[SciPy-User] [SciPy-user] help about lfilter ????

[josef.pktd at gmail.com]

On Thu, Jan 28, 2010 at 7:51 PM, persia <m.abdollahi at gmail.com> wrote:
>
> lfilter is the linear filtering function in the scipy.signal module. can
> someone tell me what is wrong with this function please ?
>
> I used this function in the following way and it gave me that error !! :
>
> In [204]:
> lfilter(array([1,2,3]),array([-4,5,6]),array([9,8,7,6,5,4,3,2,1]),zi=array([0,0]))
> ---------------------------------------------------------------------------
> <type 'exceptions.ValueError'>            Traceback (most recent call last)
>
>
>
> <type 'exceptions.ValueError'>: linear_filter not available for this type
>
>
> why is it so ??!!!! what is wrong with this "type " ? what is it meant by
> this type anyway ?! i tried with the same arguments in matlab with the
> corresponding function and it worked !!
> and if i multiply the second argument vector say by .4, this will happen :
>
> lfilter(array([1,2,3]),.4*array([-4,5,6]),array([9,8,7,6,5,4,3,2,1]),zi=array([0,0]))
>
>
> (array([   -5.625     ,   -23.28125   ,   -68.7890625 ,  -148.40820312,
>        -312.44384766,  -633.16711426, -1276.37466431, -2557.71900177,
>       -5120.46074867]),
>  array([-10242.1544385 ,  -7682.56612301]))
>
>
> see ? no problem then ?!! isnt this wierd ? please help !

it looks like lfilter doesn't like integers

>>> signal.lfilter(array([1,2,3]),array([-4.,5,6]),array([9,8,7,6,5,4,3,2,1]),zi=array([0,0]))
(array([   -2.25      ,    -9.3125    ,   -27.515625  ,   -59.36328125,
        -124.97753906,  -253.2668457 ,  -510.54986572, -1023.08760071,
       -2048.18429947]), array([-4096.8617754, -3073.0264492]))

note I only changed a 4 to 4.

can you file a ticket?

Josef


>
> thanx
> --
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