# 2D and 3D convolutions using numpy

This post will share some knowledge of 2D and 3D convolutions in a convolution neural network (CNN), and 3 implementations all done using pure numpy and scipy.
Contents

# The goal

This post will share some knowledge of 2D and 3D convolutions in a convolution neural network (CNN). Depending on the implementation, the computational efficiency of a 2D/3D convolution can differ by a great amount. We will be covering 3 different implementations, all done using pure numpy and scipy, and comparing their speeds. Some of the results will be used as building blocks for a numpy + scipy implementation of a convolution neural network, introduced in a later post.

# Recap on convolution

If you need a recap on what 2D convolution is, here is another post where I covered some aspects of 2D convolution, the numpy and scipy implementations, and a Fortran implementation that deals with missing values.

A few points that are worth reminding:

1. First and foremost, there are two similar and related operations in mathematics: convolution and cross-correlation. The difference between them is that in convolution, one flips the kernel (e.g. kernel = kernel[::-1, ::-1]), and this flipping is not done in cross-correlation. However, in the context of machine learning and convolution neural networks, people typically use the term convolution to refer to the operation that is mathematically a cross-correlation, i.e. without the kernel flipping. Therefore, without specific notices, we will be following this convention and using the term convolution to refer the operation that does NOT involve kernel flipping.

2. We will be using the term kernel and filter interchangeably.

3. In convolution neural networks, convolution is typically done in 3D, with the extra 3rd dimension being depth. More will be covered in a later post centered around CNN. For a concrete example, suppose the input image has dimensions of (100, 100, 3), where 3 corresponds to the RGB channels. The kernel/filter also needs to have a depth of 3, e.g. (5, 5, 3). Then the convolution is done in 3D: each time a dot product is computed between a data cube of (5, 5, 3) from the image, and the cube of the kernel itself. The kernel still strides across the image column by column and row by row, but each time a volume (rather than a slab) of data is involved in the dot product computation.

4. Following the above point, we assume the image data all have a shape of (h, w) or (h, w, c), where h and w being height and width of the image, and c is the depth dimension.

5. convolution can be transformed into a Fourier transform/inverse transform problem using the convolution theorem. By doing so we can leverage the powerful FFT module to speed up the computation.

6. In this context, we assume no missing data are present, so no need to worry about nan values.

# Padding, stride and convolution sizes

In a 2D convolution, the size of the output is $$h_o \times w_o$$, where $$h_o$$ is the output height, and $$w_o$$ the width. All measured in number of pixels.

The output size is determined as:

$\left\{\begin{matrix} h_{o} & = [(h_i + 2 p – f)/s] + 1 \\ w_{o} & = [(w_i + 2 p – f)/s] + 1 \\ \end{matrix}\right.$

(1)

where:

• $$h_i$$ and $$w_i$$ are the input sizes. For a convolution network, most of the time the images have equal heights and widths, so $$h = w$$.
• $$p$$ is the padding size, i.e. number of empty pixels padded onto each side of the image.
• $$f$$ is the size of the convolution kernel. Again, most of the time square kernels are used, therefore the kernel width = kernel height.
• $$s$$ is the stride, in both the row and column direction. A stride means the kernel makes a jump of $$s$$ pixels as it travels across the columns/rows of the input image. When $$s=1$$, it is a standard convolution. If $$s>1$$, we are skipping some columns/rows.
• $$[\,]$$ is the floor function.

As a concrete example, suppose:

• input image has shape $$h_i = w_i = 28$$.
• kernel size is $$f = 3$$.
• padding is $$p = 0$$.
• stride is $$s = 1$$.

Then the output size is:

$$h_o = w_o = [(28 + 0 – 3)/1] + 1 = 26$$.

Another example:

• input image has shape $$h_i = w_i = 56$$.
• kernel size is $$f = 5$$.
• padding is $$p = 1$$.
• stride is $$s = 2$$.

Then the output size is:

$$h_o = w_o = [(56 + 2 – 5)/2] + 1 = 23$$.

# Convolution modes

In numerical implementations, we often see that convolution has different "modes". E.g. the numpy.convolve() function has the following doc-string:

numpy.convolve(a, v, mode='full')
...

