大熊猫数据框列的插值

Question

I need to do interpolation between 2 columns of pandas.DataFrame , to fill the column between them. Here are a few rows of my data frame , the column to be filled is col2 :

col1  col2  col3
2.35    1   2.37
2.47    1   2.49
2.51    1   2.53
2.57    1   2.58
2.54    1   2.57

So for interpolation I want to use numpy.interp(x,xp,fp) , but I can't figure out how to organize my data so that I will be able to use it. That is because the interpolation should be between col1 and col3 for each row. For example, for the first row I need it to look like that:

xp=[1,3]
fp=[2.47,2.49]
x=2
y=numpy.interp(x,xp,fp)

and then fill first row of col2 with y . And I need to do that again and again for each row. How ?

Answer 1

This will get you to iterate over every row, replacing the value between two cells. But the interpolation does not seem to be working. I don't have much experience with it, so I couldn't find an easy fix online. That's the only line not doing the change of values. (Idon't know what xp or x do, so I kept them)

xp=[1,3]
x = 2
for rowNr in range(len(df.index)):
    fp=[df.iat[rowNr, 0], df.iat[rowNr, 2]]
    df.iat[rowNr, 1] = numpy.interp(x, xp, fp)

Answer 2

As written, the x-values are static (unless I misunderstand your problem) with values of 1 and 3. You want to do a linear interpolation between these values and two y-values that change. You simply average the y-values and that is the linear-interpolated value. Don't overlook simple/obvious solutions for something fancy (advice I try to remember all the time).

df.col2 = df[["col1", "col3"]].mean(axis=1)

BEGIN EDIT

Andre's solution should work (haven't tested it myself, but should). However, this requires iterating over every row, which can be slow. Further, there is a simple mathematical solution that allows you to operate on arrays, which should be faster.

Linear interpolation follows the general form of:

y = y0 + (x - x0) * (y1 - y0) / (x1 - x0)

Putting this in terms of dataframes/code:

df.col2 = df.col1 + (x - xp[0]) * (df.col2 - df.col1) / (xp[1] - xp[0])

I think that got translated correctly, but the formula above holds. Just implement it in your code or loop through each row and call the numpy.interp function. Either way, you should be fine.