You can achieve this by creating a third array of dtype uint8, plus a bool array (which together are more memory efficient that one uint16 array). np.putmask is useful for avoiding a temp array. a = np.array([100, 200, 250], dtype=np.uint8) b = np.array([50, 50, 50], dtype=np.uint8) c = 255 – b # a temp uint8 … Read more
In [27]: x = np.arange(16).reshape((4,2,2)) In [28]: x.reshape(2,2,2,2).swapaxes(1,2).reshape(4,-1) Out[28]: array([[ 0, 1, 4, 5], [ 2, 3, 6, 7], [ 8, 9, 12, 13], [10, 11, 14, 15]]) I’ve posted more general functions for reshaping/unshaping arrays into blocks, here.
Clearly, e.shape == (3, 2, 2) Sum over an axis is a reduction operation so the specified axis disappears. Hence, e.sum(axis=0).shape == (2, 2) e.sum(axis=1).shape == (3, 2) e.sum(axis=2).shape == (3, 2) Intuitively, we are “squashing” the array along the chosen axis, and summing the numbers that get squashed together.
numpy.pad with constant mode does what you need, where we can pass a tuple as second argument to tell how many zeros to pad on each size, a (2, 3) for instance will pad 2 zeros on the left side and 3 zeros on the right side: Given A as: A = np.array([1,2,3,4,5]) np.pad(A, (2, … Read more
Correct functions: import numpy as np from scipy import fftpack as scipy_fftpack from scipy import fft as scipy # FFTPACK RFFT 2D def fftpack_rfft2d(matrix): fftRows = scipy_fftpack.fft(matrix, axis=1) fftCols = scipy_fftpack.fft(fftRows, axis=0) return fftCols # FFTPACK IRFFT 2D def fftpack_irfft2d(matrix): ifftRows = scipy_fftpack.ifft(matrix, axis=1) ifftCols = scipy_fftpack.ifft(ifftRows, axis=0) return ifftCols.real You calculated the 2D FFT … Read more
Try numpy.array_split. From the documentation: >>> x = np.arange(8.0) >>> np.array_split(x, 3) [array([ 0., 1., 2.]), array([ 3., 4., 5.]), array([ 6., 7.])] Identical to numpy.split, but won’t raise an exception if the groups aren’t equal length. If number of chunks > len(array) you get blank arrays nested inside, to address that – if your … Read more
Your can use NumPy’s array indexing: def unison_shuffled_copies(a, b): assert len(a) == len(b) p = numpy.random.permutation(len(a)) return a[p], b[p] This will result in creation of separate unison-shuffled arrays.
The purpose of meshgrid is to create a rectangular grid out of an array of x values and an array of y values. So, for example, if we want to create a grid where we have a point at each integer value between 0 and 4 in both the x and y directions. To create … Read more
You might be interested in the SciPy Stats package. It has the percentile function you’re after and many other statistical goodies. percentile() is available in numpy too. import numpy as np a = np.array([1,2,3,4,5]) p = np.percentile(a, 50) # return 50th percentile, e.g median. print p 3.0 This ticket leads me to believe they won’t … Read more
As @hpaulj mentioned in the comments, this behaviour is because of mixing basic slicing and advanced indexing: a = np.arange(120).reshape(4,5,3,2) b = a[:,[1,2,3,4,0],:,0] In the above code snippet, what happens is the following: when we do basic slicing along last dimension, it triggers a __getitem__ call. So, that dimension is gone. (i.e. no singleton dimension) … Read more