Hi Josef,
thanks for the summary! I am responding below, later I will make an
enhancement ticket.
josef.p...@gmail.com wrote:
On Sat, Jun 6, 2009 at 4:42 AM, Neil Crighton neilcrigh...@gmail.com wrote:
Robert Cimrman cimrman3 at ntc.zcu.cz writes:
Anne Archibald wrote:
1. add a keyword
Robert Cimrman wrote:
Hi Josef,
thanks for the summary! I am responding below, later I will make an
enhancement ticket.
Done, see http://projects.scipy.org/numpy/ticket/1133
r.
___
Numpy-discussion mailing list
Numpy-discussion@scipy.org
Robert Cimrman cimrman3 at ntc.zcu.cz writes:
Anne Archibald wrote:
1. add a keyword argument to intersect1d assume_unique; if it is not
present, check for uniqueness and emit a warning if not unique
2. change the warning to an exception
Optionally:
3. change the meaning of the
On Sat, Jun 6, 2009 at 4:42 AM, Neil Crighton neilcrigh...@gmail.com wrote:
Robert Cimrman cimrman3 at ntc.zcu.cz writes:
Anne Archibald wrote:
1. add a keyword argument to intersect1d assume_unique; if it is not
present, check for uniqueness and emit a warning if not unique
2. change
josef.p...@gmail.com wrote:
On Fri, Jun 5, 2009 at 1:48 AM, Robert Cimrman cimrm...@ntc.zcu.cz wrote:
josef.p...@gmail.com wrote:
On Thu, Jun 4, 2009 at 4:30 PM, Gael Varoquaux
gael.varoqu...@normalesup.org wrote:
On Thu, Jun 04, 2009 at 10:27:11PM +0200, Kim Hansen wrote:
in(b) or
On 4-Jun-09, at 4:38 PM, Anne Archibald wrote:
It seems to me that this is the basic source of the problem. Perhaps
this can be addressed? I realize maintaining compatibility with the
current behaviour is necessary, so how about a multistage deprecation:
1. add a keyword argument to
a[(a==b[:,None]).sum(axis=0,dtype=bool)]
hth,
Alan Isaac
___
Numpy-discussion mailing list
Numpy-discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Thu, Jun 4, 2009 at 8:23 AM, Alan G Isaac ais...@american.edu wrote:
a[(a==b[:,None]).sum(axis=0,dtype=bool)]
this is my preferred way when b is small and has unique elements.
if the elements in b are not unique, then be can be replaced by np.unique(b)
If b is large this creates a huge
On Thu, Jun 4, 2009 at 8:23 AM, Alan G Isaac ais...@american.edu wrote:
a[(a==b[:,None]).sum(axis=0,dtype=bool)]
On 6/4/2009 8:35 AM josef.p...@gmail.com apparently wrote:
If b is large this creates a huge intermediate array
True enough, but one could then use fromiter:
setb = set(b)
itr =
On Thu, Jun 4, 2009 at 10:13 AM, Alan G Isaac ais...@american.edu wrote:
On Thu, Jun 4, 2009 at 8:23 AM, Alan G Isaac ais...@american.edu wrote:
a[(a==b[:,None]).sum(axis=0,dtype=bool)]
On 6/4/2009 8:35 AM josef.p...@gmail.com apparently wrote:
If b is large this creates a huge intermediate
On Thu, Jun 4, 2009 at 10:13 AM, Alan G Isaac ais...@american.edu wrote:
Or if a stable order is not important (I don't
recall if the OP specified), one could just
np.intersect1d(a, np.unique(b))
On 6/4/2009 10:50 AM josef.p...@gmail.com apparently wrote:
This requires that also `a` has only
On Thu, Jun 4, 2009 at 11:12 AM, Alan G Isaac ais...@american.edu wrote:
On Thu, Jun 4, 2009 at 10:13 AM, Alan G Isaac ais...@american.edu wrote:
Or if a stable order is not important (I don't
recall if the OP specified), one could just
np.intersect1d(a, np.unique(b))
On 6/4/2009 10:50 AM
On 6/4/2009 10:50 AM josef.p...@gmail.com apparently wrote:
intersect1d gives set intersection if both arrays have
only unique elements (i.e. are sets). I thought the
naming is pretty clear:
intersect1d(a,b) set intersection if a and b with unique elements
intersect1d_nu(a,b) set
Alan G Isaac wrote:
On 6/4/2009 10:50 AM josef.p...@gmail.com apparently wrote:
intersect1d gives set intersection if both arrays have
only unique elements (i.e. are sets). I thought the
naming is pretty clear:
intersect1d(a,b) set intersection if a and b with unique elements
On Thu, Jun 4, 2009 at 11:19 AM, Alan G Isaac ais...@american.edu wrote:
On 6/4/2009 10:50 AM josef.p...@gmail.com apparently wrote:
intersect1d gives set intersection if both arrays have
only unique elements (i.e. are sets). I thought the
naming is pretty clear:
intersect1d(a,b) set
On 6/4/2009 11:29 AM josef.p...@gmail.com apparently wrote:
intersect1d is the intersection between sets (which are stored as
arrays), just like in the mathematical definition the two sets only
have unique elements
Hmmm. OK, I see you and Robert believe this.
