resulting matrix in the respective dimension of the output.
def get_cross_corr_2d(W, X, pad = 'valid'):
if(pad == 'same'):
pr = int((W.shape[0] - 1)/2)
pc = int((W.shape[1] - 1)/2)
conv_2d = np.zeros((X.shape[0], X.shape[1]))
X_pad = np.zeros((X.shape[0] + 2*pr, X.shape[1] + 2*pc))
X_pad[pr:pr+X.shape[0], pc:pc+X.shape[1]] = X
for r in range(conv_2d.shape[0]):
for c in range(conv_2d.shape[1]):
conv_2d[r,c] = np.sum(np.inner(W, X_pad[r:r+W.shape[0], c:c+W.shape[1]]))
return conv_2d
else:
pr = W.shape[0] - 1
pc = W.shape[1] - 1
conv_2d = np.zeros((X.shape[0] - W.shape[0] + 2*pr + 1,
X.shape[1] - W.shape[1] + 2*pc + 1))
X_pad = np.zeros((X.shape[0] + 2*pr, X.shape[1] + 2*pc))
X_pad[pr:pr+X.shape[0], pc:pc+X.shape[1]] = X
for r in range(conv_2d.shape[0]):
for c in range(conv_2d.shape[1]):
conv_2d[r,c] = np.sum(np.multiply(W, X_pad[r:r+W.shape[0], c:c+W.shape[1]]))
return conv_2d
This code incorrect:
for r in range(nr):
data[r,:] = np.convolve(data[r,:], H_r, 'same')
for c in range(nc):
data[:,c] = np.convolve(data[:,c], H_c, 'same')
See Nussbaumer transformation from multidimentional convolution to one dimentional.
I have big binary 3D data and I want to re-arrange the data such as it is a sequence of values in order achieved by parsing the original data as sub-arrays of size (4x4x4).
For example, if the data is 2D and I want to re-arrange the data from 2x2 sub-arrays
example image
I used simple loops for this but just iterating over the loops took way more times, I am trying to to use some numpy functions to do so but I am new to SciPy
My code looks like this
x,y,z = 1200,800,400
data = np.fromfile(file_name, dtype=np.float32)
data.shape = (z,y,x)
new_data = np.empty(shape=x*y*z, dtype = np.float32)
index = 0
for zz in range(0,z,4):
for yy in range(0,y,4):
for xx in range(0,x,4):
for zShift in range(4):
for yShift in range(4):
for xShift in range(4):
new_data[index] = data[zz+zShift][yy+yShift][xx+xShift]
index+=1
new_data.tofile(output)
However, this takes a lot of time, any better implementation ideas?
As I said, the code works as intended, however, I need a smarter, pythonic way to achieve my output
Thank you!
x,y,z = 1200,800,400
data = np.empty([x,y,z])
# numpy calculates the shape of -1
out = data.reshape(-1, 4, 4, 4)
out.shape
>>> (6000000, 4, 4, 4)
Perform the following test, for smaller data and block size:
x, y, z = 4, 4, 4 # Dimensions
stp = 2 # Block size (in each dimension)
# Create the test array
arr = np.arange(x * y * z).reshape((x, y, z))
And to create a list of "blocks", run:
new_data = []
for xx in range(0, x, stp):
for yy in range(0, y, stp):
for zz in range(0, z, stp):
print('Index:', xx, yy, zz)
obj = arr[xx:xx+stp, yy:yy+stp, zz:zz+stp].copy()
print(obj)
new_data.append(obj)
In the target version of your code:
restore original values of x, y and z,
read the array from your source,
change stp back to 4,
drop test printouts.
Note also that your code adds individual elements to new_data,
only iterating over blocks of size 4 * 4 * 4,
whereas you wrote that you want a sequence of smaller arrays
(i.e. slices) of size 4 * 4 * 4, what my code does.
So if you need a list of slices (smaller arrays), not a single
4-D array, use my code.
