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caculating the sum of elements around an element in a numpy array [duplicate]

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.

How to vectorize a code with python numpy.bincount, using apply along axis

I'm trying to vectorize a code with numpy, to run it using multiprocessing, but i can't understand how numpy.apply_along_axis works. This is an example of the code, vectorized using map
import numpy
from scipy import sparse
import multiprocessing
from matplotlib import pyplot
#first i build a matrix of some x positions vs time datas in a sparse format
matrix = numpy.random.randint(2, size = 100).astype(float).reshape(10,10)
x = numpy.nonzero(matrix)[0]
times = numpy.nonzero(matrix)[1]
weights = numpy.random.rand(x.size)
#then i define an array of y positions
nStepsY = 5
y = numpy.arange(1,nStepsY+1)
#now i build an image using x-y-times coordinates and x-times weights
def mapIt(ithStep):
ncolumns = 80
image = numpy.zeros(ncolumns)
yTimed = y[ithStep]*times
positions = (numpy.round(x-yTimed)+50).astype(int)
values = numpy.bincount(positions,weights)
values = values[numpy.nonzero(values)]
positions = numpy.unique(positions)
image[positions] = values
return image
image = list(map(mapIt, range(nStepsY)))
image = numpy.array(image)
a = pyplot.imshow(image, aspect = 10)
Here the output plot
I tried to use numpy.apply_along_axis, but this function allows me to iterate only along the rows of image, while i need to iterate along the ithStep index too. E.g.:
#now i build an image using x-y-times coordinates and x-times weights
nrows = nStepsY
ncolumns = 80
matrix = numpy.zeros(nrows*ncolumns).reshape(nrows,ncolumns)
def applyIt(image):
image = numpy.zeros(ncolumns)
yTimed = y[ithStep]*times
positions = (numpy.round(x-yTimed)+50).astype(int)
values = numpy.bincount(positions,weights)
values = values[numpy.nonzero(values)]
positions = numpy.unique(positions)
image[positions] = values
return image
imageApplied = numpy.apply_along_axis(applyIt,1,matrix)
a = pyplot.imshow(imageApplied, aspect = 10)
It obviously return only the firs row nrows times, since nothing iterates ithStep:
And here the wrong plot
There is a way to iterate an index, or to use an index while numpy.apply_along_axis iterates?
Here the code with only matricial operations: it's quite faster than map or apply_along_axis but uses so much memory.
(in this function i use a trick with scipy.sparse, which works more intuitively than numpy arrays when you try to sum numbers on a same element)
def fullmatrix(nRows, nColumns):
y = numpy.arange(1,nStepsY+1)
image = numpy.zeros((nRows, nColumns))
yTimed = numpy.outer(y,times)
x3d = numpy.outer(numpy.ones(nStepsY),x)
weights3d = numpy.outer(numpy.ones(nStepsY),weights)
y3d = numpy.outer(y,numpy.ones(x.size))
positions = (numpy.round(x3d-yTimed)+50).astype(int)
matrix = sparse.coo_matrix((numpy.ravel(weights3d), (numpy.ravel(y3d), numpy.ravel(positions)))).todense()
return matrix
image = fullmatrix(nStepsY, 80)
a = pyplot.imshow(image, aspect = 10)
This way is simplier and very fast! Thank you so much.
nStepsY = 5
nRows = nStepsY
nColumns = 80
y = numpy.arange(1,nStepsY+1)
image = numpy.zeros((nRows, nColumns))
fakeRow = numpy.zeros(positions.size)
def itermatrix(ithStep):
yTimed = y[ithStep]*times
positions = (numpy.round(x-yTimed)+50).astype(int)
matrix = sparse.coo_matrix((weights, (fakeRow, positions))).todense()
matrix = numpy.ravel(matrix)
missColumns = (nColumns-matrix.size)
zeros = numpy.zeros(missColumns)
matrix = numpy.concatenate((matrix, zeros))
return matrix
for i in numpy.arange(nStepsY):
image[i] = itermatrix(i)
#or, without initialization of image:
imageMapped = list(map(itermatrix, range(nStepsY)))
imageMapped = numpy.array(imageMapped)

