I'm really confused. Summing from zero should be what cumsum() does now. ``` >>> np.__version__ '1.22.4' >>> 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?
Ben Root On Fri, Aug 11, 2023 at 8:55 AM Juan Nunez-Iglesias <j...@fastmail.com> wrote: > 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.daw...@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) > > ``` > > _______________________________________________ > > NumPy-Discussion mailing list -- numpy-discussion@python.org > > To unsubscribe send an email to numpy-discussion-le...@python.org > > https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ > > Member address: j...@fastmail.com > _______________________________________________ > NumPy-Discussion mailing list -- numpy-discussion@python.org > To unsubscribe send an email to numpy-discussion-le...@python.org > https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ > Member address: ben.v.r...@gmail.com >
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