scipy.interpolate.interp1d在以十进制值开头x值时是否有问题? [英] Does scipy.interpolate.interp1d have problems with decimal values leading the x values?

问题描述

我试图通过scipy使用 interp1d()来插值一些数据，但是我一直遇到超限或范围错误.经过数小时的谷歌搜索，我现在知道x值未按升序排列会导致出现相同的错误，但我已经确定这不是问题.据我所知，看起来 interp1d()不喜欢第一个值中的小数.我想念什么吗?

I'm trying to use interp1d() from scipy to interpolate some data, but I keep hitting an out or range error. After hours of Googling, I now know that x values not in increasing order will cause the same error I'm getting but I've already made sure that's not the problem. As far as I can tell, it looks like interp1d() doesn't like decimals in the first value. Am I missing something?

我的问题的简化版本:

以下运行正常.

import numpy as np
from scipy.interpolate import interp1d

interp1d(np.array([1, 2, 3, 4, 5, 6]),
         np.array([2, 4, 6, 8, 10, 12]), kind='cubic')(np.linspace(1, 6, num=40))

但是，这:

interp1d(np.array([1.1, 2.2, 3.3, 4.4, 5.5, 6.6]),
     np.array([2, 4, 6, 8, 10, 12]), kind='cubic')(np.linspace(1, 6, num=40))

返回:

ValueError:x_new中的值低于插值范围.

ValueError: A value in x_new is below the interpolation range.

但是，这很好.

interp1d(np.array([1.0, 2.2, 3.3, 4.4, 5.5, 6.6]),
         np.array([2, 4, 6, 8, 10, 12]), kind='cubic')(np.linspace(1, 6, num=40))

推荐答案

inter1d返回一个函数，该函数可让您在数据域内内插 .当您使用

inter1d returns a function which allows you to interpolate within the domain of the data. When you use

interp1d(np.array([1.1, 2.2, 3.3, 4.4, 5.5, 6.6]),
     np.array([2, 4, 6, 8, 10, 12]), kind='cubic')(np.linspace(1, 6, num=40))

数据域是间隔[1.1, 6.6]，它是x-values的最小值到最大值.

the domain of the data is the interval [1.1, 6.6], the minimum to maximum values of the x-values.

由于1在np.linspace(1, 6, num=40)中，并且1位于[1.1, 6.6]之外，所以interp1d举起

Since 1 is in np.linspace(1, 6, num=40) and 1 lies outside [1.1, 6.6], interp1d raises

ValueError: A value in x_new is below the interpolation range.

当您尝试在x-values np.linspace(1, 6, num=40)处插值数据时.

when you try to interpolate the data at the x-values np.linspace(1, 6, num=40).

将数据的x-values更改为

np.array([1.0, 2.2, 3.3, 4.4, 5.5, 6.6])

然后将数据域扩展到[1.0, 6.6]，现在包括1.因此，现在可以在np.linspace(1, 6, num=40)处进行插值了.

then the domain of the data is expanded to [1.0, 6.6], which now includes 1. Hence, interpolating at np.linspace(1, 6, num=40) now works.

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