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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'd like to randomly rotate an image tensor (B, C, H, W) around it's center (2d rotation I think?). I would like to avoid using NumPy and Kornia, so that I basically only need to import from the torch module. I'm also not using torchvision.transforms, because I need it to be autograd compatible. Essentially I'm trying to create an autograd compatible version of torchvision.transforms.RandomRotation() for visualization techniques like DeepDream (so I need to avoid artifacts as much as possible).
import torch
import math
import random
import torchvision.transforms as transforms
from PIL import Image
# Load image
def preprocess_simple(image_name, image_size):
Loader = transforms.Compose([transforms.Resize(image_size), transforms.ToTensor()])
image = Image.open(image_name).convert('RGB')
return Loader(image).unsqueeze(0)
# Save image
def deprocess_simple(output_tensor, output_name):
output_tensor.clamp_(0, 1)
Image2PIL = transforms.ToPILImage()
image = Image2PIL(output_tensor.squeeze(0))
image.save(output_name)
# Somehow rotate tensor around it's center
def rotate_tensor(tensor, radians):
...
return rotated_tensor
# Get a random angle within a specified range
r_degrees = 5
angle_range = list(range(-r_degrees, r_degrees))
n = random.randint(angle_range[0], angle_range[len(angle_range)-1])
# Convert angle from degrees to radians
ang_rad = angle * math.pi / 180
# test_tensor = preprocess_simple('path/to/file', (512,512))
test_tensor = torch.randn(1,3,512,512)
# Rotate input tensor somehow
output_tensor = rotate_tensor(test_tensor, ang_rad)
# Optionally use this to check rotated image
# deprocess_simple(output_tensor, 'rotated_image.jpg')
Some example outputs of what I'm trying to accomplish:
So the grid generator and the sampler are sub-modules of the Spatial Transformer (JADERBERG, Max, et al.). These sub-modules are not trainable, they let you apply a learnable, as well as non-learnable, spatial transformation.
Here I take these two submodules and use them to rotate an image by theta using PyTorch's functions torch.nn.functional.affine_grid and torch.nn.functional.affine_sample (these functions are implementations of the generator and the sampler, respectively):
import torch
import torch.nn.functional as F
import numpy as np
import matplotlib.pyplot as plt
def get_rot_mat(theta):
theta = torch.tensor(theta)
return torch.tensor([[torch.cos(theta), -torch.sin(theta), 0],
[torch.sin(theta), torch.cos(theta), 0]])
def rot_img(x, theta, dtype):
rot_mat = get_rot_mat(theta)[None, ...].type(dtype).repeat(x.shape[0],1,1)
grid = F.affine_grid(rot_mat, x.size()).type(dtype)
x = F.grid_sample(x, grid)
return x
#Test:
dtype = torch.cuda.FloatTensor if torch.cuda.is_available() else torch.FloatTensor
#im should be a 4D tensor of shape B x C x H x W with type dtype, range [0,255]:
plt.imshow(im.squeeze(0).permute(1,2,0)/255) #To plot it im should be 1 x C x H x W
plt.figure()
#Rotation by np.pi/2 with autograd support:
rotated_im = rot_img(im, np.pi/2, dtype) # Rotate image by 90 degrees.
plt.imshow(rotated_im.squeeze(0).permute(1,2,0)/255)
In the example above, assume we take our image, im, to be a dancing cat in a skirt:
rotated_im will be a 90-degrees CCW rotated dancing cat in a skirt:
And this is what we get if we call rot_img with theta eqauls to np.pi/4:
And the best part that it's differentiable w.r.t the input and has autograd support! Hooray!
With torchvision it should be simple:
import torchvision.transforms.functional as TF
angle = 30
x = torch.randn(1,3,512,512)
out = TF.rotate(x, angle)
For example if x is:
out with a 30 degree rotation is (NOTE: counterclockwise):
There is a pytorch function for that:
x = torch.tensor([[0, 1],
[2, 3]])
x = torch.rot90(x, 1, [0, 1])
>> tensor([[1, 3],
[0, 2]])
Here are the docs: https://pytorch.org/docs/stable/generated/torch.rot90.html
I'm trying to convert an image from RGB to LMS -and vice versa- using OpenCV in Python. From what I understand, I am supposed to use a given 3x3 transformation matrix and multiply it to a 3x1 RGB/LMS matrix. The transformation matrices used can be found here.
