9.3 KiB
9.3 KiB
title | id | challengeType | videoUrl | localeTitle |
---|---|---|---|---|
Hash join | 5956795bc9e2c415eb244de1 | 5 | 哈希加入 |
Description
内连接是一种操作,它根据匹配的列值将两个数据表组合到一个表中。实现此操作的最简单方法是嵌套循环连接算法,但更可扩展的替代方法是散列连接算法。
实现“散列连接”算法,并演示它通过下面列出的测试用例。
您应该将表表示为在编程语言中感觉自然的数据结构。
“散列连接”算法包含两个步骤:
哈希阶段:从两个表中的一个表创建一个多图,从每个连接列值映射到包含它的所有行。多图必须支持基于散列的查找,它比简单的线性搜索更好地扩展,因为这是该算法的重点。理想情况下,我们应该为较小的表创建多图,从而最小化其创建时间和内存大小。加入阶段:扫描另一个表,通过查看之前创建的多图来查找匹配的行。在伪代码中,算法可以表示如下:
让A =第一个输入表(或理想情况下,更大的输入表) 设B =第二个输入表(或理想情况下,较小的输入表) 令j A =表A的连接列ID 令j B =表B的连接列ID 让M B =一个多图,用于从单个值映射到表B的多行(从空白开始) 让C =输出表(从空开始) 对于表B中的每一行b: 将b放在密钥b(j B )下的多映射M B中 对于表A中的每一行a: 对于a(j A )项下多图M B中的每一行b: 设c =第a行和第b行的串联 将行c放在表C中测试用例
输入
|
产量
A.Age | 一个名字 | B.Character | B.Nemesis |
---|---|---|---|
27 | 约拿 | 约拿 | 鲸鱼 |
27 | 约拿 | 约拿 | 蜘蛛 |
18 | 艾伦 | 艾伦 | 鬼 |
18 | 艾伦 | 艾伦 | 植物大战僵尸 |
28 | 荣耀 | 荣耀 | 巴菲 |
28 | 艾伦 | 艾伦 | 鬼 |
28 | 艾伦 | 艾伦 | 植物大战僵尸 |
输出表中行的顺序并不重要。
如果你使用数字索引数组来表示表行(而不是按名称引用列),你可以用[[27, "Jonah"], ["Jonah", "Whales"]]
的形式表示输出行。 。
Instructions
Tests
tests:
- text: <code>hashJoin</code>是一个函数。
testString: 'assert(typeof hashJoin === "function", "<code>hashJoin</code> is a function.");'
- text: '<code>hashJoin([{ age: 27, name: "Jonah" }, { age: 18, name: "Alan" }, { age: 28, name: "Glory" }, { age: 18, name: "Popeye" }, { age: 28, name: "Alan" }], [{ character: "Jonah", nemesis: "Whales" }, { character: "Jonah", nemesis: "Spiders" }, { character: "Alan", nemesis: "Ghosts" }, { character:"Alan", nemesis: "Zombies" }, { character: "Glory", nemesis: "Buffy" }, { character: "Bob", nemesis: "foo" }])</code>应该返回<code>[{"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Whales"}, {"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Spiders"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}, {"A_age": 28,"A_name": "Glory", "B_character": "Glory", "B_nemesis": "Buffy"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}]</code>'
testString: 'assert.deepEqual(hashJoin(hash1, hash2), res, "<code>hashJoin([{ age: 27, name: "Jonah" }, { age: 18, name: "Alan" }, { age: 28, name: "Glory" }, { age: 18, name: "Popeye" }, { age: 28, name: "Alan" }], [{ character: "Jonah", nemesis: "Whales" }, { character: "Jonah", nemesis: "Spiders" }, { character: "Alan", nemesis: "Ghosts" }, { character:"Alan", nemesis: "Zombies" }, { character: "Glory", nemesis: "Buffy" }, { character: "Bob", nemesis: "foo" }])</code> should return <code>[{"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Whales"}, {"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Spiders"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}, {"A_age": 28,"A_name": "Glory", "B_character": "Glory", "B_nemesis": "Buffy"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}]</code>");'
Challenge Seed
function hashJoin (hash1, hash2) {
// Good luck!
return [];
}
After Test
console.info('after the test');
Solution
// solution required