[SciPy-dev] Is this a bugfix for scipy.hilbert?
josef.pktd@gmai...
josef.pktd@gmai...
Fri Jan 15 01:10:47 CST 2010
On Fri, Jan 15, 2010 at 1:44 AM, Ariel Rokem <arokem@berkeley.edu> wrote:
> Yes - looks good. Except I would prefer to eventually set the axis to
> default to -1, to be consistent with signal.fft (and also np.fft.fft) which
> has axis=-1.
I'm indifferent to the default axis, from a quick look and my
experience there are not many functions with axis arguments in signal.
So I'm fine with switching to axis=-1. We should do it with this
bugfix, since until now the function wasn't correct anyway for 2d.
>
> As for whether it's doing what it's supposed to do, for what it's worth - it
> seems to do similar things to what Matlab's 'hilbert' function does on a few
> simple examples I tried out.
I was reading briefly on wikipedia, and checked with fftpack.hilbert,
which returns the same array as signal.hilbert(a).imag, but I didn't
manage to figure out why fftpack.hilbert only allows 1d (i got lost
starting at convolve.pyf)
Could you write a simple test case compared to matlab, e.g. 10by3 as
in my example, for both axis, or 10by6 if 10by3 doesn't make sense?
If nobody objects, I can commit the change with axis=-1.
Josef
>
> Cheers,
>
> Ariel
>
>
>
> On Thu, Jan 14, 2010 at 8:53 PM, <josef.pktd@gmail.com> wrote:
>>
>> On Thu, Jan 14, 2010 at 11:27 PM, <josef.pktd@gmail.com> wrote:
>> > On Thu, Jan 14, 2010 at 11:02 PM, <josef.pktd@gmail.com> wrote:
>> >> On Thu, Jan 14, 2010 at 10:54 PM, <josef.pktd@gmail.com> wrote:
>> >>> On Thu, Jan 14, 2010 at 10:24 PM, Ariel Rokem <arokem@berkeley.edu>
>> >>> wrote:
>> >>>> Hi everyone,
>> >>>>
>> >>>> I have been trying to use scipy.signal.hilbert and I got the
>> >>>> following
>> >>>> puzzling result:
>> >>>>
>> >>>> In [22]: import scipy
>> >>>>
>> >>>> In [23]: scipy.__version__ #I have r6182
>> >>>> Out[23]: '0.8.0.dev'
>> >>>>
>> >>>> In [24]: import scipy.signal as signal
>> >>>>
>> >>>> In [25]: a = np.random.rand(100,100)
>> >>>>
>> >>>> In [26]: np.abs(signal.hilbert(a[-1]))
>> >>>> Out[26]:
>> >>>> array([ 0.57567681, 0.25918624, 0.50207097, 0.51834052,
>> >>>> 0.24293389,
>> >>>> 0.5779464 , 0.6515758 , 0.89973173, 1.00275444,
>> >>>> 0.37352935,
>> >>>> 0.62332717, 0.93599749, 0.40651376, 0.65088756, 0.8332281
>> >>>> ,
>> >>>> 0.5770101 , 0.9288512 , 0.46671906, 0.41536055,
>> >>>> 0.71418068,
>> >>>> 0.81250913, 0.07652627, 0.72939072, 0.26755626,
>> >>>> 0.36396146,
>> >>>> 0.59725999, 1.02264694, 0.41227986, 0.98122853,
>> >>>> 0.71906675,
>> >>>> 0.58582611, 0.77288117, 0.3217015 , 0.65261394,
>> >>>> 0.11947618,
>> >>>> 0.75632703, 0.43432935, 0.52182485, 1.0277177 ,
>> >>>> 1.01104986,
>> >>>> 0.3023265 , 0.6024772 , 0.69257548, 0.55418735,
>> >>>> 0.46259052,
>> >>>> 0.25832231, 0.38278355, 0.45508532, 0.26215872,
>> >>>> 0.34207947,
>> >>>> 0.80704729, 0.80755477, 0.95317178, 0.97458885,
>> >>>> 0.58762294,
