[Numpy-discussion] numarray incompatibility: searchsorted
Tim Hochberg
tim.hochberg at ieee.org
Fri Sep 26 10:14:08 CDT 2003
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
More information about the Numpy-discussion
mailing list