deleted guest = guests.pop(len(guests)-1) guests.insert(2,deleted_guest) del(guests[1]) print(deleted guest) print(guests)

As an AI language model, I cannot execute the code you provided, but I can explain what it does.

This code performs the following actions:

1. Deletes the last guest from a list of guests using the `pop()` method and assigns it to a variable called `deleted_guest`.
2. Inserts the `deleted_guest` variable at index 2 in the list using the `insert()` method.
3. Deletes the guest at index 1 from the list using the `del` keyword.
4. Prints the `deleted_guest` variable.
5. Prints the updated list of guests.

Overall, this code is manipulating a list of guests by removing one guest, inserting it at a different index, and deleting another guest.

相关问题

优化SQL select round( ohbmc.after_actual_amount/zz,0) cost_moneyi ,count(distinct case when ddp.orig_plan_rid = -1 then null else ddp.orig_plan_rid end) AS orig_num ,array_agg (dlt.state) AS loading_state ,count(DISTINCT CASE WHEN ddp.sale_planid = -1 THEN NULL ELSE ddp.sale_planid END) AS saleid_num--销地已计划数量 ,array_agg(dto.state) AS saletransport_state from ( SELECT id AS origin_planid , unnest(cabinet_rule_id) cabinet_rule_id -- 判断 next_plan_id 本身是空和 next_plan_id 为 {} ,unnest(case when (next_plan_id is null or next_plan_id[1] is null) then ARRAY[-1]::integer[] else next_plan_id end) as sale_planid --销地计划 , case when dp.plan_receiver_id is null then -1 else dp.plan_receiver_id end orig_plan_rid --产地计划 FROM ods.ods_durian_delivery_plan as dp left join ods.ods_hl_commodity_category as hcc on hcc.category_id = dp.category_id WHERE dp.type = 'ORIGIN' AND dp.deleted = 99 AND dp.tenant_id = 1 and cabinet_rule_id='{8}'or cabinet_rule_id='{9}'or cabinet_rule_id='{10000005}'---取白心火龙果 AND hcc.category_name = '火龙果') as ddp LEFT JOIN ods.ods_durian_loading_task AS dlt ON dlt.plan_id = ddp.origin_planid and dlt.plan_type='ORIGIN' AND dlt.deleted = 99 LEFT JOIN ods.ods_durian_transport_order AS dto ON dto.plan_id = ddp.sale_planid AND dto.deleted = 99 LEFT JOIN ods.ods_durian_receipt_task AS drt ON drt.plan_id = ddp.sale_planid AND drt.deleted = 99 LEFT JOIN ods.ods_durian_transport_order AS dto1 ON dto1.plan_id = ddp.sale_planid AND dto1.sort_no = 1 AND dto1.deleted = 99 left join (select odlsi.plan_id,sum(odlsi.quantity) zz from ods.ods_durian_loading_sku_item odlsi group by 1) odlsi on odlsi.plan_id=dlt.plan_id left join (select *, unnest(case when ( odbr.bill_main_id is null or odbr.bill_main_id is null) then ARRAY[-1]::integer[] else odbr.bill_main_id end) bill_main_id_r from ods.ods_durian_bill_rel odbr) odbr on odbr.data_id= dlt.plan_id and odbr.data_type='ORIGIN_FEE' left join ods.ods_hl_bill_main_currency ohbmc on ohbmc.bill_main_id=odbr.bill_main_id_r and ohbmc.deleted=99 group by 1;

这个SQL查询语句比较复杂，可以考虑以下几种优化方式：

1. 使用JOIN语句代替子查询：可以把子查询转换成JOIN语句，这样可以提高查询的效率。

2. 合理使用索引：根据实际情况，选择合适的索引可以大大提升查询效率。可以对WHERE子句中的条件表达式所涉及的列添加索引。

3. 减少DISTINCT关键字的使用：DISTINCT关键字会增加查询的开销，如果不必要，可以考虑去掉。

4. 减少聚合函数的使用：聚合函数是一种开销较大的操作，如果可以不用聚合函数，可以考虑去掉。

5. 优化WHERE子句中的条件表达式：可以优化WHERE子句中的条件表达式，尽量使用简单的条件表达式，减少复杂的逻辑判断。

SELECT s.student_num, s.student_name, '计算机学院' AS college_name, c.level, c.class_name FROM student_info s JOIN classes c ON s.class_id = c.id JOIN volunteer_time_table_item t ON s.student_num = t.student_num INNER JOIN student_info s ON c.id = s.college_id INNER JOIN volunteer_time_table_item t ON s.student_num = t.student_num INNER JOIN volunteer_time_table vtt ON vtt.id = t.volunteer_time_table_id WHERE c.college_id = 1 AND s.level = 2021 AND vtt.volunteer_status = 5 AND vtt.deleted = 0 AND t.deleted = 0 AND s.deleted = 0 AND c.deleted = 0 ORDER BY t.volunteer_time_value DESC LIMIT 5;哪里有问题

这个 SQL 查询语句有以下问题：

1. 语法错误：在SELECT语句中，没有为 '计算机学院' 添加引号，应该在前后添加单引号。

2. 重复声明表别名：在JOIN语句中，重复声明了表别名 s 和 t，应该将它们改成不同的别名。

3. 错误的表连接方式：在JOIN语句中，没有为 student_info 表声明别名，应该将其改为 s 或其他合适的别名。此外，使用 INNER JOIN 连接 volunteer_time_table_item 表和 student_info 表时，应该使用 ON 关键字指定连接条件。

4. SELECT语句中重复声明列名：在SELECT语句中，声明了 student_num, student_name, level, class_name 列名，但是其中的 level 和 class_name 列名已经在 JOIN 语句中声明过了，应该删除重复的声明。

5. 语法错误：在 ORDER BY 语句中，应该为 t.volunteer_time_value 添加列名，否则会报 Unknown column 'volunteer_time_value' in 'order clause' 的错误。

修改后的 SQL 查询语句如下：

SELECT s.student_num, s.student_name, '计算机学院' AS college_name, c.level, c.class_name
FROM student_info s
JOIN classes c ON s.class_id = c.id
JOIN volunteer_time_table_item t ON s.student_num = t.student_num
INNER JOIN student_info si ON c.id = si.college_id
INNER JOIN volunteer_time_table_item ti ON si.student_num = ti.student_num
INNER JOIN volunteer_time_table vtt ON vtt.id = ti.volunteer_time_table_id
WHERE c.college_id = 1
AND s.level = 2021
AND vtt.volunteer_status = 5
AND vtt.deleted = 0
AND ti.deleted = 0
AND si.deleted = 0
AND c.deleted = 0
ORDER BY ti.volunteer_time_value DESC LIMIT 5;