[Numpy-discussion] numarray incompatibility: searchsorted

[Todd Miller]

I logged this on Source Forge.  Thanks for the comments.

Todd

On Fri, 2003-09-26 at 13:06, Tim Hochberg wrote:
> 
> There are some compatibility and doc issues and perhaps a bug in 
> numarray.searchsorted. The compatibility issue is that 
> Numeric.searchsorted(a, v) accepts either a sequence or scalar value for 
> v. Numarray.searchsorted accepts only sequence values.
> 
> Second, the doc issue. The docstring for numarray.searchsorted states::
> 
>   searchsorted(bins, values)
>     searchsort(bins, values) returns the array of greatest indices 'i'
>     such that each values[i] <= bins[i].
> 
> I assume that should really read something more like::
> 
>   searchsorted(bins, values)
>     searchsort(bins, values) returns the array A[j] of greatest indices 'i'
>     such that each values[j] <= bins[i].
> 
> Third, the possible bug:
> 
> # na = numarray, np = NumPy
>  >>> na.searchsorted([1,2,3,4], [2.5, 3.5])
> array([1, 2])
>  >>> np.searchsorted([1,2,3,4], [2.5, 3.5])
> array([2, 3])
> 
> Hmmm. It appears that numarray result does match the numarray docs, (at 
> least as I read them), but I like the Numeric behaviour better.  The 
> Numeric behaviour also matches the behaviour of the bisect module, which 
> is described as::
> 
>     bisect = bisect_right(a, x, lo=0, hi=None)
>         Return the index where to insert item x in list a, assuming a is 
> sorted.
> 
>         The return value i is such that all e in a[:i] have e <= x, and 
> all e in
>         a[i:] have e > x.  So if x already appears in the list, i points 
> just
>         beyond the rightmost x already there.
> 
>         Optional args lo (default 0) and hi (default len(a)) bound the
>         slice of a to be searched.
> 
> I'd recomend matching the behaviour of the two existing modules (bisect 
> and Numeric).
> 
> -tim
> 
> 
> 
> 
> 
> 
> 
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-- 
Todd Miller 			jmiller at stsci.edu
STSCI / ESS / SSB