It feels like attempting to use map or apply_along_axis is obscuring the essentially iteration of the problem.
I rewrote your code as an explicit loop on y:
nStepsY = 5
y = numpy.arange(1,nStepsY+1)
image = numpy.zeros((nStepsY, 80))
for i, yi in enumerate(y):
yTimed = yi*times
positions = (numpy.round(x-yTimed)+50).astype(int)
values = numpy.bincount(positions,weights)
values = values[numpy.nonzero(values)]
positions = numpy.unique(positions)
image[i, positions] = values
a = pyplot.imshow(image, aspect = 10)
pyplot.show()
Looking at the code, I think I could calculate positions for all y values making a (y.shape[0],times.shape[0]) array. But the rest, the bincount and unique still have to work row by row.
apply_along_axis when working with a 2d array, and axis=1 essentially does:
res = np.zeros_like(arr)
for i in range....:
res[i,:] = func1d(arr[i,:])
If the input array has more dimensions it constructs a more elaborate indexing object [i,j,k,:]. And it can handle cases where func1d returns a different size array than the input. But in any case it is just a generalized iteration tool.
Moving the initial positions creation outside the loop:
yTimed = y[:,None]*times
positions = (numpy.round(x-yTimed)+50).astype(int)
image = numpy.zeros((positions.shape[0], 80))
for i, pos in enumerate(positions):
values = numpy.bincount(pos,weights)
values = values[numpy.nonzero(values)]
pos = numpy.unique(pos)
image[i, pos] = values
Now I can cast this as an apply_along_axis problem, with an applyIt that takes a positions vector (with all the yTimed information) rather than blank image vector.
def applyIt(pos, size, weights):
acolumn = numpy.zeros(size)
values = numpy.bincount(pos,weights)
values = values[numpy.nonzero(values)]
pos = numpy.unique(pos)
acolumn[pos] = values
return acolumn
image = numpy.apply_along_axis(applyIt, 1, positions, 80, weights)
Timing wise I expect it's a bit slower than my explicit iteration. It has to do more setup work, including a test call applyIt(positions[0,:],...) to determine the size of its return array (i.e image has different shape than positions.)
def csrmatrix(y, times, x, weights):
yTimed = numpy.outer(y,times)
n=y.shape[0]
x3d = numpy.outer(numpy.ones(n),x)
weights3d = numpy.outer(numpy.ones(n),weights)
y3d = numpy.outer(y,numpy.ones(x.size))
positions = (numpy.round(x3d-yTimed)+50).astype(int)
#print(y.shape, weights3d.shape, y3d.shape, positions.shape)
matrix = sparse.csr_matrix((numpy.ravel(weights3d), (numpy.ravel(y3d), numpy.ravel(positions))))
#print(repr(matrix))
return matrix
# one call
image = csrmatrix(y, times, x, weights)
# iterative call
alist = []
for yi in numpy.arange(1,nStepsY+1):
alist.append(csrmatrix(numpy.array([yi]), times, x, weights))
def mystack(alist):
# concatenate without offset
row, col, data = [],[],[]
for A in alist:
A = A.tocoo()
row.extend(A.row)
col.extend(A.col)
data.extend(A.data)
print(len(row),len(col),len(data))
return sparse.csr_matrix((data, (row, col)))
vimage = mystack(alist)

How to implement Matlab bwmorph(bw,'remove') in Python

I am trying to implement the Matlab function bwmorph(bw,'remove') in Python. This function removes interior pixels by setting a pixel to 0 if all of its 4-connected neighbor pixels are 1. The resulting image should return the boundary pixels. I've written a code but I'm not sure if this is how to do it.
# neighbors() function returns the values of the 4-connected neighbors
# bwmorph() function returns the input image with only the boundary pixels
def neighbors(input_matrix,input_array):
indexRow = input_array[0]
indexCol = input_array[1]
output_array = []
output_array[0] = input_matrix[indexRow - 1,indexCol]
output_array[1] = input_matrix[indexRow,indexCol + 1]
output_array[2] = input_matrix[indexRow + 1,indexCol]
output_array[3] = input_matrix[indexRow,indexCol - 1]
return output_array
def bwmorph(input_matrix):
output_matrix = input_matrix.copy()
nRows,nCols = input_matrix.shape
for indexRow in range(0,nRows):
for indexCol in range(0,nCols):
center_pixel = [indexRow,indexCol]
neighbor_array = neighbors(output_matrix,center_pixel)
if neighbor_array == [1,1,1,1]:
output_matrix[indexRow,indexCol] = 0
return output_matrix

