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informix最新版本_sql权威指南 pdf

English(CZ6QNEN) Informix International Language Supplement 3.50 MC4 Multiplatform Multilingual(CZB2CML...) Informix Client SDK V3.50.TC6 Windows 32 Bit English(CZ6VWEN) Informix Client SDK V3.50.FC6 Windows...Edition V11.50.FC5 Linux x86 64 English(CZ5IZEN) Informix Client SDK V3.50.FC5 Windows 64 bit English...Informix Client SDK V3.50.TC5 Windows 32 bit English(CZ4HKEN) Informix International Language...Supplement 3.50.MC3 Englisth(C1TB2EN) pdksh-5.2.14-30.6.i386.rpm fjs-f4gl

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R语言排序问题

0.200 setosa 5 4.70 3.20 1.30 0.200 setosa 6 5.50 3.50...0.200 setosa 8 4.40 3.20 1.30 0.200 setosa 9 5.00 3.50...0.200 setosa 5 4.70 3.20 1.30 0.200 setosa 6 5.50 3.50...0.200 setosa 8 4.40 3.20 1.30 0.200 setosa 9 5.00 3.50

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浮点数环境 cfenv（fenv.h）

printfRounding(); printf ( "rint (2.49) = %.1f\n", rint(2.49) ); printf ( "rint (3.50...) = %.1f\n", rint(3.50) ); fesetround(FE_TOWARDZERO); printfRounding(); printf ( "rint...(2.49) = %.1f\n", rint(2.49) ); printf ( "rint (3.50) = %.1f\n", rint(3.50) ); return 0;...运行结果： [cpp] view plain copy now rounding using: to-nearest rint (2.49) = 2.0 rint (3.50...) = 4.0 now rounding using: toward-zero rint (2.49) = 2.0 rint (3.50) = 3.0 浮点数环境函数

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2022-11-23：分数排名。输出结果和表的sql如下。请写出输出结果的sql语句？ +-------+------+ | score | rank | +

|+-------+------+| 4.00 | 1 || 4.00 | 1 || 3.85 | 2 || 3.65 | 3 || 3.65 | 3 || 3.50...NOT NULL, PRIMARY KEY (id)) ENGINE=InnoDB DEFAULT CHARSET=latin1;INSERT INTO scores VALUES ('1', '3.50

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列转行-lateral view outer posexplode及posexplode_outer多列对应转行

13.00,15.00,5.00,9.50,7.00 | | r003 | 0003,0007 | 4.12,8.11 | 3.50,8.00...| 9.50 | | r002 | 0010 | 6.05 | 7.00 | | r003 | 0003 | 4.12 | 3.50...| 9.50 | | r002 | 0010 | 6.05 | 7.00 | | r003 | 0003 | 4.12 | 3.50...| 9.50 | | r002 | 0010 | 6.05 | 7.00 | | r003 | 0003 | 4.12 | 3.50...| 9.50 | | r002 | 0010 | 6.05 | 7.00 | | r003 | 0003 | 4.12 | 3.50

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浮点数环境 cfenv（fenv.h）

printfRounding(); printf ( "rint (2.49) = %.1f\n", rint(2.49) ); printf ( "rint (3.50...) = %.1f\n", rint(3.50) ); fesetround(FE_TOWARDZERO); printfRounding(); printf ( "rint...(2.49) = %.1f\n", rint(2.49) ); printf ( "rint (3.50) = %.1f\n", rint(3.50) ); return 0;...运行结果： [cpp] view plain copy now rounding using: to-nearest rint (2.49) = 2.0 rint (3.50...) = 4.0 now rounding using: toward-zero rint (2.49) = 2.0 rint (3.50) = 3.0 浮点数环境函数

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行转列-多列一一对应

13.00 | | r003 | 0003 | 食品 | 2024-01-01 09:15:00 | c004 | 4.12 | 3.50...13.00,15.00,5.00,9.50,7.00 | | r003 | 0003,0007 | 4.12,8.11 | 3.50,8.00...00,0002,3.01,13.00 | | r003 | 2024-01-01 09:15:00 | 0003 | 2024-01-01 09:15:00,0003,4.12,3.50..."2024-01-01 13:30:00","0010","6.05","7.00"]] | | r003 | [["2024-01-01 09:15:00","0003","4.12","3.50...'2024-01-01 08:43:00','c003',3.01,13.00), ('r003', '0003', '食品', '2024-01-01 09:15:00','c004',4.12,3.50

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Leetcode No.178 分数排名

+----+-------+ | Id | Score | +----+-------+ | 1 | 3.50 | | 2 | 3.65 | | 3 | 4.00 | | 4 | 3.85...----+------+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 2 | | 3.65 | 3 | | 3.65 | 3 | | 3.50