Parameters

a(N,) array_like

First one-dimensional input array.
v(M,) array_like

Second one-dimensional input array.

mode{‘full’, ‘valid’, ‘same’}, optional
‘full’:
By default, mode is ‘full’. This returns the convolution at each point of overlap, with an output shape of (N+M-1,). At the end-points of the convolution, the signals do not overlap completely, and boundary effects may be seen.
‘same’:
Mode ‘same’ returns output of length max(M, N). Boundary effects are still visible.
‘valid’:
Mode ‘valid’ returns output of length max(M, N) - min(M, N) + 1. The convolution product is only given for points where the signals overlap completely. Values outside the signal boundary have no effect.

Relating back to the size computation Equation 1 above (note that (s=1) here):

valid mode (see Figure 1 a,d for an illustration) has $$p = 0$$, and $$h_o = h_i – f + 1$$. Note that the kernel never stays outside of the “valid” data domain of the input image.
same mode (Figure 1 b,e) has $$p = (f – 1)/2$$, and $$h_o = h_i + 2 p – f + 1 = h_i$$, i.e. input and output have the same size. Note that some padding (labeled as p) are required to get the values around the edges of the image.
full mode (Figure 1 c,f) does not have direct correspondence to Equation 1. However, note that the output in this mode has a size of $$h_o = h_i + f – 1$$, larger than the input. The valid convolution of this output using the same kernel has a size of $$h_i + f – 1 – f + 1 = h_i$$, i.e. we get back to the original size. This last observation is useful when dealing with the error back-propagation computation in a convolution neural network.

Before moving on to the convolution implementation, let us first create a function that does the padding. See code below:

def padArray(var, pad1, pad2=None):

Args:
var (ndarray): 2d or 3d ndarray. Padding is done on the first 2 dimensions.
Keyword Args:
Returns:
var_pad (ndarray): 2d or 3d ndarray with 0s padded along the first 2
dimensions.
'''
return var

return var_pad

Note that the function handles 2D or 3D array padding. For instance, var1 has shape (50, 60), then padArray(var1, 1, 1) gives an output with shape (52, 62).

If var1 has a shape of (50, 60, 3) (3 may corresponds to the RGB channels of an image), then padArray(var1, 1, 1) gives an output with shape (52, 62, 3).

# 1st conv3D() implementation: the FFT approach

Our 1st convolution implementation is based on the convolution theorem and utilizes the powerful FFT module.

The idea of this approach is:

1. do the padding ourselves using the padArray() function above.
2. perform a valid-mode convolution using scipy‘s fftconvolve() function. Internally, fftconvolve() handles the convolution using FFT.
3. To get the correct results when stride is not 1, we only need to pick out the result from fftconvolve() at the same stride.

See code below:

def checkShape(var, kernel):
'''Check shapes for convolution

Args:
var (ndarray): 2d or 3d input array for convolution.
kernel (ndarray): 2d or 3d convolution kernel.
Returns:
kernel (ndarray): 2d kernel reshape into 3d if needed.
'''
var_ndim = np.ndim(var)
kernel_ndim = np.ndim(kernel)

if var_ndim not in [2, 3]:
raise Exception("<var> dimension should be in 2 or 3.")
if kernel_ndim not in [2, 3]:
raise Exception("<kernel> dimension should be in 2 or 3.")
if var_ndim < kernel_ndim:
raise Exception("<kernel> dimension > <var>.")
if var_ndim == 3 and kernel_ndim == 2:
kernel = np.repeat(kernel[:, :, None], var.shape[2], axis=2)

return kernel

def pickStrided(var, stride):
'''Pick sub-array by stride

Args:
var (ndarray): 2d or 3d ndarray.
stride (int): stride/step along the 1st 2 dimensions to pick
elements from <var>.
Returns:
result (ndarray): 2d or 3d ndarray picked at <stride> from <var>.
'''
if stride < 0:
raise Exception("<stride> should be >=1.")
if stride == 1:
result = var
else:
result = var[::stride, ::stride, ...]
return result

'''3D convolution using scipy.signal.fftconvolve.

Args:
var (ndarray): 2d or 3d array to convolve along the first 2 dimensions.
kernel (ndarray): 2d or 3d kernel to convolve. If <var> is 3d and <kernel>
is 2d, create a dummy dimension to be the 3rd dimension in kernel.
Keyword Args:
stride (int): stride along the 1st 2 dimensions. Default to 1.
Returns:
conv (ndarray): convolution result.
'''
stride = int(stride)
kernel = checkShape(var, kernel)
else:

if stride > 1:
conv = pickStrided(conv, stride)

return conv

Here, checkShape() is doing some input sanity checks, and pickStrided() picks out the desired output at given strides.

