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How to take nth order discrete sum from numpy array (total numpy.diff equivalent)

I know it is possible to take the nth order discrete difference of a numpy array using the numpy function numpy.diff()

, but is there a way to do the same with nth order discrete sum?

Let's say we have an array of numpy, A = np.arange(10)

. Expected output for 1st order discrete sum:

array([  1.,   3.,   5.,   7.,   9.,  11.,  13.,  15.,  17.])

which i can get:

N = A.shape[0]
B = np.zeros(N-1)

for i in range(N-1):
    B[i] = A[i+1] + A[i]

But is there a function to avoid using a for loop?

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1 answer


A[i+1]

for for i in range(N-1)

will be covered in a A[1:]

similar way A[i]

for the same iterative facility A[:-1]

. So, in principle, you can sum these two versions of the vectorized input array in B

, for example:



B = A[:-1] + A[1:]
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