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
Hi,
On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE chaoyue...@gmail.com 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
2011/10/14 Matthew Brett matthew.br...@gmail.com
Hi,
On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE chaoyue...@gmail.com 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]:
On Fri, Oct 14, 2011 at 11:53 AM, Olivier Delalleau sh...@keba.be wrote:
2011/10/14 Matthew Brett matthew.br...@gmail.com
Hi,
On Fri, Oct 14, 2011 at 4:33 AM, Chao YUE chaoyue...@gmail.com wrote:
Dear all,
is there any difference between np.nan, np.NaN and np.NAN? they really
confuse
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 =
On Fri, Oct 14, 2011 at 7:04 AM, Neal Becker ndbeck...@gmail.com 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
What about
a=arange(len(v))
w=u[v,a]
?
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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 the
index
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
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 selects items from u
On Fri, Oct 14, 2011 at 10:24 AM, Alan G Isaac alan.is...@gmail.com 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
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 alan.is...@gmail.com 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]
Hi Rense (cross-posting to the numpy mailing list because these guys
are awesome),
On Oct 13, 10:01 pm, Rense Lange rense.la...@gmail.com wrote:
I have potentially millions of tuples v1,v2,v3 ..., observation and
I want to create frequency distributions conditional on the values of
discrete
On Fri, Oct 14, 2011 at 11:56 AM, Fabrice Silva si...@lma.cnrs-mrs.fr 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 alan.is...@gmail.com wrote:
As a simple example, if I have y0 and a white noise series e,
On Fri, Oct 14, 2011 at 12:49 PM, Alan G Isaac alan.is...@gmail.com 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,
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
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On Fri, Oct 14, 2011 at 1:26 PM, Alan G Isaac alan.is...@gmail.com 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
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 2:18 PM, Alan G Isaac alan.is...@gmail.com 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
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac alan.is...@gmail.com 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
On Fri, Oct 14, 2011 at 2:39 PM, Skipper Seabold jsseab...@gmail.com wrote:
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac alan.is...@gmail.com 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
On Fri, Oct 14, 2011 at 2:29 PM, josef.p...@gmail.com wrote:
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac alan.is...@gmail.com 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
On Fri, Oct 14, 2011 at 2:59 PM, josef.p...@gmail.com wrote:
On Fri, Oct 14, 2011 at 2:29 PM, josef.p...@gmail.com wrote:
On Fri, Oct 14, 2011 at 2:18 PM, Alan G Isaac alan.is...@gmail.com wrote:
On 10/14/2011 1:42 PM, josef.p...@gmail.com wrote:
If I remember correctly, signal.lfilter
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