>> >>>> 0.82540618, 0.62005585, 0.82494646, 1.04221293,
>> >>>> 0.14983027,
>> >>>> 1.01571579, 0.99381328, 0.24158714, 0.84256569,
>> >>>> 0.53418924,
>> >>>> 0.24067628, 0.90489883, 1.02217747, 0.34988034, 0.5310065
>> >>>> ,
>> >>>> 0.48135002, 1.03020269, 0.6013679 , 0.46062485, 0.3918485
>> >>>> ,
>> >>>> 0.21554545, 0.31704519, 0.04868385, 0.1787766 ,
>> >>>> 0.37361852,
>> >>>> 0.21977912, 0.7649772 , 0.77867281, 0.37684278,
>> >>>> 0.64432638,
>> >>>> 0.77494951, 0.87106309, 0.77611484, 0.52666801,
>> >>>> 0.88683667,
>> >>>> 0.69164967, 0.98618191, 0.84811375, 0.35934198,
>> >>>> 0.32650478,
>> >>>> 0.1752677 , 0.60574454, 0.5109132 , 0.52332287,
>> >>>> 0.99777805])
>> >>>>
>> >>>> In [27]: np.abs(signal.hilbert(a))[-1]
>> >>>> Out[27]:
>> >>>> array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
>> >>>> 0.,
>> >>>> 0., 0., 0., 0., 0., 0., 0., 0., 0.])
>> >>>>
>> >>>>
>> >>>>
>> >>>> ----------------------------------------------------------------------
>> >>>>
>> >>>> I was expecting both of these to have the same values - am I missing
>> >>>> something?
>> >>>>
>> >>>> I think that the following solves this issue, but now I am not that
>> >>>> sure
>> >>>> whether it does what it is supposed to do and I couldn't find a test
>> >>>> for
>> >>>> this in test_signaltools.py. Does anyone know of a good test-case for
>> >>>> the
>> >>>> analytic signal, that I could create for this?
>> >>>>
>> >>>> Index: scipy/signal/signaltools.py
>> >>>> ===================================================================
>> >>>> --- scipy/signal/signaltools.py (revision 6182)
>> >>>> +++ scipy/signal/signaltools.py (working copy)
>> >>>> @@ -1062,13 +1062,13 @@
>> >>>> """
>> >>>> x = asarray(x)
>> >>>> if N is None:
>> >>>> - N = len(x)
>> >>>> + N = x.shape[-1]
>> >>>> if N <=0:
>> >>>> raise ValueError, "N must be positive."
>> >>>> if iscomplexobj(x):
>> >>>> print "Warning: imaginary part of x ignored."
>> >>>> x = real(x)
>> >>>> - Xf = fft(x,N,axis=0)
>> >>>> + Xf = fft(x,N,axis=-1)
>> >>>> h = zeros(N)
>> >>>> if N % 2 == 0:
>> >>>> h[0] = h[N/2] = 1
>> >>>> @@ -1078,7 +1078,7 @@
>> >>>> h[1:(N+1)/2] = 2
>> >>>>
>> >>>> if len(x.shape) > 1:
>> >>>> - h = h[:, newaxis]
>> >>>> + h = h[newaxis,:]
>> >>>> x = ifft(Xf*h)
>> >>>> return x
>> >>>
>> >>> I think your change would break the currently advertised behavior,
>> >>> axis=0 (The transformation is done along the first axis)
>> >>>
>> >>> but fft and ifft have default axis=-1
>> >>>
>> >>> fft in hilbert uses axis=0 as in docstring
>> >>> but ifft uses default axis=-1
>> >>>
>> >>> so, I would think the fix should be x = ifft(Xf*h, axis=0)
>> >>>
>> >>> But as it currently looks like the axis argument doesn't work anyway,
>> >>> there wouldn't be much breakage if the axis would be included as an
>> >>> argument and default to -1.
>> >>> However, I don't know what the "standard" for scipy.signal is for
>> >>> default axis.