I've explored previously asked questions on this site but unfortunately they're in C++, a language I have yet to be proficient in and I have difficulty in understanding how exactly they've solved their problems.
Here is my code so far: [Solved as of 2019-05-19]
import numpy as np
import cv2
#Transformation Matrix#
MsRGB = np.zeros((3,3), dtype='float')
MHPE = np.zeros((3,3), dtype='float')
MsRGB = np.array([[0.4124564, 0.3575761, 0.1804375],
[0.2126729, 0.7151522, 0.0721750],
[0.0193339, 0.1191920, 0.9503041]])
MHPE = np.array([[ 0.4002, 0.7076, -0.0808],
[-0.2263, 1.1653, 0.0457],
[ 0, 0, 0.9182]])
Trgb2lms = MHPE # MsRGB
Tlms2rgb = np.linalg.inv(Trgb2lms)
imgpath = "(insert file directory here)"
imgIN = cv2.imread(imgpath,cv2.IMREAD_UNCHANGED)
imgINrgb = cv2.cvtColor(imgIN, cv2.COLOR_BGR2RGB)
x,y,z = imgINrgb.shape
imgLMS = np.zeros((x,y,z), dtype='float')
imgReshaped = imgINrgb.transpose(2, 0, 1).reshape(3,-1)
imgLMS = Trgb2lms # imgReshaped #Convert to LMS
imgOUT = Tlms2rgb # imgLMS #Convert back to RGB
imgLMS = imgLMS.reshape(z, x, y).transpose(1, 2, 0).astype(np.uint8)
imgOUT = imgOUT.reshape(z, x, y).transpose(1, 2, 0).astype(np.uint8)
imgOUT = cv2.cvtColor(imgOUT, cv2.COLOR_RGB2BGR)
cv2.imshow('Input', imgIN)
cv2.imshow('LMS', imgLMS)
cv2.imshow('Output', imgOUT)
cv2.waitKey(0)
cv2.destroyAllWindows()
The code is now able to perform linear transformation on a given RGB image using a given transformation matrix. Results can be found here.
There are a few errors given the context of your question:
T is not defined. Judging from the context of your code, this should be Trgb2lms instead so we need to change those.
From what I can gather from the question, you are applying a linear transformation to all pixels in the image. To do this, you want to reshape the matrix so that we have three rows where each row corresponds to a single pixel followed by an unravelling of all pixels along the columns. In that case, the reshape method is incorrect. You need not only shuffle the dimensions so that the last dimension is first but you'll also need to set the last dimension of the reshape so that it's -1. This means that we will automatically fill up the columns so that it contains the total number of pixels in the image.
Finally, once you do the linear transformation, you need to reshape the matrix back to the original image size. You can use a final reshape call and use x, y and z from the original call you made to infer the image dimensions. Remember that when we reshape, the channels come first so we'll have to permute the dimensions again. You'll also want to go back to unsigned 8-bit precision after we do the transformation.
Also to compare, let's run this through the inverse transformation to make sure we have the original.