# 2nd conv3D() implementation: the sub-matrix approach

This approach is inspired by a Stackoverflow answer. The name "sub-matrix" is informal (the author of the above SO answer called it specialconvolve), and is chosen here due to its sub-matrix slicing operations.

As a concrete example, suppose the input matrix is (10, 10), and the convolution kernel is (3, 3). It is easy to work out that the output will have a shape of (8, 8). See the Figure below for an illustration of the process.

To get this output, we make 3 * 3 = 9 sub-matrix slices from the original input, each has a size of (8, 8), and multiple that slab with one value from the kernel. More formally:

1. set i = 0 and j = 0, the value from the kernel at index (i, j) is denoted kernel[i, j] (labeled x in Figure 2a).
2. take a slice from the input image as image[i : i+8, j : j+8], and multiple it with kernel[i, j] to give slabij = image[i : i+8, j : j+8] * kernel[i, j] (red slab on the right in Figure 2a).
3. add this slabij to an accumulator variable: result = result + slabij.
4. increment j by 1 and repeat steps 2-3, do this till j == 2. (First row in Figure 2).
5. increment i by 1 and repeat steps 2-3-4, do this till i == 2.

Notice that during this process, the 9 cells in the top left corner are contributing to the final convolution output, labeled yellow on the right slab. Th number of dot product computations is 3 * 3 = 9, instead of 8 * 8 = 64. Each time, a bigger (8, 8) slab is involved in the dot product, rather than a smaller one as big as the kernel ((3, 3)). This better utilizes the vectorized matrix operation of numpy and saves a considerable amount of time. The author of the SO answer claimed a 1000 times speed up compared with scipy.signal.convolve2d(), and I also noticed that it works particularly well when the kernel size is very small. More on speed comparisons will be given in a later section.

The code is given below:

def conv3D2(var, kernel, stride=1, pad=0):
'''3D convolution by sub-matrix summing.

Args:
var (ndarray): 2d or 3d array to convolve along the first 2 dimensions.
kernel (ndarray): 2d or 3d kernel to convolve. If <var> is 3d and <kernel>
is 2d, create a dummy dimension to be the 3rd dimension in kernel.
Keyword Args:
stride (int): stride along the 1st 2 dimensions. Default to 1.
Returns:
result (ndarray): convolution result.
'''
var_ndim = np.ndim(var)
ny, nx = var.shape[:2]
ky, kx = kernel.shape[:2]
result = 0
else:

for ii in range(ky*kx):
yi, xi = divmod(ii, kx)
if var_ndim == 3:
slabii = slabii.sum(axis=-1)
result += slabii

if stride > 1:
result = pickStrided(result, stride)

return result

Notice that again, we do the padding ourselves using the padArray() function, and pick out the strided values using pickStrided(). Also, it works for either 2D or 3D data.

# 3rd conv3D() implementation: the stride-view approach

The 3rd approach uses a fairly hidden function in numpynumpy.lib.stride_tricks.as_strided() — to achieve a vectorized computation of all the dot product operations in a 2D or 3D convolution.

Recall that in a 2D convolution, we slide the kernel across the input image, and at each location, compute a dot product and save the output.

For instance, the input image has a shape of (10, 10), and the kernel is (3, 3). Then at each row, there is 10 - 3 + 1 = 8 locations where the kernel overlaps with the image, giving 8 dot product computations. After iterating through all rows, the output has a shape of (8, 8).

Now imagine for each of these 8 * 8 = 64 locations, we obtain a 3 * 3 slab from the input image, and stack them all up in a big pile and name it view. See Figure 3 below for an illustration of the first 6 slabs in the pile.

This pile will have a shape of (64, 3, 3). You may already know that in numpy, when multiplying one variable of shape (64, 3, 3) and one with shape (3, 3), a broadcasting of the latter variable automatically happens such that it is like multiplying (element-wise) two (64, 3, 3) variables together.

To express this process using some Python code, suppose:

• view is an ndarray with shape (64, 3, 3).
• kernel is an ndarray with shape (3, 3).

