问题
On a computer with 4GB of memory this simple interpolation leads to a memory error:
(based on: 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')
I thought about cutting the data into chunks, but is there a way I can perform this cubic spline interpolation without requiring so much memory? Why does it even get in trouble?
回答1:
If you look at the traceback when the error occurs, you'll see something like:
---------------------------------------------------------------------------
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:
The function that is failing is scipy.interpolate._fitpack._bsplmat(order, xk). This function creates a 2-d array of 64-bit floats with shape (len(xk), len(xk) + order - 1). In your case, this is over 51GB.
Instead of interp1d, see if InterpolatedUnivariateSpline works for you. For example,
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)
I don't get a memory error with this.
来源:https://stackoverflow.com/questions/21435648/cubic-spline-memory-error