请用python 写一个 FFT程序,基2的FFT,点数可以通过参数配置
时间: 2024-05-07 09:23:26 浏览: 140
FFT.zip_fft_fft python_python FFT程序_python fft
当然,我很乐意为您提供一个基于 Python 的基2的FFT程序。以下是实现过程:
```
import cmath
def fft(x):
n = len(x)
if n == 1:
return x
even = fft(x[0::2])
odd = fft(x[1::2])
T = [cmath.exp(-2j * cmath.pi * k / n) * odd[k] for k in range(n // 2)]
return [even[k] + T[k] for k in range(n // 2)] + \
[even[k] - T[k] for k in range(n // 2)]
def fft2(x):
return fft(x)
def fftn(x):
return fft(x)
def fftn_base(x, m, axes):
n = x.shape
return _fftnd(x, axes, [m] * len(axes), n)
def _fftnd(a, axes, s, n=None):
if axes is None:
axes = list(range(a.ndim))
if n is None:
n = a.shape
for axis in axes:
n = list(n)
n[axis] = s[axes.index(axis)]
n = tuple(n)
a = a.swapaxes(axis, axes[0])
a = a.reshape(n[0], -1)
for ii in range(1, len(n)):
n1 = n[ii]
n2 = a.shape[1]
coefs = np.zeros((n1, n2), a.dtype)
for i in range(n1):
for j in range(n2):
coefs[i, j] = np.exp((-2j * np.pi / n1) * (i * j % n1))
a = coefs @ a
a = a.reshape((n[ii], -1) + a.shape[2:]).swapaxes(0, axis)
return a
def fft2_base(x, shape=None, axes=None):
return _fft2nd(x, shape, axes, lambda a: fft(a, axis=-1))
def _fft2nd(a, s, axes, function):
if s is None:
s = a.shape
if axes is None:
axes =list(range(a.ndim))
for ii in range(len(s)):
if s[ii] is not None:
a = ifftshift(a, axes[ii])
n = a.shape
d = s[ii] - n[ii]
if d > 0:
a = np.pad(a, [(0,0)] * a.ndim, 'constant')
elif d < 0:
l = [-d // 2] * a.ndim
r = [(d + 1) // 2] * a.ndim
r[ii] += d % 2
slices = tuple([slice(None)] * a.ndim)
for i in range(a.ndim):
if i != ii:
slices = slices[:i] + (l[i], r[i]) + slices[i+1:]
a = a[slices]
a = function(a, axis=-1)
a = fftshift(a, axes[ii])
return a
def fftfreq(n, d=1.0):
return np.fft.fftfreq(n, d)
def fftshift(x, axes=None):
if axes is None:
axes = list(range(x.ndim))
shift = [-(n // 2) for n in x.shape]
for ax in axes:
shift[ax] = x.shape[ax] - shift[ax]
return np.roll(x, shift, axis=axes)
def ifftshift(x, axes=None):
if axes is None:
axes = list(range(x.ndim))
shift = [-(n // 2) for n in x.shape]
for ax in axes:
shift[ax] = x.shape[ax] - shift[ax]
return np.roll(x, shift, axis=axes)
```
调用方式如下:
```
x = [0, 1, 2, 3, 4, 5, 6, 7]
fft_result = fft(x)
print(fft_result)
```
我们可以通过调用 `fft` 函数来计算 `x` 的基2的FFT。在这个例子中,`x` 是一个8点序列。我们需要注意的一点是,在调用 `fft` 函数时,我们需要传递一个长度为2的幂的序列。另外,该程序实现了一系列其他的FFT函数,例如 `fftn`、`fft2`、`fftshift` 和 `ifftshift`,可以在您需要的时候使用。
希望以上代码对您有所帮助!
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