linear interpolation using list comprehension-CodePudding

Consider the following list

import numpy as np
import pandas as pd

l = [1,4,6,np.NaN,20,np.Nan,24]

I know I can replace the nan values using simple linear interpolation using pandas interpolate as follows

pd.Series([1,4,6,np.NaN,20,np.NaN,24]).interpolate()
Out[38]: 
0     1.0
1     4.0
2     6.0
3    13.0
4    20.0
5    22.0
6    24.0
dtype: float64

My question is: how can I get the same result by only using list comprehensions, standard numpy functions, but no built-in interpolation function (pd.interpolate() or numpy.interp()`)? That is, using directly the formula for linear interpolation between two points.

CodePudding user response：

l = [1,4,6,np.nan,20,np.nan,24]

res = [l[i] if not np.isnan(l[i]) else (l[i-1] l[i 1])/2 for i in range(len(l))]
print(res)

CodePudding user response：

Not sure if it is really a fit for this question since it is not just a list comprehension, but I've figured out a solution that works for gaps with more than 1 consecutive nan:

import numpy as np

l = [1,4,6,np.nan,20,np.nan,24, 30, 31, np.nan, np.nan, 70, 75]

# 1 -> entry is nan
nans = np.isnan(l)

# 1 -> from number to nan, -1 -> from nan to number
diffs = np.diff(list(map(int, nans)))

# get "gap of nans" begin and end indices
gap_starts = np.where(diffs == 1)[0]
gap_ends = np.where(diffs == -1)[0]


for begin, end in zip(gap_starts, gap_ends):
    # number of nans in the gap
    nans_n = end - begin
    # difference of gap extrema
    nan_diff = abs(l[begin] - l[end 1])
    # step to add at each nan
    step = round(nan_diff / (nans_n   1))
    # interpolate section from begin to end
    filling = [l[begin]   (step * n) for n in range(1, nans_n   1)]
    # fix l with interpolated values
    l[begin 1:end 1] = filling
    
    
print(l)

produces

[1, 4, 6, 13, 20, 22, 24, 30, 31, 44, 57, 70, 75]