scipy interp1d外插方法fill_value =元组不工作

我想外推函数拟合。 scipy.interpolate.interp1d应该可以做到这一点（请参阅doc片段）。相反，我得到“ValueError：x_new中的值低于插值范围。”scipy interp1d外插方法fill_value =元组不工作

使用：蟒蛇2.7.12，numpy的1.13.3，SciPy的0.19.1

fill_value : array-like or (array-like, array_like) or "extrapolate", optional - if a ndarray (or float), this value will be used to fill in for requested points outside of the data range. If not provided, then the default is NaN. The array-like must broadcast properly to the dimensions of the non-interpolation axes. - If a two-element tuple, then the first element is used as a fill value for x_new < x[0] and the second element is used for x_new > x[-1] . Anything that is not a 2-element tuple (e.g., list or ndarray, regardless of shape) is taken to be a single array-like argument meant to be used for both bounds as below, above = fill_value, fill_value .

import numpy as np 
from scipy.interpolate import interp1d 
# make a time series 
nobs = 10 
t = np.sort(np.random.random(nobs)) 
x = np.random.random(nobs) 
# compute linear interp (with ability to extrapolate too) 
f1 = interp1d(t, x, kind='linear', fill_value='extrapolate') # this works 
f2 = interp1d(t, x, kind='linear', fill_value=(0.5, 0.6)) # this doesn't

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2017-10-21 John Mahoney

按照documentation，interp1d默认为外插养ValueError时fill_value='extrapolate'，或者指定bounds_error=False除。

In [1]: f1 = interp1d(t, x, kind='linear', fill_value=(0.5, 0.6), bounds_error=False) 

In [2]: f1(0) 
Out[2]: array(0.5)

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2017-10-21 01:55:08 Craig

谢谢，克雷格。我认为给fill_value提供一个元组值会导致它被使用（可能除非你由于某种原因明确地设置了bounds_error = False）。显然这不是interp1d的行为。 –

scipy interp1d外插方法fill_value =元组不工作

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