Since you are using NumPy arrays, one suggestion I have is to change the if statement to use numpy.all to check if all values are nonzero for the neighbours. In addition, you should make sure that your input is a single channel image. Because grayscale images in colour share all of the same values in all channels, just extract the first channel. Your comments indicate a colour image so make sure you do this. You are also using the output matrix which is being modified in the loop when checking. You need to use an unmodified version. This is also why you're getting a blank output.
def bwmorph(input_matrix):
output_matrix = input_matrix.copy()
# Change. Ensure single channel
if len(output_matrix.shape) == 3:
output_matrix = output_matrix[:, :, 0]
nRows,nCols = output_matrix.shape # Change
orig = output_matrix.copy() # Need another one for checking
for indexRow in range(0,nRows):
for indexCol in range(0,nCols):
center_pixel = [indexRow,indexCol]
neighbor_array = neighbors(orig, center_pixel) # Change to use unmodified image
if np.all(neighbor_array): # Change
output_matrix[indexRow,indexCol] = 0
return output_matrix
In addition, a small grievance I have with your code is that you don't check for out-of-boundary conditions when determining the four neighbours. The test image you provided does not throw an error as you don't have any border pixels that are white. If you have a pixel along any of the borders, it isn't possible to check all four neighbours. However, one way to mitigate this would be to perhaps wrap around by using the modulo operator:
def neighbors(input_matrix,input_array):
(rows, cols) = input_matrix.shape[:2] # New
indexRow = input_array[0]
indexCol = input_array[1]
output_array = [0] * 4 # New - I like pre-allocating
# Edit
output_array[0] = input_matrix[(indexRow - 1) % rows,indexCol]
output_array[1] = input_matrix[indexRow,(indexCol + 1) % cols]
output_array[2] = input_matrix[(indexRow + 1) % rows,indexCol]
output_array[3] = input_matrix[indexRow,(indexCol - 1) % cols]
return output_array

Numpy compare 2 array shape, if different, append 0 to match shape

I am comparing 2 numpy arrays, and want to add them together. but, before doing so, i need to make sure they are the same size. If the size are not same, then take the smaller sized one and fill the last rows with zero to match the shape.
Both array have 16 columns and N rows. I am assuming it should be pretty straight forward, but I can't get my head around it. So far I am able to compare the 2 array shape.
import csv
import numpy as np
import sys
data = np.genfromtxt('./test1.csv', dtype=float, delimiter=',')
data_sys = np.genfromtxt('./test2.csv', dtype=float, delimiter=',')
print data.shape
print data_sys.shape
if data.shape != data_sys.shape:
print "we have an error"
This is the output I got:
=============New file.csv============
(603, 16)
(604, 16)
we have an error
I want the fill the last row of "data" array with 0 so that I can add the 2 arrays.
Thanks for your help.

You can use vstack(array1, array2) from numpy which stacks arrays vertically. For example:
A = np.random.randint(2, size = (2, 16))
B = np.random.randint(2, size = (5, 16))
print A.shape
print B.shape
if A.shape[0] < B.shape[0]:
A = np.vstack((A, np.zeros((B.shape[0] - A.shape[0], 16))))
elif A.shape[0] > B.shape[0]:
B = np.vstack((B, np.zeros((A.shape[0] - B.shape[0], 16))))
print A.shape
print A
In your case:
if data.shape[0] < data_sys.shape[0]:
data = np.vstack((data, np.zeros((data_sys.shape[0] - data.shape[0], 16))))
elif data.shape[0] > data_sys.shape[0]:
data_sys = np.vstack((data_sys, np.zeros((data.shape[0] - data_sys.shape[0], 16))))
I assume that your matrices have always the same number of columns, if not you can similarly use hstack to stack them horizontally.

