numpy-discussion

[Numpy-discussion] linspace upper bound not met?

[Bill B. <wb...@gm...>]

I was porting some code over from matlab in which I relied on the upper
bound of linspace to be met exactly.

It turns out that it isn't always exactly met in numpy.

In [390]: filter(lambda x: x[1]!=0.0, [ (i,1.0-numpy.linspace(0,1,i)[-1])
for i in range(2,200) ])
Out[390]:
[(50, 1.1102230246251565e-016),
 (99, 1.1102230246251565e-016),
 (104, 1.1102230246251565e-016),
 (108, 1.1102230246251565e-016),
 (162, 1.1102230246251565e-016),
 (188, 1.1102230246251565e-016),
 (197, 1.1102230246251565e-016),
 (198, 1.1102230246251565e-016)]


I know it's not a good idea to count on floating point equality in general,
but still it doesn't seem too much to expect that the first and last values
returned by linspace are exactly the values asked for if they both have
exact floating point representations.

--bb

Re: [Numpy-discussion] linspace upper bound not met?

[Alan G I. <ai...@am...>]

The definition of linspace is:
def linspace(start, stop, num=3D50, endpoint=3DTrue, retstep=3DFalse):
    """Return evenly spaced numbers.

    Return 'num' evenly spaced samples from 'start' to 'stop'.  If
    'endpoint' is True, the last sample is 'stop'. If 'retstep' is
    True then return the step value used.
    """
    num =3D int(num)
    if num <=3D 0:
        return array([], float)
    if endpoint:
        if num =3D=3D 1:
            return array([float(start)])
        step =3D (stop-start)/float((num-1))
    else:
        step =3D (stop-start)/float(num)
    y =3D _nx.arange(0, num) * step + start
    if retstep:
        return y, step
    else:
        return y

The simplest way to achieve this goal is to add right after=20
the assignment to y two new lines:
   if endpoint:
       y[-1] =3D float(stop)

Cheers,
Alan Isaac

PS I'll take this opportunity to state again my opinion that
in the denerate case num=3D1 that if endpoint=3DTrue then=20
linspace should return stop rather than start.  (Otherwise=20
endpoint is ignored. But I do not expect anyone to agree.)

Re: [Numpy-discussion] linspace upper bound not met?

[Bill B. <wb...@gm...>]

Out of curiosity I checked on what matlab does.
It does explicity set the last value to 'stop' to avoid the roundoff issue.
In numpy terms, it does something like
   y = r_[start+r_[0:num-1]*(stop-start)/(num-1.0),  stop]

But for numpy it's probably more efficient to just do the 'y[-1] = stop'
like you say.


--bb

On 8/22/06, Alan G Isaac <ai...@am...> wrote:
>
> The definition of linspace is:
> def linspace(start, stop, num=50, endpoint=True, retstep=False):
>     """Return evenly spaced numbers.
>
>     Return 'num' evenly spaced samples from 'start' to 'stop'.  If
>     'endpoint' is True, the last sample is 'stop'. If 'retstep' is
>     True then return the step value used.
>     """
>     num = int(num)
>     if num <= 0:
>         return array([], float)
>     if endpoint:
>         if num == 1:
>             return array([float(start)])
>         step = (stop-start)/float((num-1))
>     else:
>         step = (stop-start)/float(num)
>     y = _nx.arange(0, num) * step + start
>     if retstep:
>         return y, step
>     else:
>         return y
>
> The simplest way to achieve this goal is to add right after
> the assignment to y two new lines:
>    if endpoint:
>        y[-1] = float(stop)
>
> Cheers,
> Alan Isaac
>
> PS I'll take this opportunity to state again my opinion that
> in the denerate case num=1 that if endpoint=True then
> linspace should return stop rather than start.  (Otherwise
> endpoint is ignored. But I do not expect anyone to agree.)
>
>
>
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