I have used interp2 in Matlab, such as the following code, that is part of #rayryeng's answer in: Three dimensional (3D) matrix interpolation in Matlab:
d = size(volume_image)
[X,Y] = meshgrid(1:1/scaleCoeff(2):d(2), 1:1/scaleCoeff(1):d(1));
for ind = z
%Interpolate each slice via interp2
M2D(:,:,ind) = interp2(volume_image(:,:,ind), X, Y);
end
Example of Dimensions:
The image size is 512x512 and the number of slices is 133. So:
volume_image(rows, columns, slices in 3D dimenson) : 512x512x133 in 3D dimenson
X: 288x288
Y: 288x288
scaleCoeff(2): 0.5625
scaleCoeff(1): 0.5625
z = 1 up to 133 ,hence z: 1x133
ind: 1 up to 133
M2D(:,:,ind) finally is 288x288x133 in 3D dimenson
Aslo, Matlabs syntax for size: (rows, columns, slices in 3rd dimenson) and Python syntax for size: (slices in 3rd dim, rows, columns).
However, after convert the Matlab code to Python code occurred an error, ValueError: Invalid length for input z for non rectangular grid:
for ind in range(0, len(z)+1):
M2D[ind, :, :] = interpolate.interp2d(X, Y, volume_image[ind, :, :]) # ValueError: Invalid length for input z for non rectangular grid
What is wrong? Thank you so much.
In MATLAB, interp2 has as arguments:
result = interp2(input_x, input_y, input_z, output_x, output_y)
You are using only the latter 3 arguments, the first two are assumed to be input_x = 1:size(input_z,2) and input_y = 1:size(input_z,1).
In Python, scipy.interpolate.interp2 is quite different: it takes the first 3 input arguments of the MATLAB function, and returns an object that you can call to get interpolated values:
f = scipy.interpolate.interp2(input_x, input_y, input_z)
result = f(output_x, output_y)
Following the example from the documentation, I get to something like this:
from scipy import interpolate
x = np.arange(0, volume_image.shape[2])
y = np.arange(0, volume_image.shape[1])
f = interpolate.interp2d(x, y, volume_image[ind, :, :])
xnew = np.arange(0, volume_image.shape[2], 1/scaleCoeff[0])
ynew = np.arange(0, volume_image.shape[1], 1/scaleCoeff[1])
M2D[ind, :, :] = f(xnew, ynew)
[Code not tested, please let me know if there are errors.]
You might be interested in scipy.ndimage.zoom. If you are interpolating from one regular grid to another, it is much faster and easier to use than scipy.interpolate.interp2d.
See this answer for an example:
https://stackoverflow.com/a/16984081/1295595
You'd probably want something like:
import scipy.ndimage as ndimage
M2D = ndimage.zoom(volume_image, (1, scaleCoeff[0], scaleCoeff[1])
I have a set of 3d coordinates that was generated using meshgrid(). I want to be able to rotate these about the 3 axes.
I tried unraveling the meshgrid and doing a rotation on each point but the meshgrid is large and I run out of memory.
This question addresses this in 2d with einsum(), but I can't figure out the string format when extending it to 3d.
I have read several other pages about einsum() and its format string but haven't been able to figure it out.
EDIT:
I call my meshgrid axes X, Y, and Z, each is of shape (213, 48, 37). Also, the actual memory error came when I tried to put the results back into a meshgrid.
When I attempted to 'unravel' it to do point by point rotation I used the following function:
def mg2coords(X, Y, Z):
return np.vstack([X.ravel(), Y.ravel(), Z.ravel()]).T
I looped over the result with the following:
def rotz(angle, point):
rad = np.radians(angle)
sin = np.sin(rad)
cos = np.cos(rad)
rot = [[cos, -sin, 0],
[sin, cos, 0],
[0, 0, 1]]
return np.dot(rot, point)
After the rotation I will be using the points to interpolate onto.
