Is there a way to reduce scipy/numpy precision to

问题:

On my 64-bit Debian/Lenny system (4GByte RAM + 4GByte swap partition) I can successfully do:

v=array(10000*random([512,512,512]),dtype=np.int16)
f=fftn(v)

but with f being a np.complex128 the memory consumption is shocking, and I can't do much more with the result (e.g modulate the coefficients and then f=ifftn(f) ) without a MemoryError traceback.

Rather than installing some more RAM and/or expanding my swap partitions, is there some way of controlling the scipy/numpy "default precision" and have it compute a complex64 array instead ?

I know I can just reduce it afterwards with f=array(f,dtype=np.complex64); I'm looking to have it actually do the FFT work in 32-bit precision and half the memory.

回答1:

It doesn't look like there's any function to do this in scipy's fft functions ( see http://www.astro.rug.nl/efidad/scipy.fftpack.basic.html ).

Unless you're able to find a fixed point FFT library for python, it's unlikely that the function you want exists, since your native hardware floating point format is 128 bits. It does look like you could use the rfft method to get just the real-valued components (no phase) of the FFT, and that would save half your RAM.

I ran the following in interactive python:

>>> from numpy import *
>>>  v = array(10000*random.random([512,512,512]),dtype=int16)
>>> shape(v)
(512, 512, 512)
>>> type(v[0,0,0])
<type 'numpy.int16'>

At this point the RSS (Resident Set Size) of python was 265MB.

f = fft.fft(v)

And at this point the RSS of python 2.3GB.

>>> type(f)
<type 'numpy.ndarray'>
>>> type(f[0,0,0]) 
<type 'numpy.complex128'>
>>> v = []

And at this point the RSS goes down to 2.0GB, since I've free'd up v.

Using "fft.rfft(v)" to compute real-values only results in a 1.3GB RSS. (almost half, as expected)

Doing:

>>> f = complex64(fft.fft(v))

Is the worst of both worlds, since it first computes the complex128 version (2.3GB) and then copies that into the complex64 version (1.3GB) which means the peak RSS on my machine was 3.6GB, and then it settled down to 1.3GB again.

I think that if you've got 4GB RAM, this should all work just fine (as it does for me). What's the issue?

回答2:

Scipy 0.8 will have single precision support for almost all the fft code (The code is already in the trunk, so you can install scipy from svn if you need the feature now).