Interpolate only if single NaN

Question

Is there a way in pandas to interpolate only single missing data points? That is, if there is 2+ consecutive NaN\'s, I\'d like to leave them alone.

so, as an example:

Answer 1

My opinion is that this would be a great capability to include in interpolate.
That said, this boils down to masking the places where more than one np.nan exist. I'll wrap that up with some numpy logic in a handy function.

def cnan(s):
    v = s.values
    k = v.size
    n = np.append(np.isnan(v), False)
    m = np.empty(k, np.bool8)
    m.fill(True)
    i = np.where(n[:-1] & n[1:])[0] + np.arange(2)
    m[i[i < k]] = False
    return m

s.interpolate().where(cnan(s))

0    1.0
1    1.5
2    2.0
3    3.0
4    NaN
5    NaN
6    4.5
dtype: float64

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For those interested in a general solution using advanced numpy techniques

import pandas as pd
import numpy as np
from numpy.lib.stride_tricks import as_strided as strided

def mask_knans(a, x):
    a = np.asarray(a)
    k = a.size
    n = np.append(np.isnan(a), [False] * (x - 1))
    m = np.empty(k, np.bool8)
    m.fill(True)

    s = n.strides[0]
    i = np.where(strided(n, (k + 1 - x, x), (s, s)).all(1))[0][:, None]
    i = i + np.arange(x)
    i = pd.unique(i[i < k])

    m[i] = False

    return m

demo

a = np.array([1, np.nan, np.nan, np.nan, 3, np.nan, 4, 5, np.nan, np.nan, 6, 7])

print(mask_knans(a, 3))

[ True False False False  True  True  True  True  True  True  True  True]

Answer 2

s[(s.shift(-1).notnull()) & (s.shift(1).notnull())] = (s.shift(-1) + s.shift(1))/2

Actually,

s[s.isnull()] = (s.shift(-1) + s.shift(1))/2

works as well, if you are doing simple interpolation.