Pandas remove rows with less than n consecutive negative values in a column

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I have the following DataFrame:

Time:     Value:

 1000     4.6     - Keep because +ve
 2000     3.2     - Keep because +ve
 3000    -1.1     - Remove because -ve AND in consecutive group of negatives of size < 3
 4000    -0.4     - Remove because -ve AND in consecutive group of negatives of size < 3
 5000     0.8     - Keep because +ve
 6000    -1.5     - Keep because -ve AND in consecutive group of negatives of size >= 3
 7000    -2.1     - Keep because -ve AND in consecutive group of negatives of size >= 3
 8000    -3.4     - Keep because -ve AND in consecutive group of negatives of size >= 3
 9000    -1.5     - Keep because -ve AND in consecutive group of negatives of size >= 3
10000    -0.3     - Keep because -ve AND in consecutive group of negatives of size >= 3
11000     1.6     - Keep because +ve
12000     2.8     - Keep because +ve
13000     4.0     - Keep because +ve

I want to produce the following DataFrame from it: (Removing small groups of negative values where the group size is less than n, n=3 in this eg.)

Time:     Value:

 1000     4.6
 2000     3.2
 5000     0.8
 6000    -1.5
 7000    -2.1
 8000    -3.4
 9000    -1.5
10000    -0.3
11000     1.6 
12000     2.8
13000     4.0

In the absence of a clever Pandas solution, I am planning to write a element-wise loop which iterates through each row marking consecutive negative values (in a new column), and immediately removing those marked rows when the -ve sequence is interrupted by a +ve value. (Deletion will not occur if the -ve sequence reaches the minimum size of n. The example above shows n=3).

Having deleted the marked rows, I will carry on from where I left off, until the end of the original frame is reached.

I know the proposed solution is not elegant in the Pandas world(!), but cannot figure out how a purist Pandas solution would work. Maybe something using groups or shift?

pandas dataframe group-by
1 Answers

group the negative values together and get their number, use this to form a mask for boolean indexing:

n = 3

# non-negative values
m1 = df['Value'].ge(0)
# count of negative values per group of successive ones
m2 = (~m).groupby(m1.cumsum()).transform('sum').ge(n)

out = df[m1|m2]

output:

     Time  Value
0    1000    4.6
1    2000    3.2
4    5000    0.8
5    6000   -1.5
6    7000   -2.1
7    8000   -3.4
8    9000   -1.5
9   10000   -0.3
10  11000    1.6
11  12000    2.8
12  13000    4.0

intermediates:

     Time  Value     m1     m2  m1|m2
0    1000    4.6   True  False   True
1    2000    3.2   True  False   True
2    3000   -1.1  False  False  False
3    4000   -0.4  False  False  False
4    5000    0.8   True   True   True
5    6000   -1.5  False   True   True
6    7000   -2.1  False   True   True
7    8000   -3.4  False   True   True
8    9000   -1.5  False   True   True
9   10000   -0.3  False   True   True
10  11000    1.6   True  False   True
11  12000    2.8   True  False   True
12  13000    4.0   True  False   True
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