But it does not match the
On Thu, Jun 4, 2009 at 12:32 PM, Alan G Isaac ais...@american.edu wrote:
On 6/4/2009 11:29 AM josef.p...@gmail.com apparently wrote:
intersect1d is the intersection between sets (which are stored as
arrays), just like in the mathematical definition the two sets only
have unique elements
On 6/4/2009 1:27 PM josef.p...@gmail.com apparently wrote:
Note: there are two versions of the docs for np.intersect1d, the
currently published docs which describe the actual behavior (for the
non-unique case), and the new docs on the doc editor
On Thu, Jun 4, 2009 at 2:58 PM, Alan G Isaac ais...@american.edu wrote:
On 6/4/2009 1:27 PM josef.p...@gmail.com apparently wrote:
Note: there are two versions of the docs for np.intersect1d, the
currently published docs which describe the actual behavior (for the
non-unique case), and the new
Concerning the name setmember1d_nu, I personally find it quite verbose
and not the name I would expect as a non-insider coming to numpy and
not knowing all the names of the more special hidden-away functions
and not being a python-wiz either.
I think ain(a,b) would be the name I had expected as
On Thu, Jun 04, 2009 at 10:27:11PM +0200, Kim Hansen wrote:
in(b) or in_iterable(b) method, such that you could do a.in(b)
which would return a boolean array of the same shape as a with
elements true if the equivalent a members were members in the iterable
b.
That would really by what I would
2009/6/4 josef.p...@gmail.com:
intersect1d should throw a domain error if you give it arrays with
non-unique elements, which is not done for speed reasons
It seems to me that this is the basic source of the problem. Perhaps
this can be addressed? I realize maintaining compatibility with the
On Thu, Jun 4, 2009 at 4:30 PM, Gael Varoquaux
gael.varoqu...@normalesup.org wrote:
On Thu, Jun 04, 2009 at 10:27:11PM +0200, Kim Hansen wrote:
in(b) or in_iterable(b) method, such that you could do a.in(b)
which would return a boolean array of the same shape as a with
elements true if the
On Thu, Jun 04, 2009 at 04:43:39PM -0400, josef.p...@gmail.com wrote:
Just using in might promise more than it does, eg. it works only for
one dimensional arrays, maybe in1d. With in,
Then 'in_1d'
I found arraysetops because of unique1d, but I didn't figure out what
the subpackage really
On Thu, Jun 4, 2009 at 4:52 PM, Gael Varoquaux
gael.varoqu...@normalesup.org wrote:
On Thu, Jun 04, 2009 at 04:43:39PM -0400, josef.p...@gmail.com wrote:
Just using in might promise more than it does, eg. it works only for
one dimensional arrays, maybe in1d. With in,
Then 'in_1d'
No, if the
josef.p...@gmail.com wrote:
On Thu, Jun 4, 2009 at 2:58 PM, Alan G Isaac ais...@american.edu wrote:
On 6/4/2009 1:27 PM josef.p...@gmail.com apparently wrote:
Note: there are two versions of the docs for np.intersect1d, the
currently published docs which describe the actual behavior (for the
Kim Hansen wrote:
Concerning the name setmember1d_nu, I personally find it quite verbose
and not the name I would expect as a non-insider coming to numpy and
not knowing all the names of the more special hidden-away functions
and not being a python-wiz either.
To explain the naming: those
Anne Archibald wrote:
2009/6/4 josef.p...@gmail.com:
intersect1d should throw a domain error if you give it arrays with
non-unique elements, which is not done for speed reasons
It seems to me that this is the basic source of the problem. Perhaps
this can be addressed? I realize
josef.p...@gmail.com wrote:
On Thu, Jun 4, 2009 at 4:30 PM, Gael Varoquaux
gael.varoqu...@normalesup.org wrote:
On Thu, Jun 04, 2009 at 10:27:11PM +0200, Kim Hansen wrote:
in(b) or in_iterable(b) method, such that you could do a.in(b)
which would return a boolean array of the same shape as a
On Fri, Jun 5, 2009 at 1:48 AM, Robert Cimrman cimrm...@ntc.zcu.cz wrote:
josef.p...@gmail.com wrote:
On Thu, Jun 4, 2009 at 4:30 PM, Gael Varoquaux
gael.varoqu...@normalesup.org wrote:
On Thu, Jun 04, 2009 at 10:27:11PM +0200, Kim Hansen wrote:
in(b) or in_iterable(b) method, such that you
Hi, I want to extract elements of an array (say, a) that are contained in
another array (say, b). That is, if a=array([1,1,2,3,3,4]), b=array([1,4]),
then I want array([1,1,4]).
I did the following but the speed is very slow (maybe because a is very
long):
c=array([])
for x in b:
On Wed, Jun 3, 2009 at 8:29 PM, Ning Sean nings...@gmail.com wrote:
Hi, I want to extract elements of an array (say, a) that are contained in
another array (say, b). That is, if a=array([1,1,2,3,3,4]), b=array([1,4]),
then I want array([1,1,4]).
I did the following but the speed is very slow
Thanks! Tried it and it is about twice as fast as my approach.
-Ning
On Wed, Jun 3, 2009 at 7:45 PM, josef.p...@gmail.com wrote:
On Wed, Jun 3, 2009 at 8:29 PM, Ning Sean nings...@gmail.com wrote:
Hi, I want to extract elements of an array (say, a) that are contained in
another array (say,
33 matches
Mail list logo
