Python 用后续值填充数组

在我的dataframe中，我将得到一个只有很少非nan值的列。我想将非nan值用作前面所有包含nan值的行的分组变量。为了模拟它，我制作了以下数组：

count = np.array([np.NaN,np.NaN,np.NaN,3,np.NaN,np.NaN,6,np.NaN,np.NaN,np.NaN,np.NaN,np.NaN,12])
count = Series(count)

对于这个数组，我能够创建一个填充函数

def pad_expsamp_time(array):
    sect = np.zeros(array.size) # create array filled with zeros
    inds = array.index[array.notnull()] # select the non-zero values
    rev_inds = inds[::-1] # sort high to low
    # fill array with value until index of value. Repeat for lower values. 
    for i in rev_inds: 
        sect[:i] = i
    return Series(sect)

当可以假定非nan值的索引等于实际值时，此函数起作用。但是，当索引不等于内容时，如何填充数组？

---

例如，如果数组计数为：

count = np.array([np.NaN,np.NaN,np.NaN,1,np.NaN,np.NaN,2,np.NaN,np.NaN,np.NaN,np.NaN,np.NaN,3])

并且期望的输出是

count = np.array([1,1,1,1,2,2,2,3,3,3,3,3,3]

阵列末端可能有NAN。我希望这些都是NAN，这样数据帧就会忽略它们

count = np.array([np.NaN,np.NaN,np.NaN,1,np.NaN,np.NaN,2,np.NaN,np.NaN,3,np.NaN,np.NaN]) 
# Will become:
count = np.array([1,1,1,1,2,2,2,3,3,3,np.nan,np.nan]

---

这是一种矢量化方法-

# Append False at either sides of NaN mask as we try to find start &
# stop of each NaN interval by looking for rising and falling edges
mask = np.hstack((False,np.isnan(count),False))
start = np.flatnonzero(mask[1:] > mask[:-1])
stop = np.flatnonzero(mask[1:] < mask[:-1])
lens = stop - start

# Account for NaNs if any at the end of input that might throw off stop values
stop = stop.clip(max=count.size-1)

# Assign values
count[mask[1:-1]] = count[stop].repeat(lens)

案例2：

案例3：

---

IIUC您可以简单地使用以下方法：

您的样本：

In [89]: s = pd.Series(np.array([np.nan,np.nan,np.nan,1,np.nan,np.nan,2,np.nan,np.nan,3,np.nan,np.nan]))

In [90]: s
Out[90]:
0     NaN
1     NaN
2     NaN
3     1.0
4     NaN
5     NaN
6     2.0
7     NaN
8     NaN
9     3.0
10    NaN
11    NaN
dtype: float64

In [91]: s.bfill()
Out[91]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    NaN
11    NaN
dtype: float64

Divakar的样品：

In [81]: s = pd.Series(array([ nan,  nan,  nan,   6.,  nan,  nan,   5.,  nan,  nan,  nan,  nan, nan,   2.]))

In [82]: s
Out[82]:
0     NaN
1     NaN
2     NaN
3     6.0
4     NaN
5     NaN
6     5.0
7     NaN
8     NaN
9     NaN
10    NaN
11    NaN
12    2.0
dtype: float64

In [83]: s.bfill()
Out[83]:
0     6.0
1     6.0
2     6.0
3     6.0
4     5.0
5     5.0
6     5.0
7     2.0
8     2.0
9     2.0
10    2.0
11    2.0
12    2.0
dtype: float64

In [84]: s = pd.Series(array([ nan,  nan,  nan,   1.,  nan,  nan,   2.,  nan,  nan,  nan,  nan, nan, 3.]))

In [85]: s.bfill()
Out[85]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    3.0
11    3.0
12    3.0
dtype: float64

In [86]: s = pd.Series(array([ nan, nan, nan, 1., nan, nan, 2., nan, nan, 3., nan, nan]))

In [87]: s.bfill()
Out[87]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    NaN
11    NaN
dtype: float64

---

最后一个元素是否可以是

NaN

？@Divakar是的，它可以被索引。那么，您需要它填充一些东西吗？如果是这样，我们应该用什么来填充它？也许可以添加一个示例案例？太棒了！非常感谢你@Divakar，依我看，这是相当标准的熊猫操作，所以那里没有魔法；）哇！真正地所有的麻烦都是徒劳的哈哈。要是我早知道就好了。现在我也可以停止编写正向填充函数了。。。

In [114]: count
Out[114]: 
array([ nan,  nan,  nan,   1.,  nan,  nan,   2.,  nan,  nan,   3.,  nan,
        nan])

In [115]:   # Listed code ...

In [116]: count
Out[116]: 
array([  1.,   1.,   1.,   1.,   2.,   2.,   2.,   3.,   3.,   3.,  nan,
        nan])

In [89]: s = pd.Series(np.array([np.nan,np.nan,np.nan,1,np.nan,np.nan,2,np.nan,np.nan,3,np.nan,np.nan]))

In [90]: s
Out[90]:
0     NaN
1     NaN
2     NaN
3     1.0
4     NaN
5     NaN
6     2.0
7     NaN
8     NaN
9     3.0
10    NaN
11    NaN
dtype: float64

In [91]: s.bfill()
Out[91]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    NaN
11    NaN
dtype: float64

In [81]: s = pd.Series(array([ nan,  nan,  nan,   6.,  nan,  nan,   5.,  nan,  nan,  nan,  nan, nan,   2.]))

In [82]: s
Out[82]:
0     NaN
1     NaN
2     NaN
3     6.0
4     NaN
5     NaN
6     5.0
7     NaN
8     NaN
9     NaN
10    NaN
11    NaN
12    2.0
dtype: float64

In [83]: s.bfill()
Out[83]:
0     6.0
1     6.0
2     6.0
3     6.0
4     5.0
5     5.0
6     5.0
7     2.0
8     2.0
9     2.0
10    2.0
11    2.0
12    2.0
dtype: float64

In [84]: s = pd.Series(array([ nan,  nan,  nan,   1.,  nan,  nan,   2.,  nan,  nan,  nan,  nan, nan, 3.]))

In [85]: s.bfill()
Out[85]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    3.0
11    3.0
12    3.0
dtype: float64

In [86]: s = pd.Series(array([ nan, nan, nan, 1., nan, nan, 2., nan, nan, 3., nan, nan]))

In [87]: s.bfill()
Out[87]:
0     1.0
1     1.0
2     1.0
3     1.0
4     2.0
5     2.0
6     2.0
7     3.0
8     3.0
9     3.0
10    NaN
11    NaN
dtype: float64