SciPy的，fftpack和float64 [英] scipy, fftpack and float64

问题描述

我想用DCT功能从scipy.fftpack与numpy的float64的数组。然而，现在看来，这只是实施了np.float32。有没有快速的解决办法，我可以做的就是这个工作？我赶紧看了进去，但我不知道所有的依赖。所以，一切都搞乱之前，我想我会问这里的技巧！

我已经发现迄今这个唯一的一点是此链接： http://mail.scipy.org/pipermail/scipy-svn/2010-September/004197.html

在此先感谢。

下面是它提出了一个ValueError错误：

  -------------------------------------- -------------------------------------
ValueError错误回溯（最新最后调用）
＆LT; IPython中输入-12-f09567c28e37＆GT;与＆lt;模块＆GT;（）
----＆GT; 1 scipy.fftpack.dct（C [100]）

/usr/local/Cellar/python/2.7.3/lib/python2.7/site-packages/scipy/fftpack/realtransforms.pyc的DCT（X，类型，N，轴，规范，overwrite_x）
    118升NotImplementedError（
    119正交化尚不支持DCT-I）
 - ＆GT; 120回_dct（X，类型，N，轴，规范=常态，overwrite_x = overwrite_x）
    121
    122高清IDCT（X，类型= 2，N =无，轴= -1，规范=无，overwrite_x = 0）：

/usr/local/Cellar/python/2.7.3/lib/python2.7/site-packages/scipy/fftpack/realtransforms.pyc在_dct（X，类型，N，轴，overwrite_x，标准化）
    215升ValueError错误（类型％D不理解％型）
    216其他：
 - ＆GT; 217升ValueError错误（DTYPE％S不支持％tmp.dtype）
    218
    219，如果规范化：

ValueError错误：DTYPE＆GT;不支持F8
 

解决方案

现在的问题是不是双precision。双precision当然是支持的。问题是，你有一点点端计算机和（也许从文件加载一个文件？）有大端数据，请注意＆GT; 在 DTYPE＆GT;不支持F8 。看来你只需将它转换为本地翻一番自己。如果你知道它的双precision，你可能只是想everytiong一旦转换成你的本地命令：

  C = c.astype（浮点）
 

虽然我猜你也可以检查 c.dtype.byteorder ，我想应该是 ='，如果，开关...沿东西：

 如果c.dtype.byteorder ='='！
    C = c.astype（c.dtype.newbyteorder（'='））
 

这应该也是，如果你碰巧有一个precision或整数...

I would like to use the dct functionality from the scipy.fftpack with an array of numpy float64. However, it seems it is only implemented for np.float32. Is there any quick workaround I could do to get this done? I looked into it quickly but I am not sure of all the dependencies. So, before messing everything up, I thought I'd ask for tips here!

The only thing I have found so far about this is this link : http://mail.scipy.org/pipermail/scipy-svn/2010-September/004197.html

Thanks in advance.

Here is the ValueError it raises:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-12-f09567c28e37> in <module>()
----> 1 scipy.fftpack.dct(c[100])

/usr/local/Cellar/python/2.7.3/lib/python2.7/site-packages/scipy/fftpack/realtransforms.pyc in dct(x, type, n, axis, norm, overwrite_x)
    118         raise NotImplementedError(
    119               "Orthonormalization not yet supported for DCT-I")
--> 120     return _dct(x, type, n, axis, normalize=norm, overwrite_x=overwrite_x)
    121 
    122 def idct(x, type=2, n=None, axis=-1, norm=None, overwrite_x=0):

/usr/local/Cellar/python/2.7.3/lib/python2.7/site-packages/scipy/fftpack/realtransforms.pyc in _dct(x, type, n, axis, overwrite_x, normalize)
    215             raise ValueError("Type %d not understood" % type)
    216     else:
--> 217         raise ValueError("dtype %s not supported" % tmp.dtype)
    218 
    219     if normalize:

ValueError: dtype >f8 not supported

解决方案

The problem is not the double precision. Double precision is of course supported. The problem is that you have a little endian computer and (maybe loading a file from a file?) have big endian data, note the > in dtype >f8 not supported. It seems you will simply have to cast it to native double yourself. If you know its double precision, you probably just want to convert everytiong to your native order once:

c = c.astype(float)

Though I guess you could also check c.dtype.byteorder which I think should be '=', and if, switch... something along:

if c.dtype.byteorder != '=':
    c = c.astype(c.dtype.newbyteorder('=')) 

Which should work also if you happen to have single precision or integers...

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