On Thu, Feb 25, 2010 at 10:20, Bruno Santos wrote:
> This is the same example we discuss yesterday.
I think I can help you this time, but when we ask for complete code,
we mean complete, self-contained code that we can run immediately, not
a fragment of code that needs variables to be initialized
This is the same example we discuss yesterday.
The working code is this one:
lsPhasedValues = [aLoci[i] for i in xrange(length) if i%21==0 and
aLoci[i]>0]
I was able to get the same result after a while:
aAux =aLoci[index_nSize]
lsPhasedValues = numpy.unique1d(aAux[numpy.where(aAux>0)[0]])
I could
On Thu, Feb 25, 2010 at 07:51, Bruno Santos wrote:
> I just realized that the line lsPhasedValues =
> numpy.unique1d(aLoci[numpy.where(aLoci[index_nSize]>0)]) does not work
> properly.
> How can I get the unique values of an array based on their indexes?
I don't know what that sentence means. Ple
I just realized that the line lsPhasedValues =
numpy.unique1d(aLoci[numpy.where(aLoci[index_nSize]>0)]) does not work
properly.
How can I get the unique values of an array based on their indexes?
2010/2/25 Bruno Santos
> After implementation all the possibilities we discuss yesterday mi fastest
After implementation all the possibilities we discuss yesterday mi fastest
version is this one:
index_nSize=numpy.arange(0,length,nSize)
lsPhasedValues = numpy.unique1d(aLoci[numpy.where(aLoci[index_nSize]>0)])
...
bigaLoci = (aLoci>=r)
k = (aLoci>=r).sum()
This is taking around 0.12s for my tes
On Wed, Feb 24, 2010 at 12:38, Bruno Santos wrote:
> This is probably me just being stupid. But what is the reason for this peace
> of code not to be working:
> index_nSize=numpy.arange(0,length,nSize)
> lsPhasedValues = set([aLoci[i] for i in xrange(length) if (i%nSize==0 and
> aLoci[i]>0)])
> ls
This is probably me just being stupid. But what is the reason for this peace
of code not to be working:
index_nSize=numpy.arange(0,length,nSize)
lsPhasedValues = set([aLoci[i] for i in xrange(length) if (i%nSize==0 and
aLoci[i]>0)])
lsPhasedValues1 = numpy.where(aLoci[index_nSize]>0)
print aLoci[in
2010/2/24 Chris Colbert
> In [4]: %timeit a = np.random.randint(0, 20, 100)
> 10 loops, best of 3: 4.32 us per loop
>
> In [5]: %timeit (a>=10).sum()
> 10 loops, best of 3: 7.32 us per loop
>
> In [8]: %timeit np.where(a>=10)
> 10 loops, best of 3: 5.36 us per loop
>
>
> am i missing
On Wed, Feb 24, 2010 at 11:50, Bruno Santos wrote:
> In both versions your lsPhasedValues contains the number of positions in the
> array that match a certain criteria. What I need in that step is the unique
> values and not their positions.
Oops!
lsPhasedValues = np.unique1d(aLoci[j_nSize_mask
In [4]: %timeit a = np.random.randint(0, 20, 100)
10 loops, best of 3: 4.32 us per loop
In [5]: %timeit (a>=10).sum()
10 loops, best of 3: 7.32 us per loop
In [8]: %timeit np.where(a>=10)
10 loops, best of 3: 5.36 us per loop
am i missing something?
On Wed, Feb 24, 2010 at 12:50 PM
In both versions your lsPhasedValues contains the number of positions in the
array that match a certain criteria. What I need in that step is the unique
values and not their positions.
2010/2/24 Robert Kern
> On Wed, Feb 24, 2010 at 11:19, Bruno Santos wrote:
> > It seems that the python 2.6.4
On Wed, Feb 24, 2010 at 11:19, Bruno Santos wrote:
> It seems that the python 2.6.4 has a more efficient implementation of the
> lists. It runs faster on this version and slower on 2.5.4 on the same
> machine with debian. A lot faster in fact.
> I was trying to change my headche for the last coupl
It seems that the python 2.6.4 has a more efficient implementation of the
lists. It runs faster on this version and slower on 2.5.4 on the same
machine with debian. A lot faster in fact.
I was trying to change my headche for the last couple of weeks. But you
migth give me a lot more optimizations
On Wed, Feb 24, 2010 at 10:40, Bruno Santos wrote:
> Funny. Which version of python are you using?
Python 2.5.4 on OS X.
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an
Funny. Which version of python are you using? My python is still better for
small lists. But you are rigth it gets better with size, here how the same
code performs on mine computer:
In [1]: N = 100
In [2]: import numpy as np
In [3]: A = np.random.randint(0, 21, N)
...:
In [4]: L = A.tolist
On Wed, Feb 24, 2010 at 10:21, Bruno Santos wrote:
>> The idiomatic way of doing this for numpy arrays would be:
>>
>> def test2(arrx):
>> return (arrx >= 10).sum()
>>
> Even this versions takes more time to run than my original python version
> with arrays.
Works fine for me, and gets bette
>
>
>
> The idiomatic way of doing this for numpy arrays would be:
>
> def test2(arrx):
>return (arrx >= 10).sum()
>
> Even this versions takes more time to run than my original python version
> with arrays.
>>> def test3(listx):
... return (listx>=10).sum()
>>> t = timeit.Timer("test3(l
On Wed, Feb 24, 2010 at 09:55, Bruno Santos wrote:
> Hello everyone,
> I am using numpy arrays whenever I demand performance from my
> algorithms. Nevertheless, I am having a performance issue at the moment
> mainly because I am iterating several times over numpy arrays. Fot that
> reason I decide
Hello everyone,
I am using numpy arrays whenever I demand performance from my
algorithms. Nevertheless, I am having a performance issue at the moment
mainly because I am iterating several times over numpy arrays. Fot that
reason I decided to use timeit to see the performance of different versions
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