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Another way is to use numpy.dstack.Supposing that you want to repeat the matrix a num_repeats times:. import numpy as np b = np.dstack([a]*num_repeats) The trick is to wrap the matrix a into a list of a single element, then using the * operator to duplicate the elements in this list num_repeats times.

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Method 2: built in numpy.where. This is much shorted and probably faster to compute. NumPy has a nice function that returns the indices where your criteria are met in some arrays: condition_1 = (a == 1) condition_2 = (b == 1) Now we can combine the operation by saying "and" - the binary operator version: &.

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Suppose you have a 2d numpy array and you want to remove duplicate rows (or columns). In this blog post, I'll show you a trick you can use to do this more efficiently than using np.unique(A, axis=0) . This algorithm has time complexity O(max(nlogn,nm)).

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Reshaping of arrays: Changing the shape of a given array. Joining and splitting of arrays: Combining multiple arrays into one, and splitting one array into Array Indexing: Accessing Single Elements¶. If you are familiar with Python's standard list indexing, indexing in NumPy will feel quite familiar.I am trying to apply a function to every element of numpy array along with few parameters. Result : [0 0]. Can somebody point out what am I doing wrong!!!! Thanks. Answer 1. Try passing the array as np.float32. import numpy as np def wcdf(U, A, k): return np.array([1 - np.exp(-(u/A)**(k)) if u >0 else 0...

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Suppose you have a 2d numpy array and you want to remove duplicate rows (or columns). In this blog post, I'll show you a trick you can use to do this more efficiently than using np.unique(A, axis=0) . This algorithm has time complexity O(max(nlogn,nm)).

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Let’s say I have a NumPy array, a: a = np.array([ [1, 2, 3], [2, 3, 4] ]) And I would like to add a column of zeros to get an array, b: b = np.array([ [1, 2, 3, 0 ...