Question

我有一个mysql样本数据，如下表所示

main_cat| sub_cat | number | org_id
Career  | school  | 5      | A
Career  | college | 3      | A
Career  | higher  | 4      | A 
Job     | Blr     | 6      | A
Job     | Hyd     | 11     | A
Job     | Chennai | 12     | A 
Career  | school  | 15     | B
Career  | college | 30     | B
Career  | higher  | 5      | B 
Job     | Blr     | 5      | B
Career  | college | 8      | C
Job     | Chennai | 4      | C

我想为每个组织打印前2 main_cat。对于每个前2 main_cat，我想为每个组织打印前2 sub_cat。因此，对于每个main_cat

，每个组织应该有4个或更少的记录，前2个sub_cat和前2个main_cat

请帮帮我。

Answer 1

您可以使用pandas将查询处理为read_sql的数据框：

import pandas as pd
df = pd.read_sql(connection,query)
result = df.groupby(['org_id', 'main_cat', 'sub_cat'])['number'].head(2)

变量connection是您与数据库的连接，query是您的SELECT字符串。

Answer 2

对于分组，python提供了itertools.groupby()，它按给定的键功能对排序输入进行分组。

在这种情况下，输入需要先按org_id排序，然后按main_cat排序，然后按number按降序排序，例如：如果您的数据列表如下所示：

data = [
 ['Career', 'school', 5, 'A'],
 ['Career', 'college', 3, 'A'],
 ['Career', 'higher', 4, 'A'],
 ['Job', 'Blr', 6, 'A'],
 ['Job', 'Hyd', 11, 'A'],
 ['Job', 'Chennai', 12, 'A'],
 ['Career', 'school', 15, 'B'],
 ['Career', 'college', 30, 'B'],
 ['Career', 'higher', 5, 'B'],
 ['Job', 'Blr', 5, 'B'],
 ['Career', 'college', 8, 'C'],
 ['Job', 'Chennai', 4, 'C']
]

然后你会这样排序：

data.sort(key = lambda x: (x[3], x[0], -x[2]))

或通过改变你的sql语句来包含ORDER BY main_cat, sub_cat, number DESC，然后你就可以从数据库中以正确的顺序得到它。

现在，您可以使用groupby进行分组，islice可以限制每个分组类别的结果数量：

from itertools import groupby, islice
from operator import itemgetter

# already sorted data
data = [
 ['Career', 'school', 5, 'A'],
 ['Career', 'higher', 4, 'A'],
 ['Career', 'college', 3, 'A'],
 ['Job', 'Chennai', 12, 'A'],
 ['Job', 'Hyd', 11, 'A'],
 ['Job', 'Blr', 6, 'A'],
 ['Career', 'college', 30, 'B'],
 ['Career', 'school', 15, 'B'],
 ['Career', 'higher', 5, 'B'],
 ['Job', 'Blr', 5, 'B'],
 ['Career', 'college', 8, 'C'],
 ['Job', 'Chennai', 4, 'C']
]

data.sort(key = lambda x: (x[3], x[0], -x[2]))

for org, by_org in groupby(data, key=itemgetter(3)):
    print("org:", org)
    for cat, by_cat in islice(groupby(by_org, key=itemgetter(0)), 2):
        print("  cat:", cat)
        for subcat, by_subcat in islice(groupby(by_cat, key=itemgetter(1)), 2):
            print("    subcat:", subcat, " = ", list(by_subcat))

输出：

org: A
  cat: Career
    subcat: school  =  [['Career', 'school', 5, 'A']]
    subcat: higher  =  [['Career', 'higher', 4, 'A']]
  cat: Job
    subcat: Chennai  =  [['Job', 'Chennai', 12, 'A']]
    subcat: Hyd  =  [['Job', 'Hyd', 11, 'A']]
org: B
  cat: Career
    subcat: college  =  [['Career', 'college', 30, 'B']]
    subcat: school  =  [['Career', 'school', 15, 'B']]
  cat: Job
    subcat: Blr  =  [['Job', 'Blr', 5, 'B']]
org: C
  cat: Career
    subcat: college  =  [['Career', 'college', 8, 'C']]
  cat: Job
    subcat: Chennai  =  [['Job', 'Chennai', 4, 'C']]

如何使用基于mysql结果数据的python中的条件打印数据

2 个答案: