I'm very sensitive to the issues of adding to the already bloated numpy API, but I would definitely find use in this function. I literally made this error (thinking that the first element of cumsum should be 0) just a couple of days ago! What are the plans for the "extended" NumPy API after 2.0? Is there a good place for these variants?
On Fri, 11 Aug 2023, at 2:07 AM, john.dawson@camlingroup.com wrote:
> `cumsum` computes the sum of the first k summands for every k from 1.
> Judging by my experience, it is more often useful to compute the sum of
> the first k summands for every k from 0, as `cumsum`'s behaviour leads
> to fencepost-like problems.
> https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error
> For example, `cumsum` is not the inverse of `diff`. I propose adding a
> function to NumPy to compute cumulative sums beginning with 0, that is,
> an inverse of `diff`. It might be called `cumsum0`. The following code
> is probably not the best way to implement it, but it illustrates the
> desired behaviour.
>
> ```
> def cumsum0(a, axis=None, dtype=None, out=None):
> """
> Return the cumulative sum of the elements along a given axis,
> beginning with 0.
>
> cumsum0 does the same as cumsum except that cumsum computes the sum
> of the first k summands for every k from 1 and cumsum, from 0.
>
> Parameters
> ----------
> a : array_like
> Input array.
> axis : int, optional
> Axis along which the cumulative sum is computed. The default
> (None) is to compute the cumulative sum over the flattened
> array.
> dtype : dtype, optional
> Type of the returned array and of the accumulator in which the
> elements are summed. If `dtype` is not specified, it defaults to
> the dtype of `a`, unless `a` has an integer dtype with a
> precision less than that of the default platform integer. In
> that case, the default platform integer is used.
> out : ndarray, optional
> Alternative output array in which to place the result. It must
> have the same shape and buffer length as the expected output but
> the type will be cast if necessary. See
> :ref:`ufuncs-output-type` for more details.
>
> Returns
> -------
> cumsum0_along_axis : ndarray.
> A new array holding the result is returned unless `out` is
> specified, in which case a reference to `out` is returned. If
> `axis` is not None the result has the same shape as `a` except
> along `axis`, where the dimension is smaller by 1.
>
> See Also
> --------
> cumsum : Cumulatively sum array elements, beginning with the first.
> sum : Sum array elements.
> trapz : Integration of array values using the composite trapezoidal rule.
> diff : Calculate the n-th discrete difference along given axis.
>
> Notes
> -----
> Arithmetic is modular when using integer types, and no error is
> raised on overflow.
>
> ``cumsum0(a)[-1]`` may not be equal to ``sum(a)`` for floating-point
> values since ``sum`` may use a pairwise summation routine, reducing
> the roundoff-error. See `sum` for more information.
>
> Examples
> --------
> >>> a = np.array([[1, 2, 3], [4, 5, 6]])
> >>> a
> array([[1, 2, 3],
> [4, 5, 6]])
> >>> np.cumsum0(a)
> array([ 0, 1, 3, 6, 10, 15, 21])
> >>> np.cumsum0(a, dtype=float) # specifies type of output value(s)
> array([ 0., 1., 3., 6., 10., 15., 21.])
>
> >>> np.cumsum0(a, axis=0) # sum over rows for each of the 3 columns
> array([[0, 0, 0],
> [1, 2, 3],
> [5, 7, 9]])
> >>> np.cumsum0(a, axis=1) # sum over columns for each of the 2 rows
> array([[ 0, 1, 3, 6],
> [ 0, 4, 9, 15]])
>
> ``cumsum(b)[-1]`` may not be equal to ``sum(b)``
>
> >>> b = np.array([1, 2e-9, 3e-9] * 1000000)
> >>> np.cumsum0(b)[-1]
> 1000000.0050045159
> >>> b.sum()
> 1000000.0050000029
>
> """
> empty = a.take([], axis=axis)
> zero = empty.sum(axis, dtype=dtype, keepdims=True)
> later_cumsum = a.cumsum(axis, dtype=dtype)
> return concatenate([zero, later_cumsum], axis=axis, dtype=dtype, out=out)
> ```
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