Thanks. I'm still struggling with learning the idiom. Too much perl
to unlearn. 8-)
On Sat, 30 Aug 2008 23:10:10 -0400
Zachary Pincus <[EMAIL PROTECTED]> wrote:
> Hi Alan,
>
> > Traceback (most recent call last):
> > File "/usr/local/lib/python2.5/site-packages/enthought.traits-2.0.4-
> > py2.
Hi Alan,
> Traceback (most recent call last):
> File "/usr/local/lib/python2.5/site-packages/enthought.traits-2.0.4-
> py2.5-linux-i686.egg/enthought/traits/trait_notifiers.py", line 325,
> in call_1
>self.handler( object )
> File "TrimMapl_1.py", line 98, in _Run_fired
>outdata = np.
[ Sorry I never sent this, I just found it in my drafts folder. Just
in case it's useful to the OP, here it is. ]
On Thu, Jul 10, 2008 at 9:38 AM, Dan Lussier <[EMAIL PROTECTED]> wrote:
> r2 = numpy.power(r2,2).sum(axis=1)
>
> r2 = numpy.extract(r2
I been beating myself up over this bit of code for far too long now - I
know I must be missing something really simple, but what is it?
TYPICALLY_UINT_COLUMNS = ['Track', 'Bin', 'code', 'horizon']
..
dtypes = [ ]
for i in range(0, len(self.var_list)) :
if TYPICALLY_
Stéfan van der Walt wrote:
> Maybe
>
> (np.sign(a) | 1) * np.maximum(np.abs(a), min_value)
>
> is a little bit easier on the eye.
nice! that does it. somehow I never think of using or.
Alan G Isaac wrote:
> idx = np.abs(a) a[idx] = min_value*(np.sign(a[idx]) + (a[idx]==0))
that's do it too,
I think the question of the timing of removing
the remaining default index inconsistencies in
NumPy was raised by not answered? Anyway, what
is the answer?
I just ran into this again, this time with ``argsort``,
which still has ``axis=-1`` as the default. But maybe
it is intentional to keep this
On Sat, Aug 30, 2008 at 7:29 AM, Keith Goodman <[EMAIL PROTECTED]> wrote:
> On Sat, Aug 30, 2008 at 6:24 AM, Alan G Isaac <[EMAIL PROTECTED]> wrote:
>> Stéfan van der Walt wrote:
>>> (np.sign(a) | 1) ...
>>
>> Ah, that's nice. How about
>> idx = np.abs(a)> a[idx] = min_value*(np.sign(a[idx]) | 1)
>
On Sat, Aug 30, 2008 at 6:24 AM, Alan G Isaac <[EMAIL PROTECTED]> wrote:
> Stéfan van der Walt wrote:
>> (np.sign(a) | 1) ...
>
> Ah, that's nice. How about
> idx = np.abs(a) a[idx] = min_value*(np.sign(a[idx]) | 1)
Or, since he asked to do it in one line,
>> min_value = 2
>> x
array([ 1, 2,
Stéfan van der Walt wrote:
> (np.sign(a) | 1) ...
Ah, that's nice. How about
idx = np.abs(a)http://projects.scipy.org/mailman/listinfo/numpy-discussion
Christopher Barker wrote:
> 0 should go to 2 --it's not, because sign(0) == 0, so the 2 gets turned
> to zero, my original problem.
Odd asymmetry. Then you can use Keith's version. Or
a[idx] = min_value*(np.sign(a[idx]) + (a[idx]==0))
Alan
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