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I know I can do np.subtract.outer(x, x). If x has shape (n,), then I end up with an array with shape (n, n). However, I have an x with shape (n, 3). I want to output something with shape (n, n, 3). How do I do this? Maybe np.einsum?
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To count each unique element's number of occurrences in the numpy array, we can use the numpy. unique() function. It takes the array as an input argument and returns all the unique elements inside the array in ascending order.
The easiest way to compare two NumPy arrays is to: Create a comparison array by calling == between two arrays. Call . all() method for the result array object to check if the elements are True.
diff(arr[, n[, axis]]) function is used when we calculate the n-th order discrete difference along the given axis. The first order difference is given by out[i] = arr[i+1] – arr[i] along the given axis. If we have to calculate higher differences, we are using diff recursively.
You can use broadcasting after extending the dimensions with None/np.newaxis to form a 3D array version of x and subtracting the original 2D array version from it, like so -
x[:, np.newaxis, :] - x
Sample run -
In [6]: x
Out[6]:
array([[6, 5, 3],
[4, 3, 5],
[0, 6, 7],
[8, 4, 1]])
In [7]: x[:,None,:] - x
Out[7]:
array([[[ 0, 0, 0],
[ 2, 2, -2],
[ 6, -1, -4],
[-2, 1, 2]],
[[-2, -2, 2],
[ 0, 0, 0],
[ 4, -3, -2],
[-4, -1, 4]],
[[-6, 1, 4],
[-4, 3, 2],
[ 0, 0, 0],
[-8, 2, 6]],
[[ 2, -1, -2],
[ 4, 1, -4],
[ 8, -2, -6],
[ 0, 0, 0]]])
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