Why is pandas nlargest slower than mine?

user_562769 0
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I have a data frame

            ID  CAT    SCORE
0            0    0  8325804
1            0    1  1484405
...        ...  ...      ...
1999980  99999    0  4614037
1999981  99999    1  1818470

Where I group the data by ID and want to know the 2 categories per ID with the highest score. I can see two solutions to that:

df2 = df.groupby('ID').apply(lambda g: g.nlargest(2, columns='SCORE'))

or manually converting it to a list of tuples, sorting the tuples, removing for each ID except for two and then converting back to a dataframe. The first one should be way faster than the second one, but I observe that the manual solution is WAY faster.

Why is manual nlargest faster than pandas solution?

MVCE

import numpy as np
import pandas as pd
import time


def create_df(n=10**5, categories=20):
    np.random.seed(0)
    df = pd.DataFrame({'ID': [id_ for id_ in range(n) for c in range(categories)],
                       'CAT': [c for id_ in range(n) for c in range(categories)],
                       'SCORE': np.random.randint(10**7, size=n * categories)})
    return df


def are_dfs_equal(df1, df2):
    columns = sorted(df1.columns)
    if len(df1.columns) != len(df2.columns):
        return False
    elif not all(el1 == el2 for el1, el2 in zip(columns, sorted(df2.columns))):
        return False
    df1_list = [tuple(x) for x in df1[columns].values]
    df1_list = sorted(df1_list, reverse=True)
    df2_list = [tuple(x) for x in df2[columns].values]
    df2_list = sorted(df2_list, reverse=True)
    is_same = df1_list == df2_list
    return is_same


def manual_nlargest(df, n=2):
    df_list = [tuple(x) for x in df[['ID', 'SCORE', 'CAT']].values]
    df_list = sorted(df_list, reverse=True)
    l = []
    current_id = None
    current_id_count = 0
    for el in df_list:
        if el[0] != current_id:
            current_id = el[0]
            current_id_count = 1
        else:
            current_id_count += 1
        if current_id_count <= n:
            l.append(el)
    df = pd.DataFrame(l, columns=['ID', 'SCORE', 'CAT'])
    return df

df = create_df()

t0 = time.time()
df2 = df.groupby('ID').apply(lambda g: g.nlargest(2, columns='SCORE'))
t1 = time.time()
print('nlargest solution: {:0.2f}s'.format(t1 - t0))

t0 = time.time()
df3 = manual_nlargest(df, n=2)
t1 = time.time()
print('manual nlargest solution: {:0.2f}s'.format(t1 - t0))
print('is_same: {}'.format(are_dfs_equal(df2, df3)))

gives

nlargest solution: 97.76s
manual nlargest solution: 4.62s
is_same: True
pandas-groupby pandas dataframe python
2 Answers

I guess you can use this :

df.sort_values(by=['SCORE'],ascending=False).groupby('ID').head(2)

This is the same as your manual solution using Sort/head functions on pandas groupby.

t0 = time.time()
df4 = df.sort_values(by=['SCORE'],ascending=False).groupby('ID').head(2)
t1 = time.time()
df4_list = [tuple(x) for x in df4[['ID', 'SCORE', 'CAT']].values]
df4_list = sorted(df4_list, reverse=True)
is_same = df3_list == df4_list
print('SORT/HEAD solution: {:0.2f}s'.format(t1 - t0))
print(is_same)

gives

SORT/HEAD solution: 0.08s
True

timeit

77.9 ms ± 7.91 ms per loop (mean ± std. dev. of 7 runs, 10 loops each).

As to why nlargest is slower than the other solutions ?, I guess calling it for each group is creating an overhead (%prun is showing 15764409 function calls (15464352 primitive calls) in 30.293 seconds).

For this solution (1533 function calls (1513 primitive calls) in 0.078 seconds)

user_52767280

Here'a faster solution than your manual solution, unless I made a mistake ;) I guess nlargest() is not the fastest way for solving this problem if speed is what you need, but it is the more readable solution.

t0 = time.time()
df4 = df.sort_values(by=['ID', 'SCORE'], ascending=[True, False])
df4['cumcount'] = df4.groupby('ID')['SCORE'].cumcount()
df4 = df4[df4['cumcount'] < 2]
t1 = time.time()
print('cumcount solution: {:0.2f}s'.format(t1 - t0))
user_34891550
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