CODEWITHSUNDEEP
pythonpandaspandas-groupby
gorjan
gorjan

Reputation: 5555

Pandas: Strip rows at the end of data-frame group based on condition

Let's say that I have the following data-frame:

df = pd.DataFrame({"id": [1, 1, 1, 2, 2, 2, 3, 3, 3, 3], "date": [pd.Timestamp(2002, 2, 2), pd.Timestamp(2003, 3, 3), pd.Timestamp(2004, 4, 4), pd.Timestamp(2005, 5, 5), pd.Timestamp(2006, 6, 6), pd.Timestamp(2007, 7, 7), pd.Timestamp(2008, 8, 8), pd.Timestamp(2009, 9, 9), pd.Timestamp(2010, 10, 10), pd.Timestamp(2011, 11, 11)], "numeric": [0.9, 0.4, 0.2, 0.6, np.nan, 0.8, 0.7, np.nan, np.nan, 0.5], "nominal": [0, 1, 0, 1, 0, 0, 0, 1, 1, 1]})

What I want to achieve is to strip rows at the end of each group (assuming that the rows are grouped by id), such that the rows will be removed until a non-nan value will appear for the numeric column. Additionally, the last row for each group will always have a non-nan value for the numeric column and the last row should always be removed. So, the resulting data-frame is:

result_df = pd.DataFrame({"id": [1, 1, 2, 3], "date": [pd.Timestamp(2002, 2, 2), pd.Timestamp(2003, 3, 3), pd.Timestamp(2005, 5, 5), pd.Timestamp(2008, 8, 8)], "numeric": [0.9, 0.4, 0.6, 0.7], "nominal": [0, 1, 1, 0]})

More explanation on how we get to the resulting data-frame:

Moreover, what I am currently doing is:

df.groupby("id", as_index=False).apply(lambda x: x.iloc[:-1]).reset_index(drop=True)

However, this only removes the last row for each group and I want to remove the last N rows based on the condition explained above.

Please let me know if you need any further information and looking forward to your answers!

Upvotes: 1

Views: 115

Answers (1)

KRKirov
KRKirov

Reputation: 4004

For the specific example you have posted just dropping the NaNs before grouping does the trick:

df = df.dropna().groupby('id').apply(lambda x: x.iloc[:-1]).reset_index(drop=True)

df
Out[58]: 
   id       date  numeric  nominal
0   1 2002-02-02      0.9        0
1   1 2003-03-03      0.4        1
2   2 2005-05-05      0.6        1
3   3 2008-08-08      0.7        0

If you have a non-contiguous NaNs and you want to remove only the last block of NaNs:

def strip_rows(X):    
    X = X.iloc[:-1, :]
    while pd.isna(X.iloc[-1, 2]):        
        X = X.iloc[:-1, :]
    return X

df_1 = pd.DataFrame({"id": [1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3],
                   "date": [pd.Timestamp(2002, 2, 2),
                            pd.Timestamp(2003, 3, 3), 
                            pd.Timestamp(2004, 4, 4), 
                            pd.Timestamp(2005, 5, 5), 
                            pd.Timestamp(2006, 6, 6),
                            pd.Timestamp(2007, 7, 7),
                            pd.Timestamp(2008, 8, 8),
                            pd.Timestamp(2009, 9, 9),
                            pd.Timestamp(2010, 10, 10), 
                            pd.Timestamp(2011, 11, 11),
                            pd.Timestamp(2011, 12, 12),
                            pd.Timestamp(2012, 1, 1)],
                    "numeric": [0.9, 0.4, 0.2, 0.6, np.nan, 0.8, 0.7, np.nan, np.nan, 0.5, np.nan, 0.3],
                    "nominal": [0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1]})

df_2 = df_1.groupby('id').apply(strip_rows).reset_index(drop=True)

df_1
Out[151]: 
    id       date  numeric  nominal
0    1 2002-02-02      0.9        0
1    1 2003-03-03      0.4        1
2    1 2004-04-04      0.2        0
3    2 2005-05-05      0.6        1
4    2 2006-06-06      NaN        0
5    2 2007-07-07      0.8        0
6    3 2008-08-08      0.7        0
7    3 2009-09-09      NaN        1
8    3 2010-10-10      NaN        1
9    3 2011-11-11      0.5        1
10   3 2011-12-12      NaN        0
11   3 2012-01-01      0.3        1

df_2
Out[152]: 
   id       date  numeric  nominal
0   1 2002-02-02      0.9        0
1   1 2003-03-03      0.4        1
2   2 2005-05-05      0.6        1
3   3 2008-08-08      0.7        0
4   3 2009-09-09      NaN        1
5   3 2010-10-10      NaN        1
6   3 2011-11-11      0.5        1

Upvotes: 2

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