How to implement linear interpolation

Question

Say I am given data as follows   x    1  2 5  3 4  5 8  6  y    2  4  5 8  4 3  4    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   I have tried looking through this Python tutorial  but I am still unable to get my head around it

User · Answer

import scipy interpolate y interp   scipy interpolate interp1d x  y  print y interp 5 0    scipy interpolate interp1d does linear interpolation by and can be customized to handle error conditions

User · Answer

Building on Lauritz  answer  here s a version with the following changes   Updated to python3  the map was causing problems for me and is unnecessary  Fixed behavior at edge values Raise exception when x is out of bounds Use   call   instead of   getitem     from bisect import bisect right  class Interpolate      def   init   self  x list  y list           if any y - x  lt   0 for x  y in zip x list  x list 1                  raise ValueError  x list must be in strictly ascending order            self x list   x list         self y list   y list         intervals   zip x list  x list 1    y list  y list 1            self slopes     y2 - y1     x2 - x1  for x1  x2  y1  y2 in intervals       def   call   self  x           if not  self x list 0   lt   x  lt   self x list -1                raise ValueError  x out of bounds            if x    self x list -1               return self y list -1          i   bisect right self x list  x  - 1         return self y list i    self slopes i     x - self x list i     Example usage    gt  gt  gt  interp   Interpolate  1  2 5  3 4  5 8  6    2  4  5 8  4 3  4    gt  gt  gt  interp 4  5 425

User · Answer

Instead of extrapolating off the ends  you could return the extents of the y list  Most of the time your application is well behaved  and the Interpolate x  will be in the x list  The  presumably  linear affects of extrapolating off the ends may mislead you to believe that your data is well behaved    Returning a non-linear result  bounded by the contents of x list and y list  your program s behavior may alert you to an issue for values greatly outside x list   Linear behavior goes bananas when given non-linear inputs   Returning the extents of the y list for Interpolate x  outside of x list also means you know the range of your output value  If you extrapolate based on x much  much less than x list 0  or x much  much greater than x list -1   your return result could be outside of the range of values you expected   def   getitem   self  x       if x  lt   self x list 0           return self y list 0      elif x  gt   self x list -1           return self y list -1      else          i   bisect left self x list  x  - 1         return self y list i    self slopes i     x - self x list i

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I thought up a rather elegant solution  IMHO   so I can t resist posting it   from bisect import bisect left  class Interpolate object       def   init   self  x list  y list           if any y - x  lt   0 for x  y in zip x list  x list 1                  raise ValueError  x list must be in strictly ascending order            x list   self x list   map float  x list          y list   self y list   map float  y list          intervals   zip x list  x list 1    y list  y list 1            self slopes     y2 - y1   x2 - x1  for x1  x2  y1  y2 in intervals       def   getitem   self  x           i   bisect left self x list  x  - 1         return self y list i    self slopes i     x - self x list i     I map to float so that integer division  python  lt   2 7  won t kick in and ruin things if x1  x2  y1 and y2 are all integers for some iterval   In   getitem   I m taking advantage of the fact that self x list is sorted in ascending order by using bisect left to  very  quickly find the index of the largest element smaller than x in self x list   Use the class like this   i   Interpolate  1  2 5  3 4  5 8  6    2  4  5 8  4 3  4     Get the interpolated value at x   4  y   i 4    I ve not dealt with the border conditions at all here  for simplicity  As it is  i x  for x  lt  1 will work as if the line from  2 5  4  to  1  2  had been extended to minus infinity  while i x  for x    1 or x  gt  6 will raise an IndexError  Better would be to raise an IndexError in all cases  but this is left as an exercise for the reader

User · Answer

def interpolate x1  float  x2  float  y1  float  y2  float  x  float        quot  quot  quot Perform linear interpolation for x between  x1 y1  and  x2 y2   quot  quot  quot       return   y2 - y1    x   x2   y1 - x1   y2     x2 - x1

User · Answer

Your solution did not work in Python 2 7  There was an error while checking for the order of the x elements  I had to change to code to this to get it to work   from bisect import bisect left class Interpolate object       def   init   self  x list  y list           if any  y - x  lt   0 for x  y in zip x list  x list 1                   raise ValueError  x list must be in strictly ascending order            x list   self x list   map float  x list          y list   self y list   map float  y list          intervals   zip x list  x list 1    y list  y list 1            self slopes     y2 - y1   x2 - x1  for x1  x2  y1  y2 in intervals      def   getitem   self  x           i   bisect left self x list  x  - 1         return self y list i    self slopes i     x - self x list i

User · Answer

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    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  Calculate the slope on this interval by  y values i2 -y values i1    x values i2 -x values i1    ie dy dx   The value at x is now the value at x1 plus the slope multiplied by the distance from x1    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   Did this help  or did you need more specific advice

[python] How to implement linear interpolation?

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