On Fri, Oct 14, 2011 at 2:59 PM, wrote:
> On Fri, Oct 14, 2011 at 2:29 PM, wrote:
>> On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac wrote:
>>> On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
If I remember correctly, signal.lfilter doesn't require stationarity,
but handling of the s
On Fri, Oct 14, 2011 at 2:29 PM, wrote:
> On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac wrote:
>> On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
>>> If I remember correctly, signal.lfilter doesn't require stationarity,
>>> but handling of the starting values is a bit difficult.
>>
>>
>> Hmm.
On Fri, Oct 14, 2011 at 2:39 PM, Skipper Seabold wrote:
> On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac wrote:
>>
>> On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
>> > If I remember correctly, signal.lfilter doesn't require stationarity,
>> > but handling of the starting values is a bit diff
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac wrote:
>
> On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
> > If I remember correctly, signal.lfilter doesn't require stationarity,
> > but handling of the starting values is a bit difficult.
>
>
> Hmm. Yes.
> AR(1) is trivial, but how do you hand
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac wrote:
> On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
>> If I remember correctly, signal.lfilter doesn't require stationarity,
>> but handling of the starting values is a bit difficult.
>
>
> Hmm. Yes.
> AR(1) is trivial, but how do you handle h
On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
> If I remember correctly, signal.lfilter doesn't require stationarity,
> but handling of the starting values is a bit difficult.
Hmm. Yes.
AR(1) is trivial, but how do you handle higher orders?
Thanks,
Alan
___
On Fri, Oct 14, 2011 at 1:26 PM, Alan G Isaac wrote:
>>> Assuming stationarity ...
>
> On 10/14/2011 1:22 PM, josef.p...@gmail.com wrote:
>> maybe ?
>
> I just meant that the MA approximation is
> not reliable for a non-stationary AR.
> E.g., http://www.jstor.org/stable/2348631
section 5: simulat
>> Assuming stationarity ...
On 10/14/2011 1:22 PM, josef.p...@gmail.com wrote:
> maybe ?
I just meant that the MA approximation is
not reliable for a non-stationary AR.
E.g., http://www.jstor.org/stable/2348631
Cheers,
Alan
___
NumPy-Discussion maili
On Fri, Oct 14, 2011 at 12:49 PM, Alan G Isaac wrote:
> On 10/14/2011 12:21 PM, josef.p...@gmail.com wrote:
>> One other way to simulate the AR is to get the (truncated)
>> MA-representation, and then convolve can be used
>
>
> Assuming stationarity ...
maybe ?
If it's integrated, then you need a
On 10/14/2011 12:21 PM, josef.p...@gmail.com wrote:
> One other way to simulate the AR is to get the (truncated)
> MA-representation, and then convolve can be used
Assuming stationarity ...
Alan
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy
On Fri, Oct 14, 2011 at 11:56 AM, Fabrice Silva wrote:
> Le vendredi 14 octobre 2011 à 10:49 -0400, josef.p...@gmail.com a
> écrit :
>> On Fri, Oct 14, 2011 at 10:24 AM, Alan G Isaac wrote:
>> > As a simple example, if I have y0 and a white noise series e,
>> > what is the best way to produces a
Hi Rense (cross-posting to the numpy mailing list because these guys
are awesome),
On Oct 13, 10:01 pm, Rense Lange wrote:
> I have potentially millions of tuples and
> I want to create frequency distributions conditional on the values of
> discrete variables v1, v2, ... (e.g. the sums for boys
Le vendredi 14 octobre 2011 à 10:49 -0400, josef.p...@gmail.com a
écrit :
> On Fri, Oct 14, 2011 at 10:24 AM, Alan G Isaac wrote:
> > As a simple example, if I have y0 and a white noise series e,
> > what is the best way to produces a series y such that y[t] = 0.9*y[t-1] +
> > e[t]
> > for t=1,2,
On Fri, Oct 14, 2011 at 10:24 AM, Alan G Isaac wrote:
> As a simple example, if I have y0 and a white noise series e,
> what is the best way to produces a series y such that y[t] = 0.9*y[t-1] + e[t]
> for t=1,2,...?