Then the result of

prod = view * kernel


is the same as these:

kernel_broadcasted = np.repeat(kernel[None, :, :], 64, axis=0)  # kernel_broadcasted is (64, 3, 3)
prod = view * kernel_broadcasted

Then it is easy to get the convolution result by summing up the last 2 dimensions:

conv = np.sum(prod, axis=(1, 2))  #  conv shape is (64,)


Lastly, we only need to reshape this conv into a shape of (8, 8):

conv = conv.reshape((8, 8))


It is noticed that the core of this trick lies in the construction of the that big "pile" of matrix slices, and this is achieved using the above mentioned numpy.lib.stride_tricks.as_strided() function. I will leave it to the reader to figure out how this trick works. Below is the code of this 3rd convolution implementation:

def asStride(arr, sub_shape, stride):
'''Get a strided sub-matrices view of an ndarray.

Args:
arr (ndarray): input array of rank 2 or 3, with shape (m1, n1) or (m1, n1, c).
sub_shape (tuple): window size: (m2, n2).
stride (int): stride of windows in both y- and x- dimensions.
Returns:
subs (view): strided window view.

'''
s0, s1 = arr.strides[:2]
m1, n1 = arr.shape[:2]
m2, n2 = sub_shape[:2]

view_shape = (1+(m1-m2)//stride, 1+(n1-n2)//stride, m2, n2)+arr.shape[2:]
strides = (stride*s0, stride*s1, s0, s1)+arr.strides[2:]
subs = np.lib.stride_tricks.as_strided(
arr, view_shape, strides=strides, writeable=False)

return subs

'''3D convolution by strided view.

Args:
var (ndarray): 2d or 3d array to convolve along the first 2 dimensions.
kernel (ndarray): 2d or 3d kernel to convolve. If <var> is 3d and <kernel>
is 2d, create a dummy dimension to be the 3rd dimension in kernel.
Keyword Args:
stride (int): stride along the 1st 2 dimensions. Default to 1.
Returns:
conv (ndarray): convolution result.
'''
kernel = checkShape(var, kernel)
else:

if np.ndim(kernel) == 2:
conv = np.sum(view*kernel, axis=(2, 3))
else:
conv = np.sum(view*kernel, axis=(2, 3, 4))

return conv

Here, the asStride() function is responsible for creating that big "pile". Using the above example, the return value of subs will have a shape of (8, 8, 3, 3).

Then in the conv3D3() function, the dot products in the convolution is computed in

conv = np.sum(view*kernel, axis=(2, 3))


giving a result conv with shape (8, 8).

Notice that when using the strided-view trick, the last reshaping from (64,) to (8, 8) array is not even necessary, because we can construct the view directly into a shape of (8, 8, 3, 3), and the dot product operation happens at the last 2 dimensions.

Also note that it also works when the stride is not 1, and when the input and kernel are 3D.

When the stride is smaller than the kernel size (which is typically the case), it is easy to notice that there are many duplicated values in the "pile". Or even more precisely, there are many mirrored locations in the strided-view, and each group of mirrored locations share the same value in memory. The number of locations in the pile will be the same as the input image only when the stride equals the kernel size (i.e. non-overlapping convolution).

Lastly, this strided-view trick can also be used during the pooling process. More on pooling will be covered in a separate post.

One additional word of warning: according to the numpy documentation, it is regarded as dangerous to write into the as_strided() output:

Furthermore, arrays created with this function often contain self
overlapping memory, so that two elements are identical. Vectorized
write operations on such arrays will typically be unpredictable. They
may even give different results for small, large, or transposed
arrays. Since writing to these arrays has to be tested and done with
great care, you may want to use writeable=False to avoid accidental
write operations.

For these reasons it is advisable to avoid as_strided when possible.


# Comparing different implementations

We have introduced 3 different implementations of 2D or 3D convolution using numpy or scipy:

1. conv3D(), using scipy.signal.fftconvolve().
2. conv3D2(), using sub-matrix slices.
3. conv3D3(), using strided-view of an ndarray.

I did some comparisons on the speeds of these 3 methods, using inputs and kernels of different sizes. The results are given in this SO answer. Here is an image showing the speed comparison results.

To summarize:

• conv3D() is in general the fastest.
• conv3D2() and conv3D3() get slow as kernel size increases, but faster again as the kernel size approaches the size of input data.
• conv3D2() has advantages for very small kernels (smaller than about 5 x 5).
• conv3D3() is the fastest when kernel size approaches input.

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