在具有 4GB 内存的计算机上,这种简单的插值会导致内存错误:
(基于:http://docs.scipy.org/doc/scipy/reference/tutorial/interpolate.html)
import numpy as np
from scipy.interpolate import interp1d
x = np.linspace(0, 10, 80000)
y = np.cos(-x**2/8.0)
f2 = interp1d(x, y, kind='cubic')
我考虑过将数据分成 block ,但有没有一种方法可以在不需要那么多内存的情况下执行此三次样条插值? 为什么它甚至会遇到麻烦?
最佳答案
如果您在错误发生时查看回溯,您会看到如下内容:
---------------------------------------------------------------------------
MemoryError Traceback (most recent call last)
<ipython-input-4-1e538e8d766e> in <module>()
----> 1 f2 = interp1d(x, y, kind='cubic')
/home/warren/local_scipy/lib/python2.7/site-packages/scipy/interpolate/interpolate.py in __init__(self, x, y, kind, axis, copy, bounds_error, fill_value)
390 else:
391 minval = order + 1
--> 392 self._spline = splmake(x, y, order=order)
393 self._call = self.__class__._call_spline
394
/home/warren/local_scipy/lib/python2.7/site-packages/scipy/interpolate/interpolate.py in splmake(xk, yk, order, kind, conds)
1754
1755 # the constraint matrix
-> 1756 B = _fitpack._bsplmat(order, xk)
1757 coefs = func(xk, yk, order, conds, B)
1758 return xk, coefs, order
MemoryError:
失败的函数是 scipy.interpolate._fitpack._bsplmat(order, xk)
。此函数创建一个 64 位 float 的二维数组,形状为 (len(xk), len(xk) + order - 1)
。在您的情况下,这超过了 51GB。
而不是 interp1d
,看看是否 InterpolatedUnivariateSpline
为你工作。例如,
import numpy as np
from scipy.interpolate import InterpolatedUnivariateSpline
x = np.linspace(0, 10, 80000)
y = np.cos(-x**2/8.0)
f2 = InterpolatedUnivariateSpline(x, y, k=3)
我没有遇到内存错误。
关于python - 三次样条内存错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21435648/