If I have z = cumsum( [ 0, 1, 2, 6, 9 ] )
, which gives me z = [ 0, 1, 3, 9, 18 ]
, how can I get back to the original array [ 0, 1, 2, 6, 9 ]
?
cumsum. Return the cumulative sum of the elements along a given axis.
cumsum() function is used when we want to compute the cumulative sum of array elements over a given axis. Syntax : numpy.cumsum(arr, axis=None, dtype=None, out=None) Parameters : arr : [array_like] Array containing numbers whose cumulative sum is desired.
z[1:] -= z[:-1].copy()
Short and sweet, with no slow Python loops. We take views of all but the first element (z[1:]
) and all but the last (z[:-1]
), and subtract elementwise. The copy makes sure we subtract the original element values instead of the values we're computing. (On NumPy 1.13 and up, you can skip the copy
call.)
You can use np.diff
to compute elements 1...N
which will take the difference between any two elements. This is the opposite of cumsum
. The only difference is that diff
will not return the first element, but the first element is the same in the original and cumsum
output so we just re-use that value.
orig = np.insert(np.diff(z), 0, z[0])
Rather than insert
, you could also use np.concatenate
orig = np.concatenate((np.array(z[0]).reshape(1,), np.diff(z)))
We could also just copy and replace elements 1...N
orig = z.copy() orig[1:] = np.diff(z)
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