>> >>>
>> >>> Josef
>> >>
>> >> after adding axis to ifft:
>> >>>>> print hilbert(aa).real
>> >> [[ 0.82584851 0.15215031 0.14767381]
>> >> [ 0.95021675 0.16803995 0.43562964]
>> >> [ 0.13033881 0.06198952 0.70729614]
>> >> [ 0.69409563 0.06962778 0.72552601]
>> >> [ 0.34297612 0.50579001 0.86463304]
>> >> [ 0.28355261 0.21626889 0.85165102]
>> >> [ 0.49481491 0.21290645 0.71416814]
>> >> [ 0.2645843 0.95783096 0.77514016]
>> >> [ 0.38735994 0.14274852 0.56344808]
>> >> [ 0.88084015 0.39879649 0.64949951]]
>> >>>>> print hilbert(aa[:,:1]).real
>> >> [[ 0.82584851]
>> >> [ 0.95021675]
>> >> [ 0.13033881]
>> >> [ 0.69409563]
>> >> [ 0.34297612]
>> >> [ 0.28355261]
>> >> [ 0.49481491]
>> >> [ 0.2645843 ]
>> >> [ 0.38735994]
>> >> [ 0.88084015]]
>> >>
>> >> but it treats a 1d array as row vector and transforms along zero axis
>> >> of length 1, and not along the length of the array.
>> >> so another fix to handle 1d arrays correctly should be done
>> >>
>> >>>>> print hilbert(aa[:,1]).real
>> >> [ 0.15215031 0.16803995 0.06198952 0.06962778 0.50579001
>> >> 0.21626889
>> >> 0.21290645 0.95783096 0.14274852 0.39879649]
>> >>>>> aa[:,1]
>> >> array([ 0.15215031, 0.16803995, 0.06198952, 0.06962778, 0.50579001,
>> >> 0.21626889, 0.21290645, 0.95783096, 0.14274852, 0.39879649])
>> >>>>>
>> >
>> > there's something wrong with my example, the real part is the same
>> > which confused me
>> >
>> > it works correctly with 1d
>> >
>> >>>> np.abs(hilbert(aa[:,0]))
>> > array([ 0.83251128, 1.04487091, 0.27702083, 0.69901499, 0.49170197,
>> > 0.31227114, 0.49505637, 0.26461488, 0.61385196, 0.90716272])
>> >
>> >>>> np.abs(hilbert(aa[:,:1])).T
>> > array([[ 0.83251128, 1.04487091, 0.27702083, 0.69901499, 0.49170197,
>> > 0.31227114, 0.49505637, 0.26461488, 0.61385196,
>> > 0.90716272]])
>> >
>> >>>> np.abs(hilbert(aa))[:,0]
>> > array([ 0.83251128, 1.04487091, 0.27702083, 0.69901499, 0.49170197,
>> > 0.31227114, 0.49505637, 0.26461488, 0.61385196, 0.90716272])
>> >
>> > besides reading the docstring, I don't know what hilbert is supposed
>> > to be good for.
>>
>> Would something like the function in the attachment do ?
>>
>>
>>
>> > Josef
>> >
>> >
>> >> Josef
>> >>
>> >>
>> >>>
>> >>>>
>> >>>>
>> >>>> Cheers,
>> >>>>
>> >>>> Ariel
>> >>>> --
>> >>>> Ariel Rokem
>> >>>> Helen Wills Neuroscience Institute
>> >>>> University of California, Berkeley
>> >>>> http://argentum.ucbso.berkeley.edu/ariel
>> >>>>
>> >>>> _______________________________________________
>> >>>> SciPy-Dev mailing list
>> >>>> SciPy-Dev@scipy.org
>> >>>> http://mail.scipy.org/mailman/listinfo/scipy-dev
>> >>>>
>> >>>>
>> >>>
>> >>
>> >
>>
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>
>
>
> --
> Ariel Rokem
> Helen Wills Neuroscience Institute
> University of California, Berkeley
> http://argentum.ucbso.berkeley.edu/ariel
>
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