Working with your definitions:
In [840]: def mg2coords(X, Y, Z):
return np.vstack([X.ravel(), Y.ravel(), Z.ravel()]).T
In [841]: def rotz(angle):
rad = np.radians(angle)
sin = np.sin(rad)
cos = np.cos(rad)
rot = [[cos, -sin, 0],
[sin, cos, 0],
[0, 0, 1]]
return np.array(rot)
# just to the rotation matrix
define a sample grid:
In [842]: X,Y,Z=np.meshgrid([0,1,2],[0,1,2,3],[0,1,2],indexing='ij')
In [843]: xyz=mg2coords(X,Y,Z)
rotate it row by row:
In [844]: xyz1=np.array([np.dot(rot,i) for i in xyz])
equivalent einsum row by row calculation:
In [845]: xyz2=np.einsum('ij,kj->ki',rot,xyz)
They match:
In [846]: np.allclose(xyz2,xyz1)
Out[846]: True
Alternatively I could collect the 3 arrays into one 4d array, and rotate that with einsum. Here np.array adds a dimension at the start. So the dot sum j dimension is 1st, and the 3d of the arrays follow:
In [871]: XYZ=np.array((X,Y,Z))
In [872]: XYZ2=np.einsum('ij,jabc->iabc',rot,XYZ)
In [873]: np.allclose(xyz2[:,0], XYZ2[0,...].ravel())
Out[873]: True
Similary for the 1 and 2.
Alternatively I could split XYZ2 into 3 component arrays:
In [882]: X2,Y2,Z2 = XYZ2
In [883]: np.allclose(X2,xyz2[:,0].reshape(X.shape))
Out[883]: True
Use ji instead of ij if you want to rotate in the other direction, i.e. use rot.T.
I am trying to perform a 2d convolution in python using numpy
I have a 2d array as follows with kernel H_r for the rows and H_c for the columns
data = np.zeros((nr, nc), dtype=np.float32)
#fill array with some data here then convolve
for r in range(nr):
data[r,:] = np.convolve(data[r,:], H_r, 'same')
for c in range(nc):
data[:,c] = np.convolve(data[:,c], H_c, 'same')
data = data.astype(np.uint8);
It does not produce the output that I was expecting, does this code look OK, I think the problem is with the casting from float32 to 8bit. Whats the best way to do this
Thanks
Maybe it is not the most optimized solution, but this is an implementation I used before with numpy library for Python:
def convolution2d(image, kernel, bias):
m, n = kernel.shape
if (m == n):
y, x = image.shape
y = y - m + 1
x = x - m + 1
new_image = np.zeros((y,x))
for i in range(y):
for j in range(x):
new_image[i][j] = np.sum(image[i:i+m, j:j+m]*kernel) + bias
return new_image
I hope this code helps other guys with the same doubt.
Regards.
Edit [Jan 2019]
#Tashus comment bellow is correct, and #dudemeister's answer is thus probably more on the mark. The function he suggested is also more efficient, by avoiding a direct 2D convolution and the number of operations that would entail.
Possible Problem
I believe you are doing two 1d convolutions, the first per columns and the second per rows, and replacing the results from the first with the results of the second.
Notice that numpy.convolve with the 'same' argument returns an array of equal shape to the largest one provided, so when you make the first convolution you already populated the entire data array.
One good way to visualize your arrays during these steps is to use Hinton diagrams, so you can check which elements already have a value.
Possible Solution
You can try to add the results of the two convolutions (use data[:,c] += .. instead of data[:,c] = on the second for loop), if your convolution matrix is the result of using the one dimensional H_r and H_c matrices like so:
Another way to do that would be to use scipy.signal.convolve2d with a 2d convolution array, which is probably what you wanted to do in the first place.
Since you already have your kernel separated you should simply use the sepfir2d function from scipy:
from scipy.signal import sepfir2d
convolved = sepfir2d(data, H_r, H_c)
On the other hand, the code you have there looks all right ...
I checked out many implementations and found none for my purpose, which should be really simple. So here is a dead-simple implementation with for loop
def convolution2d(image, kernel, stride, padding):
image = np.pad(image, [(padding, padding), (padding, padding)], mode='constant', constant_values=0)
kernel_height, kernel_width = kernel.shape
padded_height, padded_width = image.shape
output_height = (padded_height - kernel_height) // stride + 1
output_width = (padded_width - kernel_width) // stride + 1
new_image = np.zeros((output_height, output_width)).astype(np.float32)
for y in range(0, output_height):
for x in range(0, output_width):
new_image[y][x] = np.sum(image[y * stride:y * stride + kernel_height, x * stride:x * stride + kernel_width] * kernel).astype(np.float32)
return new_image
It might not be the most optimized solution either, but it is approximately ten times faster than the one proposed by #omotto and it only uses basic numpy function (as reshape, expand_dims, tile...) and no 'for' loops:
def gen_idx_conv1d(in_size, ker_size):
"""
Generates a list of indices. This indices correspond to the indices
of a 1D input tensor on which we would like to apply a 1D convolution.
