我的观点是,这将是一个很好的能力,可以包含在interpolate
.
也就是说,这归结为掩盖多个地方np.nan
存在。我会用一些东西来结束它numpy
方便的函数中的逻辑。
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
对于那些对使用高级通用解决方案感兴趣的人numpy
技巧
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]