如何实现线性插值? [英] How to implement linear interpolation?

问题描述

说我得到的数据如下:

x = [1, 2.5, 3.4, 5.8, 6]
y = [2, 4, 5.8, 4.3, 4]

我想设计一个函数，该函数将在1和2.5之间，在2.5到3.4之间进行线性插值，依此类推，等等.

I want to design a function that will interpolate linearly between 1 and 2.5, 2.5 to 3.4, and so on using Python.

我已经尝试浏览此Python教程，但是我仍然无法理解

I have tried looking through this Python tutorial, but I am still unable to get my head around it.

推荐答案

据我了解您的问题，您想编写一些函数y = interpolate(x_values, y_values, x)，该函数将在某些x上为您提供y值吗?然后，基本思路如下:

As I understand your question, you want to write some function y = interpolate(x_values, y_values, x), which will give you the y value at some x? The basic idea then follows these steps:

1. 在x_values中查找值的索引，这些值定义了包含x的间隔.例如，对于带有示例列表的x=3，包含间隔将为[x1,x2]=[2.5,3.4]，索引将为i1=1，i2=2
2. 通过(y_values[i2]-y_values[i1])/(x_values[i2]-x_values[i1])(即dy/dx)计算该间隔的斜率.
3. x处的值现在是x1处的值加上斜率乘以距x1处的距离.

1. Find the indices of the values in x_values which define an interval containing x. For instance, for x=3 with your example lists, the containing interval would be [x1,x2]=[2.5,3.4], and the indices would be i1=1, i2=2
2. Calculate the slope on this interval by (y_values[i2]-y_values[i1])/(x_values[i2]-x_values[i1]) (ie dy/dx).
3. The value at x is now the value at x1 plus the slope multiplied by the distance from x1.

您还需要确定如果x在x_values的间隔之外，会发生什么情况，这是一个错误，或者您可以插值向后"，假设斜率与第一个/最后一个间隔相同.

You will additionally need to decide what happens if x is outside the interval of x_values, either it's an error, or you could interpolate "backwards", assuming the slope is the same as the first/last interval.

有此帮助吗?还是您需要更具体的建议?

Did this help, or did you need more specific advice?

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