Therefore:
import numpy as np
import cv2
#Transformation Matrix#
MsRGB = np.zeros((3,3), dtype='float')
MHPE = np.zeros((3,3), dtype='float')
MsRGB = np.array([[0.4124564, 0.3575761, 0.1804375],
[0.2126729, 0.7151522, 0.0721750],
[0.0193339, 0.1191920, 0.9503041]])
MHPE = np.array([[ 0.4002, 0.7076, -0.0808],
[-0.2263, 1.1653, 0.0457],
[ 0, 0, 0.9182]])
Trgb2lms = MHPE # MsRGB
# Change
Tlms2rgb = np.linalg.inv(Trgb2lms)
imgpath = "(insert filename here)"
imgIN = cv2.imread(imgpath,cv2.IMREAD_UNCHANGED)
imgINrgb = cv2.cvtColor(imgIN, cv2.COLOR_BGR2RGB)
x,y,z = imgINrgb.shape
imgLMS = np.zeros((x,y,z), dtype='float')
#imgFlatten = imgINrgb.flatten()
# Change
imgReshaped = imgINrgb.transpose(2, 0, 1).reshape(3,-1)
# Change
imgLMS = Trgb2lms # imgReshaped
imgOUT = Tlms2rgb # imgLMS
# New
imgLMS = imgLMS.transpose(z, x, y).permute(1, 2, 0).astype(np.uint8)
imgOUT = imgOUT.transpose(z, x, y).permute(1, 2, 0).astype(np.uint8)
I implemented the code given by Cris Luengo for convolution in frequency in domain, however I'm not getting the intended gradient image in x direction.
Image without flipping the kernel in x and y direction:
Image after flipping the kernel:
If you notice, the second image is same as given by ImageKernel filter from the pillow library. Also, one thing to notice is I don't have to flip the kernel if I apply Sobel kernel in y direction, I get the exactly intended image.
This is my code:
import numpy as np
from scipy import misc
from scipy import fftpack
import matplotlib.pyplot as plt
from PIL import Image,ImageDraw,ImageOps,ImageFilter
from pylab import figure, title, imshow, hist, grid,show
im1=Image.open("astronaut.png").convert('L')
# im1=ImageOps.grayscale(im1)
img=np.array(im1)
# kernel = np.ones((3,3)) / 9
# kernel=np.array([[0,-1,0],[-1,4,-1],[0,-1,0]])
kernel=np.array([[-1,0,1],[-2,0,2],[-1,0,1]])
kernel=np.rot90(kernel,2)
print(kernel)
sz = (img.shape[0] - kernel.shape[0], img.shape[1] - kernel.shape[1]) # total
amount of padding
kernel = np.pad(kernel, (((sz[0]+1)//2, sz[0]//2), ((sz[1]+1)//2, sz[1]//2)),
'constant')
kernel = fftpack.ifftshift(kernel)
filtered = np.real(fftpack.ifft2(fftpack.fft2(img) *
fftpack.fft2(kernel)))+np.imag(fftpack.ifft2(fftpack.fft2(img) *
fftpack.fft2(kernel)))
filtered=np.maximum(0,np.minimum(filtered,255))
im2=Image.open("astronaut.png").convert('L')
u=im2.filter(ImageFilter.Kernel((3,3), [-1,0,1,-2,0,2,-1,0,1],
scale=1, offset=0))
fig2=figure()
ax1 = fig2.add_subplot(221)
ax2 = fig2.add_subplot(222)
ax3 = fig2.add_subplot(223)
ax1.title.set_text('Original Image')
ax2.title.set_text('After convolving in freq domain')
ax3.title.set_text('imagefilter conv')
ax1.imshow(img,cmap='gray')
ax2.imshow(filtered,cmap='gray')
ax3.imshow(np.array(u),cmap='gray')
show()
We can use np.fft module's FFT implementation too and here is how we can obtain convolution with sobel horizontal kernel in frequency domain (by the convolution theorem):
h, w = im.shape
kernel = np.array(array([[-1, 0, 1],
[-2, 0, 2],
[-1, 0, 1]])) #sobel_filter_x
k = len(kernel) // 2 # assuming odd-length square kernel, here it's 3x3
kernel_padded = np.pad(kernel, [(h//2-k-1, h//2-k), (w//2-k-1, w//2-k)])
im_freq = np.fft.fft2(im) # input image frequency
kernel_freq = np.fft.fft2(kernel_padded) # kernel frequency
out_freq = im_freq * kernel_freq # frequency domain convolution output
out = np.fft.ifftshift(np.fft.ifft2(out_freq)).real # spatial domain output
The below figure shows the input, kernel and output images in spatial and frequency domain:
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.