For instance, with a 1D input array of size 5 and a kernel of size 3, the
1D convolution product will successively looks at elements of indices [0,1,2],
[1,2,3] and [2,3,4] in the input array. In this case, the function idx_conv1d(5,3)
outputs the following array: array([0,1,2,1,2,3,2,3,4]).
args:
in_size: (type: int) size of the input 1d array.
ker_size: (type: int) kernel size.
return:
idx_list: (type: np.array) list of the successive indices of the 1D input array
access to the 1D convolution algorithm.
example:
>>> gen_idx_conv1d(in_size=5, ker_size=3)
array([0, 1, 2, 1, 2, 3, 2, 3, 4])
"""
f = lambda dim1, dim2, axis: np.reshape(np.tile(np.expand_dims(np.arange(dim1),axis),dim2),-1)
out_size = in_size-ker_size+1
return f(ker_size, out_size, 0)+f(out_size, ker_size, 1)
def repeat_idx_2d(idx_list, nbof_rep, axis):
"""
Repeats an array of indices (idx_list) a number of time (nbof_rep) "along" an axis
(axis). This function helps to browse through a 2d array of size
(len(idx_list),nbof_rep).
args:
idx_list: (type: np.array or list) a 1D array of indices.
nbof_rep: (type: int) number of repetition.
axis: (type: int) axis "along" which the repetition will be applied.
return
idx_list: (type: np.array) a 1D array of indices of size len(idx_list)*nbof_rep.
example:
>>> a = np.array([0, 1, 2])
>>> repeat_idx_2d(a, 3, 0) # repeats array 'a' 3 times along 'axis' 0
array([0, 0, 0, 1, 1, 1, 2, 2, 2])
>>> repeat_idx_2d(a, 3, 1) # repeats array 'a' 3 times along 'axis' 1
array([0, 1, 2, 0, 1, 2, 0, 1, 2])
>>> b = np.reshape(np.arange(3*4), (3,4))
>>> b[repeat_idx_2d(np.arange(3), 4, 0), repeat_idx_2d(np.arange(4), 3, 1)]
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
"""
assert axis in [0,1], "Axis should be equal to 0 or 1."
tile_axis = (nbof_rep,1) if axis else (1,nbof_rep)
return np.reshape(np.tile(np.expand_dims(idx_list, 1),tile_axis),-1)
def conv2d(im, ker):
"""
Performs a 'valid' 2D convolution on an image. The input image may be
a 2D or a 3D array.
The output image first two dimensions will be reduced depending on the
convolution size.
The kernel may be a 2D or 3D array. If 2D, it will be applied on every
channel of the input image. If 3D, its last dimension must match the
image one.
args:
im: (type: np.array) image (2D or 3D).
ker: (type: np.array) convolution kernel (2D or 3D).
returns:
im: (type: np.array) convolved image.
example:
>>> im = np.reshape(np.arange(10*10*3),(10,10,3))/(10*10*3) # 3D image
>>> ker = np.array([[0,1,0],[-1,0,1],[0,-1,0]]) # 2D kernel
>>> conv2d(im, ker) # 3D array of shape (8,8,3)
"""
if len(im.shape)==2: # if the image is a 2D array, it is reshaped by expanding the last dimension
im = np.expand_dims(im,-1)
im_x, im_y, im_w = im.shape
if len(ker.shape)==2: # if the kernel is a 2D array, it is reshaped so it will be applied to all of the image channels
ker = np.tile(np.expand_dims(ker,-1),[1,1,im_w]) # the same kernel will be applied to all of the channels
assert ker.shape[-1]==im.shape[-1], "Kernel and image last dimension must match."
