remove all rows in pandas dataframe with N or more consecutive NaNs

Question

corollary to this question: replace values in pandas column when N number of NaNs exist in another column

         a         b         c     d           e
2018-05-25  0.000381  0.264318     land    2018-05-25
2018-05-26  0.000000  0.264447     land    2018-05-26
2018-05-27  0.000000  0.264791     NaN           NaT
2018-05-28  0.000000  0.265253     NaN           NaT
2018-05-29  0.000000  0.265720     NaN           NaT
2018-05-30  0.000000  0.266066     land    2018-05-30
2018-05-31  0.000000  0.266150     NaN           NaT
2018-06-01  0.000000  0.265816     NaN           NaT
2018-06-02  0.000000  0.264892     land    2018-06-02
2018-06-03  0.000000  0.263191     NaN           NaT
2018-06-04  0.000000  0.260508     land    2018-06-04
2018-06-05  0.000000  0.256619     NaN           NaT
2018-06-06  0.000000  0.251286     NaN           NaT
2018-06-07  0.000000  0.244250     NaN           NaT
2018-06-08  0.000000  0.235231     NaN           NaT
2018-06-09  0.000000  0.223932     land    2018-06-09

I want to remove all rows where there is a NaN in the 4th column ( d ) 3 or more times. The output should be:

         a         b         c     d           e
2018-05-25  0.000381  0.264318     land    2018-05-25
2018-05-26  0.000000  0.264447     land    2018-05-26
2018-05-30  0.000000  0.266066     land    2018-05-30
2018-05-31  0.000000  0.266150     NaN           NaT
2018-06-01  0.000000  0.265816     NaN           NaT
2018-06-02  0.000000  0.264892     land    2018-06-02
2018-06-03  0.000000  0.263191     NaN           NaT
2018-06-04  0.000000  0.260508     land    2018-06-04
2018-06-09  0.000000  0.223932     land    2018-06-09

From that question, I tried this:

    threshold = 3
    mask = df.d.notna()
    df.loc[(~mask).groupby(mask.cumsum()).transform('cumsum') < threshold, 'c'] = np.nan
    df = df[np.isfinite(df['c'])]

but it does not work

Answer 1

Create helper Series a by consecutive values and transform size , last filter by boolean indexing :

mask = df.d.notna()
a = mask.ne(mask.shift()).cumsum()

df = df[(a.groupby(a).transform('size') < 3) | mask]
print (df)
             a         b         c     d           e
0   2018-05-25  0.000381  0.264318  land  2018-05-25
1   2018-05-26  0.000000  0.264447  land  2018-05-26
5   2018-05-30  0.000000  0.266066  land  2018-05-30
6   2018-05-31  0.000000  0.266150   NaN         NaT
7   2018-06-01  0.000000  0.265816   NaN         NaT
8   2018-06-02  0.000000  0.264892  land  2018-06-02
9   2018-06-03  0.000000  0.263191   NaN         NaT
10  2018-06-04  0.000000  0.260508  land  2018-06-04
15  2018-06-09  0.000000  0.223932  land  2018-06-09

Detail :

print (a)
0     1
1     1
2     2
3     2
4     2
5     3
6     4
7     4
8     5
9     6
10    7
11    8
12    8
13    8
14    8
15    9
Name: d, dtype: int32

---

print (a.groupby(a).transform('size'))
0     2
1     2
2     3
3     3
4     3
5     1
6     2
7     2
8     1
9     1
10    1
11    4
12    4
13    4
14    4
15    1
Name: d, dtype: int64

Answer 2

This should work:

df = df.groupby(pd.notnull(df.d).cumsum()).apply(lambda x: x.dropna() if pd.isnull(x.d).sum() >= 3 else x).reset_index(drop=True)

Output:

            a         b         c     d           e
0  2018-05-25  0.000381  0.264318  land  2018-05-25
1  2018-05-26  0.000000  0.264447  land  2018-05-26
2  2018-05-30  0.000000  0.266066  land  2018-05-30
3  2018-05-31  0.000000  0.266150   NaN         NaT
4  2018-06-01  0.000000  0.265816   NaN         NaT
5  2018-06-02  0.000000  0.264892  land  2018-06-02
6  2018-06-03  0.000000  0.263191   NaN         NaT
7  2018-06-04  0.000000  0.260508  land  2018-06-04
8  2018-06-09  0.000000  0.223932  land  2018-06-09

Answer 3

one way to solve this,

df['seq'] = df.groupby(df['d'].notnull().cumsum())['a'].transform(len)
df=df[(df['seq']<4)|df['d'].notnull()]

Output:

             a         b         c     d           e  seq
0   2018-05-25  0.000381  0.264318  land  2018-05-25    1
1   2018-05-26  0.000000  0.264447  land  2018-05-26    4
5   2018-05-30  0.000000  0.266066  land  2018-05-30    3
6   2018-05-31  0.000000  0.266150   NaN         NaN    3
7   2018-06-01  0.000000  0.265816   NaN         NaN    3
8   2018-06-02  0.000000  0.264892  land  2018-06-02    2
9   2018-06-03  0.000000  0.263191   NaN         NaN    2
10  2018-06-04  0.000000  0.260508  land  2018-06-04    5
15  2018-06-09  0.000000  0.223932  land  2018-06-09    1