列表中点之间的线性插值

Question

I have a list in python like this [4, 0, 0, 6, 0, 8, 0, 0, 0, 3] , and I want to convert into something like this [4, 4.67, 5.33, 6, 7, 8, 6.75, 5.5, 4.25, 3] . Basically just replacing zeroes with an interpolation of the points already in the list. Any ideas?

Answer 1

One way using pandas.Series.interpolate :

import pandas as pd

pd.Series([i  if i else np.nan for i in l]).interpolate().tolist()

Output:

[4.0,
 4.666666666666667,
 5.333333333333333,
 6.0,
 7.0,
 8.0,
 6.75,
 5.5,
 4.25,
 3.0]

Answer 2

Using accumulate from itertools, you can find the starting and ending indexes for streaks of zeros around each position. Then use these ranges to compute a linear ratio for each zero position relative to its starting and ending non-zero range boundaries:

from itertools import accumulate

n = [4, 0, 0, 6, 0, 8, 0, 0, 0, 3]
starts = accumulate(range(len(n)),lambda a,b: b if n[b] else a)
ends   = [*accumulate(reversed(range(len(n))),lambda a,b: b if n[b] else a)][::-1]
inter  = [ n[i] or n[s]+(n[e]-n[s])*(i-s)/(e-s) for i,(s,e) in enumerate(zip(starts,ends)) ]

# inter = [4, 4.666666666666667, 5.333333333333333, 6, 7.0, 8, 6.75, 5.5, 4.25, 3]

The starts list will contain the index of the previous non-zero value for each position (uses the position itself for non-zero values):

[0, 0, 0, 3, 3, 5, 5, 5, 5, 9]

The ends list contains the index of the next non-zero values

[0, 3, 3, 3, 5, 5, 9, 9, 9, 9]

Combining these two lists using zip we obtain all the information needed to compute the intermediate values:

                         start  end    range  position        Interpolation 
index value start end    value  value  size   in range     ratio        value
(i)   n[i]  (s)   (e)    n[s]   n[e]   e-s    i-s       (i-s)/(e-s)   see below
 0     4     0     0      4      4      0     0            -----          4
 1     0     0     3      4      6      3     1            0.67          4.67
 2     0     0     3      4      6      3     2            0.33          5.33
 3     6     3     3      6      6      0     0            -----          6
 4     0     3     5      6      8      2     1            0.50          7.00
 5     8     5     5      8      8      0     0            -----          8
 6     0     5     9      8      3      4     1            0.75          6.75
 7     0     5     9      8      3      4     2            0.50          5.50
 8     0     5     9      8      3      4     3            0.25          4.25
 9     3     9     9      3      3      0     0            -----          3

Keeping non-zero values where present and calculating the interpolated value as startValue + (endValue-startValue) x InterpolationRatio for zero positions.