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LeetCode 0178 - Rank Scores

. +----+-------+ | Id | Score | +----+-------+ | 1 | 3.50 | | 2 | 3.65 | | 3 | 4.00 | | 4 | 3.85...----+------+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 2 | | 3.65 | 3 | | 3.65 | 3 | | 3.50

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R语言读取文件

Petal.Width Species #> #> 1 5.10 3.50...Petal.Width Species #> #> 1 5.10 3.50...Petal.Width Species #> #> 1 5.10 3.50...Petal.Width Species #> #> 1 5.10 3.50...x 3 #> Sepal.Width Petal.Length Petal.Width #> #> 1 3.50

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开窗函数 spark sql实现累加、累积计数、累乘

| 0.00 | 0.00 | | 7 | b | 0.50 | 2025-01-03 | 0.50 | 3.50...| 3.50 | | 8 | b | 2.00 | 2025-01-03 | 2.50 | 3.50 | 3.50...| | 9 | b | 1.00 | 2025-01-03 | 3.50 | 3.50 | 3.50 | +

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Pandas 2.2 中文官方教程和指南（四）

No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...time 0 16.99 1.01 No Dinner 1 10.34 1.66 No Dinner 2 21.01 3.50...2 0.059447 1 10.34 1.66 Male No Sun Dinner 3 0.160542 2 21.01 3.50...No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50

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Postgresql自定义聚合函数入门案例

5, -1); END; $$ LANGUAGE 'plpgsql'; CREATE AGGREGATE taxi(numeric, numeric) ( INITCOND = 3.50...postgres=# SELECT trip_id, taxi(km, 2.20), 3.50 + sum(km)*2.2 AS manual FROM t_taxi GROUP BY 1; NOTICE...: prev:[3.50] curr:(3.4) outer:(2.20) return:(10.980) NOTICE: prev:[10.980] curr:(5.3) outer:(2.20)...return:(22.640) NOTICE: prev:[22.640] curr:(2.9) outer:(2.20) return:(29.020) NOTICE: prev:[3.50]

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Linux之sar命令

Linux之sar命令 01 sar命令简介 Linux中的sar命令是系统运行状态的统计命令，他讲指定的操作系统状态显示到标准的输出设备中，它的全称是system activity reporter...03 实际应用 “纸上得来终觉浅，绝知此事要躬行”，这里通过一系列的操作来了解这个命令： 1.查看内存和交换空间的使用率 [root@localhost yeyz]# sar -r Linux...2.6.32-431.el6.x86_64 (localhost.localdomain) 年月日 _x86_64_ ( CPU) 时分秒 LINUX RESTART...%nice %system %iowait %steal %idle :: PM all 3.70 0.01 3.56 3.50...0.00 89.22 Average: all 3.70 0.01 3.56 3.50 0.00 89.22

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LeetCode MySQL 178. 分数排名（dense_rank连续排名）

+----+-------+ | Id | Score | +----+-------+ | 1 | 3.50 | | 2 | 3.65 | | 3 | 4.00 | | 4 | 3.85...----+------+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 2 | | 3.65 | 3 | | 3.65 | 3 | | 3.50

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Pandas 2.2 中文官方教程和指南（五）

No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...Female No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...2 7.495 1 8.34 1.66 Male No Sun Dinner 3 4.170 2 19.01 3.50...smoker day time size 0 14.99 1.01 Female No Sun Dinner 2 2 19.01 3.50...1.01 No Sun Dinner 2 1 8.34 1.66 No Sun Dinner 3 2 19.01 3.50

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java 常量表达式,需要常量表达式？（Java switch语句）

weight > 16) price = weight * 3.95; else if (weight > 32) price = (1.20 * (weight / 16)); else price = 3.50...weight > 13) price = weight * 3.95; else if (weight > 32) price = 1.20 * (weight / 16); else price = 3.50

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Pandas 2.2 中文官方教程和指南（六）

No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...Female No Sun Dinner 2 1 10.34 1.66 Male No Sun Dinner 3 2 21.01 3.50...2 7.495 1 8.34 1.66 Male No Sun Dinner 3 4.170 2 19.01 3.50...smoker day time size 0 14.99 1.01 Female No Sun Dinner 2 2 19.01 3.50...1.01 No Sun Dinner 2 1 8.34 1.66 No Sun Dinner 3 2 19.01 3.50

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系统负载过高

示例输出：top - 10:15:23 up 2 days, 10:12, 1 user, load average: 4.50, 4.00, 3.50Tasks: 212 total, 1 running...使用 iostat 命令检查磁盘I/O情况： iostat -x 1 5示例输出：Linux 4.15.0-106-generic (hostname) 02/03/2025 _x86_64_(4 CPU

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【SQL刷题系列】：leetcode178 Rank Scores

+----+-------+ | Id | Score | +----+-------+ | 1 | 3.50 | | 2 | 3.65 | | 3 | 4.00 | | 4 | 3.85...----+------+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 2 | | 3.65 | 3 | | 3.65 | 3 | | 3.50

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