ker_x = ker.shape[0]
ker_y = ker.shape[1]
# shape of the output image
out_x = im_x - ker_x + 1
out_y = im_y - ker_y + 1
# reshapes the image to (out_x, ker_x, out_y, ker_y, im_w)
idx_list_x = gen_idx_conv1d(im_x, ker_x) # computes the indices of a 1D conv (cf. idx_conv1d doc)
idx_list_y = gen_idx_conv1d(im_y, ker_y)
idx_reshaped_x = repeat_idx_2d(idx_list_x, len(idx_list_y), 0) # repeats the previous indices to be used in 2D (cf. repeat_idx_2d doc)
idx_reshaped_y = repeat_idx_2d(idx_list_y, len(idx_list_x), 1)
im_reshaped = np.reshape(im[idx_reshaped_x, idx_reshaped_y, :], [out_x, ker_x, out_y, ker_y, im_w]) # reshapes
# reshapes the 2D kernel
ker = np.reshape(ker,[1, ker_x, 1, ker_y, im_w])
# applies the kernel to the image and reduces the dimension back to the one of original input image
return np.squeeze(np.sum(im_reshaped*ker, axis=(1,3)))
I tried to add a lot of comments to explain the method but the global idea is to reshape the 3D input image to a 5D one of shape (output_image_height, kernel_height, output_image_width, kernel_width, output_image_channel) and then to apply the kernel directly using the basic array multiplication. Of course, this methods is then using more memory (during the execution the size of the image is thus multiply by kernel_height*kernel_width) but it is faster.
To do this reshape step, I 'over-used' the indexing methods of numpy arrays, especially, the possibility of giving a numpy array as indices into a numpy array.
This methods could also be used to re-code the 2D convolution product in Pytorch or Tensorflow using the base math functions but I have no doubt in saying that it will be slower than the existing nn.conv2d operator...
I really enjoyed coding this method by only using the numpy basic tools.
One of the most obvious is to hard code the kernel.
img = img.convert('L')
a = np.array(img)
out = np.zeros([a.shape[0]-2, a.shape[1]-2], dtype='float')
out += a[:-2, :-2]
out += a[1:-1, :-2]
out += a[2:, :-2]
out += a[:-2, 1:-1]
out += a[1:-1,1:-1]
out += a[2:, 1:-1]
out += a[:-2, 2:]
out += a[1:-1, 2:]
out += a[2:, 2:]
out /= 9.0
out = out.astype('uint8')
img = Image.fromarray(out)
This example does a box blur 3x3 completely unrolled. You can multiply the values where you have a different value and divide them by a different amount. But, if you honestly want the quickest and dirtiest method this is it. I think it beats Guillaume Mougeot's method by a factor of like 5. His method beating the others by a factor of 10.
It may lose a few steps if you're doing something like a gaussian blur. and need to multiply some stuff.
Try to first round and then cast to uint8:
data = data.round().astype(np.uint8);
I wrote this convolve_stride which uses numpy.lib.stride_tricks.as_strided. Moreover it supports both strides and dilation. It is also compatible to tensor with order > 2.
