drop consecutive duplicates of groups - pandas

Question

I am removing consecutive duplicates in groups in a dataframe. I am looking for a faster way than this:
def remove_consecutive_dupes(subdf):
dupe_ids = [ "A", "B" ]
is_duped = (subdf[dupe_ids].shift(-1) == subdf[dupe_ids]).all(axis=1)
subdf = subdf[~is_duped]
return subdf
# dataframe with columns key, A, B
df.groupby("key").apply(remove_consecutive_dupes).reset_index()
Is it possible to remove these without grouping first? Applying the above function to each group individually takes a lot of time, especially if the group count is like half the row count. Is there a way to do this operation on the entire dataframe at once?
A simple example for the algorithm if the above was not clear:
input:
key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5
5 x 1 2
output:
key A B
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
5 x 1 2
Row 2 was dropped because A=1 B=2 was also the previous row in group x.
Row 5 will not be dropped because it is not a consecutive duplicate in group x.

Answer 1

According to your code, you drop only lines if they appear below each other if
they are grouped by the key. So rows with another key inbetween do not influence this logic. But doing this, you want to preserve the original order of the records.
I guess the biggest influence in the runtime is the call of your function and
possibly not the grouping itself.
If you want to avoid this, you can try the following approach:
# create a column to restore the original order of the dataframe
df.reset_index(drop=True, inplace=True)
df.reset_index(drop=False, inplace=True)
df.columns= ['original_order'] + list(df.columns[1:])
# add a group column, that contains consecutive numbers if
# two consecutive rows differ in at least one of the columns
# key, A, B
compare_columns= ['key', 'A', 'B']
df.sort_values(['key', 'original_order'], inplace=True)
df['group']= (df[compare_columns] != df[compare_columns].shift(1)).any(axis=1).cumsum()
df.drop_duplicates(['group'], keep='first', inplace=True)
df.drop(columns=['group'], inplace=True)
# now just restore the original index and it's order
df.set_index('original_order', inplace=True)
df.sort_index(inplace=True)
df
Testing this, results in:
key A B
original_order
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
If you don't like the index name above (original_order), you just need to add the following line to remove it:
df.index.name= None
Testdata:
from io import StringIO
infile= StringIO(
""" key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5"""
)
df= pd.read_csv(infile, sep='\s+') #.set_index('Date')
df