If you have only two files, and their shapes differ in just the 0th dimension, a simple check and copy is probably easiest, though it lacks generality:
import numpy as np
data = np.genfromtxt('./test1.csv', dtype=float, delimiter=',')
data_sys = np.genfromtxt('./test2.csv', dtype=float, delimiter=',')
fill_value = 0 # could be np.nan or something else instead
if data.shape[0]>data_sys.shape[0]:
temp = data_sys
data_sys = np.ones(data.shape)*fill_value
data_sys[:temp.shape[0],:] = temp
elif data.shape[0]<data_sys.shape[0]:
temp = data
data = np.ones(data_sys.shape)*fill_value
data[:temp.shape[0],:] = temp
print 'Using conditional:'
print data.shape
print data_sys.shape
if data.shape != data_sys.shape:
print "we have an error"
A much more general solution is a custom class--overkill for your two files but much easier if you have lots of files to handle. The basic idea is that static class variables sx and sy keep track of the largest widths and heights, and are used when get_data is called, to output a standard shape array. This is pre-filled with your desired fill value, and the actual data from the corresponding file are copied into the upper left corner of the standard shape array:
import numpy as np
class IsomorphicArray:
sy = 0 # static class variable
sx = 0 # static class variable
fill_value = 0.0
def __init__(self,csv_filename):
self.data = np.genfromtxt(csv_filename,dtype=float,delimiter=',')
self.instance_sy,self.instance_sx = self.data.shape
if self.instance_sy>IsomorphicArray.sy:
IsomorphicArray.sy = self.instance_sy
if self.instance_sx>IsomorphicArray.sx:
IsomorphicArray.sx = self.instance_sx
def get_data(self):
out = np.ones((IsomorphicArray.sy,IsomorphicArray.sx))*self.fill_value
out[:self.instance_sy,:self.instance_sx] = self.data
return out
isomorphic_array_list = []
for filename in ['./test1.csv','./test2.csv']:
isomorphic_array_list.append(IsomorphicArray(filename))
numpy_array_list = []
for isomorphic_array in isomorphic_array_list:
numpy_array_list.append(isomorphic_array.get_data())
print 'Using custom class:'
for numpy_array in numpy_array_list:
print numpy_array.shape

Assuming both arrays have 16 columns
len1=len(data)
len2=len(data_sys)
if len1<len2:
data=np.append(data, np.zeros((len2-len1, 16)),axis=0)
elif len2<len1:
data_sys=np.append(data_sys, np.zeros((len1-len2, 16)),axis=0)
print data.shape
print data_sys.shape
if data.shape != data_sys.shape:
print "we have an error"
else:
print "we r good"

Numpy provides an append function to add values to an array: see here for details. In multi-dimensional arrays you can define how the values should be added. As you have already the information which of your arrays is the smaller one, just add the desired number of zeroes with creating a zero filled array first by numpy.zeroes and then append it to your target array.
It might be necessary to flatten your array first and then to reshape it.

I had a similar situation. Two arrays of sizes mask_in:(n1,m1) and mask_ot:(n2,m2)that were generated through a mask of a 2D image of size (N,M) where A2 is larger than A1 and both share a common center (X0,Y0). I followed the approach suggested by #AniaG using vstack and hstack. I simply obtained the shapes of both arrays, size difference and finally account the number of missing elements at both ends.
Here is what I got:
mask_in = np.random.randint(2, size = (2, 8))
mask_ot = np.random.randint(2, size = (6, 16))
mask_in_amp = mask_in
dif_row = mask_ot.shape[0]-mask_in_amp.shape[0]
dif_col = mask_ot.shape[1]-mask_in_amp.shape[1]
complete_row = dif_row / 2
complete_col = dif_col / 2
mask_in_amp = np.vstack((mask_in_amp, np.zeros((complete_row, mask_in_amp.shape[1]))))
mask_in_amp = np.vstack((np.zeros((complete_row, mask_in_amp.data.shape[1])), mask_in_amp))
mask_in_amp = np.hstack((mask_in_amp, np.zeros((mask_in_amp.shape[0],complete_col))))
mask_in_amp = np.hstack((np.zeros((mask_in_amp.shape[0],complete_col)), mask_in_amp))

If you don't care about the exact shapes of two arrays you can also do the following:
if data.size == datasys.size:
print ('arrays have the same number of elements, and possibly shape')
else:
print ('arrays do not have the same shape for sure')

2d convolution using python and numpy

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.