Try out latest SVN. It should have this problem fixed.
Thanks for this. I've realized that for my case, using object arrays
is probably best. I still think that long term it would be good to
allow comparison functions to take different types, so that one could
compare say integer arrays with
Hi,
The following gives the wrong answer:
In [2]: A = array(['a','aa','b'])
In [3]: B = array(['d','e'])
In [4]: A.searchsorted(B)
Out[4]: array([3, 0])
The answer should be [3,3]. I've come across this while trying to come
up with an ismember function which works for strings (setmember1d
Hi,
OK, i'm using:
In [6]: numpy.__version__
Out[6]: '1.0.3'
Should I try the development version? Which version of numpy would
people generally recommend?
James
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works fine for me.
In [33]: A = numpy.array(['a','aa','b'])
In [34]: B = numpy.array(['d','e'])
In [35]: A.searchsorted(B)
Out[35]: array([3, 3])
In [36]: numpy.__version__
Out[36]: '1.0.5.dev4567'
L.
On 1/31/08, James Philbin [EMAIL PROTECTED] wrote:
Hi,
The following gives the
Hmmm. Just downloaded and installed 1.0.4 and i'm still getting this
error. Are you guys using the bleeding edge version or the official
1.0.4 tarball from the webpage?
James
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Hi,
The following gives the wrong answer:
In [2]: A = array(['a','aa','b'])
In [3]: B = array(['d','e'])
In [4]: A.searchsorted(B)
Out[4]: array([3, 0])
The answer should be [3,3]. I've come across this while trying to come
up with an ismember function which works for strings (setmember1d
It works fine also for me (numpy 1.04 gentoo linux on amd64)
Nadav
On Thu, 2008-01-31 at 15:51 +0100, lorenzo bolla wrote:
works fine for me.
In [33]: A = numpy.array(['a','aa','b'])
In [34]: B = numpy.array(['d','e'])
In [35]: A.searchsorted(B)
Out[35]: array([3, 3])
In [36]:
I use a dev version (1.0.5.dev4567).
L.
On 1/31/08, James Philbin [EMAIL PROTECTED] wrote:
Hmmm. Just downloaded and installed 1.0.4 and i'm still getting this
error. Are you guys using the bleeding edge version or the official
1.0.4 tarball from the webpage?
James
Hi,
Just tried with numpy from svn and still get this problem:
import numpy
numpy.__version__
'1.0.5.dev4763'
A = numpy.array(['a','aa','b'])
B = numpy.array(['d','e'])
A.searchsorted(B)
array([3, 0])
I guess this must be a platform-dependent bug. I'm running python version:
Python 2.5
No problem for me (also a svn version) :
Python 2.5.1 (r251:54863, Oct 30 2007, 13:54:11)
[GCC 4.1.2 20070925 (Red Hat 4.1.2-33)] on linux2
import numpy
A = numpy.array(['a','aa','b'])
B = numpy.array(['d','e'])
A.searchsorted(B)
array([3, 3])
Matthieu
2008/1/31, lorenzo bolla [EMAIL
I do get the problem with a recent(ish) svn, on OS X 10.5.1, python
2.5.1 (from python.org):
In [76]: A = array(['a','aa','b'])
In [77]: B = array(['d','e'])
In [78]: A.searchsorted(B)
Out[78]: array([3, 0])
In [79]: numpy.__version__
Out[79]: '1.0.5.dev4722'
Hi,
With my system running x86_64 SUSE10.0 AMD opteron:
Under Python 2.5.1 (Python 2.5.1 -r251:54863) and numpy 1.0.4
(download of released version) I have the same bug.
Under Python 2.4.1 (May 25 2007, 18:41:31) and numpy 1.0.3 I have no problem.
Perhaps a 32/64 bit problem?
Bruce
On Jan 31,
Am Donnerstag, 31. Januar 2008 15:35:25 schrieb James Philbin:
The following gives the wrong answer:
In [2]: A = array(['a','aa','b'])
In [3]: B = array(['d','e'])
In [4]: A.searchsorted(B)
Out[4]: array([3, 0])
The answer should be [3,3].
Heh, I got both answers in the same session (not
Problem also with Windows P3 binaries.
fwiw,
Alan Isaac
Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310
32 bit (Intel)] on win32
Type help, copyright, credits or license for more information.
import numpy
numpy.__version__
'1.0.4'
A = numpy.array(['a','aa','b'])
B =
from docstring in multiarraymodule.c
/** @brief Use bisection of sorted array to find first entries = keys.
*
* For each key use bisection to find the first index i s.t. key = arr[i].
* When there is no such index i, set i = len(arr). Return the results in
ret.
* All arrays are assumed
oops. it fails also on an SGI Altix with Suse Linux on it:
Linux pico 2.6.16.27-0.9-default #1 SMP Tue Feb 13 09:35:18 UTC 2007 ia64
ia64 ia64 GNU/Linux
In [33]: A = numpy.array(['a','aa','b'])
In [34]: B = numpy.array(['d','e'])
In [35]:
lorenzo bolla wrote:
* All arrays are assumed contiguous on entry and both arr and key must
be of-
* the same comparable type. -
A and B are not of the same type ('|S2' is not '|S1').
This should be mentioned somewhere more accessible.
It should also raise an exception --
Hi,
In particular:
* All arrays are assumed contiguous on entry and both arr and key must be
of-
* the same comparable type. -
In which case, this seems to be an overly strict implementation of
searchsorted. Surely all that should be required is that the
comparison function can
True. The problem is knowing when that is the case. The subroutine in
question is at the bottom of the heap and don't know nothin'. IIRC, it just
sits there and does the comparison by calling through a pointer with char*
arguments.
What does the comparison function actually look like for the
On Jan 31, 2008 10:49 AM, James Philbin [EMAIL PROTECTED] wrote:
Well, i've digged around in the source code and here is a patch which
makes it work for the case I wanted:
--- multiarraymodule.c.old 2008-01-31 17:42:32.0 +
+++ multiarraymodule.c 2008-01-31
On Jan 31, 2008 10:55 AM, James Philbin [EMAIL PROTECTED] wrote:
True. The problem is knowing when that is the case. The subroutine in
question is at the bottom of the heap and don't know nothin'. IIRC, it
just
sits there and does the comparison by calling through a pointer with
char*
James Philbin wrote:
I can't fathom where the comparison functions exist in the code. It
seems that the comparison signature is of the form (void*, void*,
PyArrayObject*), so it doesn't seem possible at the moment to specify
a compare function which can reason about the underlying types of the
I can't fathom where the comparison functions exist in the code. It
seems that the comparison signature is of the form (void*, void*,
PyArrayObject*), so it doesn't seem possible at the moment to specify
a compare function which can reason about the underlying types of the
two void*'s. However, I
James Philbin wrote:
I can't fathom where the comparison functions exist in the code. It
seems that the comparison signature is of the form (void*, void*,
PyArrayObject*), so it doesn't seem possible at the moment to specify
a compare function which can reason about the underlying types of the
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