为什么最大的熊猫比我的慢？

我有一个数据框

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

我在哪里对数据进行分组，ID并想知道每个ID得分最高的2个类别。我可以看到两种解决方案：

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

或手动将其转换为元组列表，对元组进行排序，为每个ID（除两个ID之外）移除，然后转换回数据框。第一个应该比第二个要快，但是我观察到手动解决方案要快得多。

为什么手动最大比熊猫解决方案更快？

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)))

给

nlargest solution: 97.76s
manual nlargest solution: 4.62s
is_same: True