import numpy as np
from numpy.lib.stride_tricks import as_strided
from im2col import im2col
def conv_view(X, F_s, dr, std):
X_s = np.array(X.shape)
F_s = np.array(F_s)
dr = np.array(dr)
Fd_s = (F_s - 1) * dr + 1
if np.any(Fd_s > X_s):
raise ValueError('(Dilated) filter size must be smaller than X')
std = np.array(std)
X_ss = np.array(X.strides)
Xn_s = (X_s - Fd_s) // std + 1
Xv_s = np.append(Xn_s, F_s)
Xv_ss = np.tile(X_ss, 2) * np.append(std, dr)
return as_strided(X, Xv_s, Xv_ss, writeable=False)
def convolve_stride(X, F, dr=None, std=None):
if dr is None:
dr = np.ones(X.ndim, dtype=int)
if std is None:
std = np.ones(X.ndim, dtype=int)
if not (X.ndim == F.ndim == len(dr) == len(std)):
raise ValueError('X.ndim, F.ndim, len(dr), len(std) must be the same')
Xv = conv_view(X, F.shape, dr, std)
return np.tensordot(Xv, F, axes=X.ndim)
%timeit -n 100 -r 10 convolve_stride(A, F)
#31.2 ms ± 1.31 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)
Super simple and fast convolution using only basic numpy:
import numpy as np
def conv2d(image, kernel):
# apply kernel to image, return image of the same shape
# assume both image and kernel are 2D arrays
# kernel = np.flipud(np.fliplr(kernel)) # optionally flip the kernel
k = kernel.shape[0]
width = k//2
# place the image inside a frame to compensate for the kernel overlap
a = framed(image, width)
b = np.zeros(image.shape) # fill the output array with zeros; do not use np.empty()
# shift the image around each pixel, multiply by the corresponding kernel value and accumulate the results
for p, dp, r, dr in [(i, i + image.shape[0], j, j + image.shape[1]) for i in range(k) for j in range(k)]:
b += a[p:dp, r:dr] * kernel[p, r]
# or just write two nested for loops if you prefer
# np.clip(b, 0, 255, out=b) # optionally clip values exceeding the limits
return b
def framed(image, width):
a = np.zeros((image.shape[0]+2*width, image.shape[1]+2*width))
a[width:-width, width:-width] = image
# alternatively fill the frame with ones or copy border pixels
return a
Run it:
Image.fromarray(conv2d(image, kernel).astype('uint8'))
Instead of sliding the kernel along the image and computing the transformation pixel by pixel, create a series of shifted versions of the image corresponding to each element in the kernel and apply the corresponding kernel value to each of the shifted image versions.
This is probably the fastest you can get using just basic numpy; the speed is already comparable to C implementation of scipy convolve2d and better than fftconvolve. The idea is similar to #Tatarize. This example works only for one color component; for RGB just repeat for each (or modify the algorithm accordingly).
Typically, Convolution 2D is a misnomer. Ideally, under the hood,
whats being done is a correlation of 2 matrices.
pad == same
returns the output as the same as input dimension
It can also take asymmetric images. In order to perform correlation(convolution in deep learning lingo) on a batch of 2d matrices, one can iterate over all the channels, calculate the correlation for each of the channel slices with the respective filter slice.
For example: If image is (28,28,3) and filter size is (5,5,3) then take each of the 3 slices from the image channel and perform the cross correlation using the custom function above and stack the resulting matrix in the respective dimension of the output.
def get_cross_corr_2d(W, X, pad = 'valid'):
if(pad == 'same'):
pr = int((W.shape[0] - 1)/2)
pc = int((W.shape[1] - 1)/2)
conv_2d = np.zeros((X.shape[0], X.shape[1]))
X_pad = np.zeros((X.shape[0] + 2*pr, X.shape[1] + 2*pc))
X_pad[pr:pr+X.shape[0], pc:pc+X.shape[1]] = X
for r in range(conv_2d.shape[0]):
for c in range(conv_2d.shape[1]):
conv_2d[r,c] = np.sum(np.inner(W, X_pad[r:r+W.shape[0], c:c+W.shape[1]]))
return conv_2d
else:
pr = W.shape[0] - 1
pc = W.shape[1] - 1
conv_2d = np.zeros((X.shape[0] - W.shape[0] + 2*pr + 1,
X.shape[1] - W.shape[1] + 2*pc + 1))
X_pad = np.zeros((X.shape[0] + 2*pr, X.shape[1] + 2*pc))
X_pad[pr:pr+X.shape[0], pc:pc+X.shape[1]] = X
for r in range(conv_2d.shape[0]):
for c in range(conv_2d.shape[1]):
conv_2d[r,c] = np.sum(np.multiply(W, X_pad[r:r+W.shape[0], c:c+W.shape[1]]))
return conv_2d
This code incorrect:
for r in range(nr):
data[r,:] = np.convolve(data[r,:], H_r, 'same')
for c in range(nc):
data[:,c] = np.convolve(data[:,c], H_c, 'same')
See Nussbaumer transformation from multidimentional convolution to one dimentional.