On 11 Aug 2023, at 7:52 pm, Robert Kern <robert.k...@gmail.com<mailto:robert.k...@gmail.com>> wrote:
>>> np.cumsum([[1, 2, 3], [4, 5, 6]]) array([ 1, 3, 6, 10, 15, 21]) ``` which matches your example in the cumsum0() documentation. Did something change in a recent release? That's not what's in his example. The example is creating a cumsum-like array of n+1 elements starting with the number 0, not array[0] – i.e. essentially just inserting 0 along every axis, so that np.diff(np.cumsum0(a)) = a Not sure if this would be too complicated to effect with the existing ufuncs either… Almost all of the documentation sounds very repetitive, so maybe implementing this via a new kwarg to cumsum would be a better option? Cheers, Derek
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