Yet another example is
```
d = np.zeros(n)
d[1:] = np.linalg.norm(np.diff(points, axis=1), axis=0)
r = d.cumsum()
```
https://github.com/WarrenWeckesser/ufunclab/blob/main/examples/linear_interp1d_demo.py#L13-L15
___
NumPy-Discussion mailing list -- numpy
Dom Grigonis wrote:
> 1. Dimension length stays constant, while cumusm0 extends length to n+1, then
> np.diff, truncates it back. This adds extra complexity, while things are very
> convenient to work with when dimension length stays constant throughout the
> code.
For n values there are n-1 di
> From my point of view, such function is a bit of a corner-case to be added to
> numpy. And it doesn’t justify it’s naming anymore. It is not one operation
> anymore. It is a cumsum and prepending 0. And it is very difficult to argue
> why prepending 0 to cumsum is a part of cumsum.
That is ba
`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#Fencepos
