This has come up before, see https://github.com/numpy/numpy/issues/6044 for the first time this came up; there were several subsequent discussions linked there. In the meantime, the data APIs consortium has been actively working on adding a `cumulative_sum` function to the array API standard, see https://github.com/data-apis/array-api/issues/597 and https://github.com/data-apis/array-api/pull/653. The proposed `cumulative_sum` function includes an `include_initial` keyword argument that gets the OP's desired behavior. I think we should probably eventually deprecate `cumsum` and `cumprod` in favor of the array API standard's `cumulative_sum` and `cumulative_product` if only because of the embarrassing naming issue. Once the array API standard has finalized the name for the keyword argument, I think it makes sense to add the keyword argument to np.cumsum, even if we don't deprecate it yet. I don't think it makes sense to add a new function just for this. On Fri, Aug 11, 2023 at 6:34 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) ``` _______________________________________________ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-leave@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: nathan12343@gmail.com