[Numpy-discussion] proposal: new logspace without the log in the argument

[Chris Barker]

On Thu, Feb 18, 2016 at 2:19 PM, Alan Isaac <alan.isaac at gmail.com> wrote:

> Would such a new function correct the apparent mistake (?) of
> `linspace` including the endpoint by default?
> Or is the current API justified by its Matlab origins?
>

I don't think so -- we don't need no stinkin' Matlab !

But I LIKE including the endpoint in the sequence -- for the common use
cases, it's often what you want, and if it didn't include the end point but
you did want that, it would get pretty ugly to figure out how to get what
you want.

On the other hand, if I had it to do over, I would have the count specify
the number of intervals, rather than the number of items. A common cae may
be: values from zero to 10 (inclusive), and I want ten steps:

In [19]: np.linspace(0, 10, 10)
Out[19]:

array([  0.        ,   1.11111111,   2.22222222,   3.33333333,
         4.44444444,   5.55555556,   6.66666667,   7.77777778,
         8.88888889,  10.        ])

HUH? I was expecting [0,1,2,3 ....] (OK, not me, this isn't my first
Rodeo), so now I need to do:
In [20]: np.linspace(0, 10, 11)
Out[20]: array([  0.,   1.,   2.,   3.,   4.,   5.,   6.,   7.,   8.,   9.,
 10.])

This gets uglier if I know what "delta" I want:

In [21]: start = 0.0; end = 9.0; delta = 1.0
In [24]: np.linspace(start, end, (end-start)/delta)
Out[24]: array([ 0.   ,  1.125,  2.25 ,  3.375,  4.5  ,  5.625,  6.75 ,
 7.875,  9.   ])

oops!

In [25]: np.linspace(start, end, (end-start)/delta + 1)
Out[25]: array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

But in any case, there is no changing it now.

-CHB


-- 

Christopher Barker, Ph.D.
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