Pandas DataFrame filtering of groups of rows on multiple columns

Question

Here's a simplified version of my dataframe:

d = {'col1': ['a1', 'a2', 'a3', 'b1', 'b2', 'b3', 'c1', 'c2', 'c3', 'd1', 'd2', 'd3'], 'col2': [1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, 1], 'col3': [-1, -1, 1, -1, -1, 1, 1, 1, 1, -1, 1, 1]}
df = pd.DataFrame(d)
df

    col1    col2    col3
0   a1       1      -1
1   a2       1      -1
2   a3       1       1
3   b1      -1      -1
4   b2      -1      -1
5   b3      -1       1
6   c1      -1       1
7   c2       1       1
8   c3       1       1
9   d1      -1      -1
10  d2       1      -1
11  d3       1       1

i would like to be able to pull out only those rows where col3 == 1 for the first time n rows after col2 == 1 for the first time , for each letter group.

so for example, if we're looking for when col3 became 1 one row after col2 became 1 (for each letter group), we'll get

    col1    col2    col3
0   d3      1       1

because for group d col2 turned from -1 to 1 at d2 and col3 turned from -1 to 1 at d3. And that hasn't happened in any other group.

if we want rows where col3 became 1 two rows after col2 became 1 (for each letter group), we'll get

    col1    col2    col3
0   a3      1       1

because for group a col2 started with 1 at a1 and col3 turned from -1 to 1 at a3.

Edit:

Here's my awkward way of doing it ... anyone got more elegant solutions?

df['newCol'] = (
           (((df['col2'].shift(n+1).isnull() | (df['col2'].shift(n+1) == -1)) &
           (df['col2'].shift(n+1).isnull() | (df['col2'].shift(n+1) == -1))) |
           (df['col1'].shift(n+1).str[0] != df['col1'].str[0])) &
           (df['col2'].shift(n) == 1) &
           (df['col3'].shift(n) == -1) &
           (df['col2'].shift(1) == 1) &
           (df['col3'].shift(1) == -1) &
           (df['col2'] == 1) &
           (df['col3'] == 1) &
           (df['col1'].shift(n).str[0] == df['col1'].str[0])) if n > 0 \
            else \
           ((((df['col2'].shift(n+1).isnull() | (df['col2'].shift(n+1) == -1)) &
           (df['col2'].shift(n+1).isnull() | (df['col2'].shift(n+1) == -1))) |
           (df['col1'].shift(n+1).str[0] != df['col1'].str[0])) &
           (df['col2'] == 1) &
           (df['col3'] == 1))
    

Answer 1

To put my last comment into an answer. Create a new column that is a lag using n , then just filter the standard way and grab the first value of col1 .

n = 2
df['newCol'] = df['col2'].shift(n)
df.loc[(df['col3'] == 1) & (df['newCol'] == 1), ['col1']].values[0]

You can wrap this into a function and make everything parameters.

Answer 2

Try this:

n=2
cond = pd.concat([(df['col2'] == 1).groupby(df['col1'].str[0]).cumsum().shift(n),
                  (df['col3'] == 1).groupby(df['col1'].str[0]).cumsum()], 
                 axis=1)\
         .eq(1)\
         .all(axis=1)
df[cond]

Output:

  col1  col2  col3
2   a3     1     1

Or more simply I think:

cond1 = (df['col2'] == 1).groupby(df['col1'].str[0]).cumsum().shift(n) == 1
cond2 = (df['col3'] == 1).groupby(df['col1'].str[0]).cumsum() == 1
df[cond1 & cond2]