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 forx_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 asbelow, 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
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)
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。 –