Consider the speed of processing to shift the image buffer with numpy.ndarray

background

When processing a video, you may want to buffer n consecutive frames. So, when you want to roll the buffer and use it, you have to shift the index of the buffer.

When processing with numpy, I think there are the following two processing methods.

  1. Use the numpy.roll method
  2. Shift the index of array and assign

I want to take the fastest processing method possible, so I will consider which of these two processes is faster.

environment

inspection result

I tried the following process on jupyter notebook.

Make the first index of array the index indicating the number of images

arr1 = np.zeros((30, 720, 1280))
# numpy.use the roll method
%timeit np.roll(arr1, 1, axis = 0)
# 147 ms ± 1.73 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

#Substitute by shifting by index
%timeit arr1[1:,:,:] = arr1[0:-1,:,:]
# 189 ms ± 7.14 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Set the last index of array as the index indicating the number of images

arr2 = np.zeros((720, 1280, 30))
# numpy.use the roll method
%timeit np.roll(arr2, 1, axis = 2)
# 186 ms ± 2.71 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

#Substitute by shifting by index
%timeit arr2[:,:, 1:] = arr2[:,:,0:-1]
# 203 ms ± 1.06 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

It seems that using numpy.roll is faster under the condition that the image size is large. Also, it seems faster to switch images at the first index.

If the image is small

When the image is very small (20x20pixel), the result is that the index shift is faster as shown below.

arr3 = np.zeros((30, 20, 20))
# numpy.use the roll method
%timeit np.roll(arr3, 1, axis = 0)
# 17.1 µs ± 551 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#Substitute by shifting by index
%timeit arr3[1:,:,:] = arr3[0:-1,:,:]
# 7.36 µs ± 90.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Points I was addicted to in terms of processing implementation

#mistake
arr[0:-2,:,:] = arr[1:-1,:,:]
arr[-1,:,:] = new_image

#Correct answer
arr[0:-1,:,:] = arr[1:,:,:]
arr[-1,:,:] = new_image

Setting the first index to -1 specifies the last index.

arr[1:-1,:,:]Then


 At first glance, it seems that you can specify from the second to the first (up to the last index).
 Actually, since the second to the penultimate is specified, the movement is that the last image does not shift to the front.



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