我正在尝试开始使用 scipy，并且在使用“线性”以外的其他类型（我尝试了“零”和“立方”）时，似乎无法从interpolation.interp1d中获取示例。
我在谷歌搜索时找不到遇到同样问题的人，所以我想这对我来说是愚蠢的。
我在 python 2.73 和 OSX 10.8 上使用 scipy 0.11
代码不起作用：
from scipy import interpolate
import numpy as np
x = np.arange(0, 10)
y = np.exp(-x/3.0)
f = interpolate.interp1d(x, y, kind="zero")
xnew = np.arange(0,9, 0.1)
ynew = f(xnew) # use interpolation function returned by `interp1d`
plt.plot(x, y, 'o', xnew, ynew, '-')
plt.show()
IndexError Traceback (most recent call last)
<ipython-input-1-23bb96a1589b> in <module>()
4 f = interpolate.interp1d(x, y, kind="zero")
5 xnew = np.arange(0,9, 0.1)
----> 6 ynew = f(xnew) # use interpolation function returned by `interp1d`
7 plt.plot(x, y, 'o', xnew, ynew, '-')
8 plt.show()
/usr/local/lib/python2.7/site-packages/scipy/interpolate/interpolate.pyc in __call__(self, x_new)
394 out_of_bounds = self._check_bounds(x_new)
395
--> 396 y_new = self._call(x_new)
397
398 # Rotate the values of y_new back so that they correspond to the
/usr/local/lib/python2.7/site-packages/scipy/interpolate/interpolate.pyc in _call_spline(self, x_new)
370 def _call_spline(self, x_new):
371 x_new =np.asarray(x_new)
--> 372 result = spleval(self._spline,x_new.ravel())
373 return result.reshape(x_new.shape+result.shape[1:])
374
/usr/local/lib/python2.7/site-packages/scipy/interpolate/interpolate.pyc in spleval((xj, cvals, k), xnew, deriv)
833 res[sl].imag = _fitpack._bspleval(xx,xj,cvals.imag[sl],k,deriv)
834 else:
--> 835 res[sl] = _fitpack._bspleval(xx,xj,cvals[sl],k,deriv)
836 res.shape = oldshape + sh
837 return res
IndexError: too many indices
在这里启动调试控制台时，我可以将其缩小到 cvals[sl] 导致错误
sl = (slice(None, None, None), 0) # <-- I don't really get the slice part here...
cvals = array([ 1. , 0.71653131, 0.51341712, 0.36787944, 0.26359714,
0.1888756 , 0.13533528, 0.09697197, 0.06948345])
有人可以重现这个还是我的机器上有问题？