>
> 1. How can I best simulate an autoregressive process using NumPy?
>
> 2. With
Fabrice Silva wrote:
> Le vendredi 14 octobre 2011 à 08:04 -0400, Neal Becker a écrit :
>> suppose I have:
>>
>> In [10]: u
>> Out[10]:
>> array([[0, 1, 2, 3, 4],
>>[5, 6, 7, 8, 9]])
>>
>> And I have a vector v:
>> v = np.array ((0,1,0,1,0))
>>
>> I want to form an output vector which
As a simple example, if I have y0 and a white noise series e,
what is the best way to produces a series y such that y[t] = 0.9*y[t-1] + e[t]
for t=1,2,...?
1. How can I best simulate an autoregressive process using NumPy?
2. With SciPy, it looks like I could do this as
e[0] = y0
signal.lfilter((1
Le vendredi 14 octobre 2011 à 08:04 -0400, Neal Becker a écrit :
> suppose I have:
>
> In [10]: u
> Out[10]:
> array([[0, 1, 2, 3, 4],
>[5, 6, 7, 8, 9]])
>
> And I have a vector v:
> v = np.array ((0,1,0,1,0))
>
> I want to form an output vector which selects items from u where v is th
What about
a=arange(len(v))
w=u[v,a]
?
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Fri, Oct 14, 2011 at 7:04 AM, Neal Becker wrote:
> suppose I have:
>
> In [10]: u
> Out[10]:
> array([[0, 1, 2, 3, 4],
> [5, 6, 7, 8, 9]])
>
> And I have a vector v:
> v = np.array ((0,1,0,1,0))
>
> I want to form an output vector which selects items from u where v is the
> index
> of t
suppose I have:
In [10]: u
Out[10]:
array([[0, 1, 2, 3, 4],
[5, 6, 7, 8, 9]])
And I have a vector v:
v = np.array ((0,1,0,1,0))
I want to form an output vector which selects items from u where v is the index
of the row of u to be selected.
In the above example, I want:
w = [0,6,2,8,4
On Fri, Oct 14, 2011 at 11:53 AM, Olivier Delalleau wrote:
> 2011/10/14 Matthew Brett
>>
>> Hi,
>>
>> On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE wrote:
>> > Dear all,
>> >
>> > is there any difference between np.nan, np.NaN and np.NAN? they really
>> > confuse me
>> > they are all Not a Numbe
2011/10/14 Matthew Brett
> Hi,
>
> On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE wrote:
> > Dear all,
> >
> > is there any difference between np.nan, np.NaN and np.NAN? they really
> > confuse me
> > they are all Not a Number?
> >
> > In [75]: np.nan==np.NaN
> > Out[75]: False
> >
> > In [77]: n
good answer
2011/10/14 Matthew Brett
> Hi,
>
> On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE wrote:
> > Dear all,
> >
> > is there any difference between np.nan, np.NaN and np.NAN? they really
> > confuse me
> > they are all Not a Number?
> >
> > In [75]: np.nan==np.NaN
> > Out[75]: False
>
Hi,
On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE wrote:
> Dear all,
>
> is there any difference between np.nan, np.NaN and np.NAN? they really
> confuse me
> they are all Not a Number?
>
> In [75]: np.nan==np.NaN
> Out[75]: False
>
> In [77]: np.NaN==np.NAN
> Out[77]: False
The nan value is not
Dear all,
is there any difference between np.nan, np.NaN and np.NAN? they really
confuse me
they are all Not a Number?
In [75]: np.nan==np.NaN
Out[75]: False
In [77]: np.NaN==np.NAN
Out[77]: False
Thanks a lot,
Chao
--
*
Thanks Josef, you're right.
Could you explain me what's the difference between
In [4]: a=np.arange(10)
In [5]: a.shape
Out[5]: (10,)
and
In [6]: a=np.arange(10).reshape(10,1)
In [7]: a.shape
Out[7]: (10, 1)
(10) means the first a is only a one-dimensional ndarray, but the (10,1)
means the s
26 matches
Mail list logo