Question

什么是T-SQL中与TOP函数等效的python？我希望将我的数据帧过滤到前50K行。我在网上看了，我找不到一个简单的例子。

Answer 1

更新： - 显示不同的pandas方法，包括：

每组前N行

前N行有偏移量

等效于SQL聚合函数：

ROW_NUMBER() / RANK() OVER(PARTITION BY ... ORDER BY ...)

样本DF：

df = pd.DataFrame({
  'dep': np.random.choice(list('ABC'), 20),
  'manager_id': np.random.randint(0, 10, 20),
  'salary': np.random.randint(5000, 5006, 20)
})

----------------------- Original DF ---------------------- -

In [2]: df
Out[2]:
   dep  manager_id  salary
0    B           5    5005
1    A           6    5001
2    C           8    5000
3    A           7    5000
4    B           0    5002
5    A           3    5003
6    A           2    5004
7    A           2    5004
8    C           3    5002
9    C           4    5001
10   A           9    5002
11   C           9    5000
12   B           8    5004
13   A           1    5003
14   C           7    5005
15   B           0    5002
16   B           2    5003
17   A           4    5000
18   B           2    5003
19   B           7    5003

------------------前5行（按原始索引排序）------------------- < / p>

In [3]: df.head(5)
Out[3]:
  dep  manager_id  salary
0   B           5    5005
1   A           6    5001
2   C           8    5000
3   A           7    5000
4   B           0    5002

---前5行（按manager_id DESC排序，dep ASC）----

In [4]: df.sort_values(by=['manager_id', 'dep'], ascending=[False,True]).head(5)
Out[4]:
   dep  manager_id  salary
10   A           9    5002
11   C           9    5000
12   B           8    5004
2    C           8    5000
3    A           7    5000

---相当于SELECT * FROM tab ORDER BY salary DESC LIMIT 5 OFFSET 3 ---

In [19]: df.nlargest(5+3, columns=['salary']).tail(5)
Out[19]:
   dep  manager_id  salary
7    A           2    5004
12   B           8    5004
5    A           3    5003
13   A           1    5003
16   B           2    5003

----每个部门的前2名工资（无重复）-----

---等同于SQL：row_number() over(partition by DEP order by SALARY desc) ---

In [7]: (df.assign(rn=df.sort_values(['salary'], ascending=False)
   ...:                 .groupby(['dep'])
   ...:                 .cumcount() + 1)
   ...:    .query('rn < 3')
   ...:    .sort_values(['dep','rn'])
   ...: )
Out[7]:
   dep  manager_id  salary  rn
6    A           2    5004   1
7    A           2    5004   2
0    B           5    5005   1
12   B           8    5004   2
14   C           7    5005   1
8    C           3    5002   2

---每个部门的前2名工资（使用＆＃34; nlargest＆＃34;）----

In [15]: df.loc[df.groupby('dep')['salary'].nlargest(2).reset_index()['level_1']]
Out[15]:
   dep  manager_id  salary
6    A           2    5004
7    A           2    5004
0    B           5    5005
12   B           8    5004
14   C           7    5005
8    C           3    5002

---每个部门的第二和第三高薪---

In [16]: (df.assign(rn=df.sort_values(['salary'], ascending=False)
   ....:                 .groupby(['dep'])
   ....:                 .cumcount() + 1)
   ....:    .query('rn >= 2 and rn <= 3')
   ....:    .sort_values(['dep','rn'])
   ....: )
Out[16]:
   dep  manager_id  salary  rn
7    A           2    5004   2
13   A           1    5003   3
12   B           8    5004   2
18   B           2    5003   3
8    C           3    5002   2
9    C           4    5001   3

---每个部门的前两名工资（有重复）----

---等同于SQL：rank() over(partition by DEP order by SALARY desc) ---

In [18]: (df.assign(rnk=df.groupby(['dep'])['salary']
   ....:                  .rank(method='min', ascending=False))
   ....:    .query('rnk < 3')
   ....:    .sort_values(['dep','rnk'])
   ....: )
Out[18]:
   dep  manager_id  salary  rnk
6    A           2    5004  1.0
7    A           2    5004  1.0
0    B           5    5005  1.0
12   B           8    5004  2.0
14   C           7    5005  1.0
8    C           3    5002  2.0

熊猫：前N行，每组前N行，相当于ROW_NUMBER OVER（PARTITION BY ... ORDER BY ...）

1 个答案: