On 1/24/07, Robert Cimrman <[EMAIL PROTECTED]> wrote:
Robert Kern wrote:
> Robert Cimrman wrote:
>> Or you could just call unique1d prior to your call to setmember1d - it
>> was meant to be used that way... you would not loose much speed that
>> way, IMHO.
>
> But that doesn't do what they want. They want a function that gives the
mask
> against their original array of the elements that are in the other
array. The
> result of
>
> setmember1d(unique1d(ar1), unique1d(ar2))
>
> is a mask against
>
> unique1d(ar1)
>
> not
>
> ar1
>
> as they want.
I see. I was thinking in terms of 'set member' - in set one assumes
unique elements. A good name for this kind of function would be
'arraymember'.
Naive pseudo-implementation:
import numpy as nm
def arraymember1d( ar1, ar2 ):
ar = ar2.copy().sort()
indx = findsorted( ar2, ar1 )
flag = nm.zeros( ar1.shape, dtype = nm.bool )
flag[indx] = True
return flag
Note that nm.searchsorted cannot be used here - a new function
('findsorted'?) would be needed.
Maybe something like
countelements(ar1, ar2) :
a = sort(ar2)
il = a.searchsorted(ar1, side='left')
ir = a.searchsorted(ar1, side='right')
return ir - il
which returns an array containing the number of times the corresponding
element of ar1 is contained in ar2.
Chuck
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