把客户表与服务商表拼在一起，难点往往不是写出一次连接，而是明确三件事：没有匹配的行要不要保留，结果需要哪一侧的列，以及两边同名的列究竟指向谁。DataFusion Python 的 join() 支持内连接、左连接、右连接、全连接、左半连接和左反连接。本文按照官方指南的全部示例，逐一说明这些选择，并处理 51.0.0 之后连接键合并行为的变化。
原文：Apache DataFusion Python 官方文档 Joins。原页未署个人作者，维护方为 Apache Software Foundation。本文为经授权的中文翻译整理，核验日期为 2026 年 10 月 5 日。以下表格是原文示例结果的整理，不是本文运行得到的输出；没有显式排序时，不应依赖行的展示顺序。
先准备两张小表
左表有三位客户，连接键叫 customer_id；右表有三家服务商，连接键叫 id。键值 1、2 同时出现在两张表里，3 只在左表，5 只在右表。这样一组数据足以看清“匹配”和“保留未匹配行”的区别。
from datafusion import SessionContext
ctx = SessionContext()
left = ctx.from_pydict(
{
"customer_id": [1, 2, 3],
"customer": ["Alice", "Bob", "Charlie"],
}
)
right = ctx.from_pylist([
{"id": 1, "name": "CityCabs"},
{"id": 2, "name": "MetroRide"},
{"id": 5, "name": "UrbanGo"},
])
内连接：只留下双方都有的键
内连接只保留连接键能够匹配的行。因此 Alice 与 CityCabs 配对，Bob 与 MetroRide 配对；Charlie 和 UrbanGo 都不进入结果。两侧连接键的名字不同，用 left_on 和 right_on 分别指定。
left.join(right, left_on="customer_id", right_on="id", how="inner")| customer_id | customer | id | name |
|---|---|---|---|
| 1 | Alice | 1 | CityCabs |
| 2 | Bob | 2 | MetroRide |
原文此处还用 join_keys 解释“分别指定两侧待匹配列”的概念，但实际示例采用的是 left_on、right_on。本文保留可直接对应示例的参数写法，不将那句说明当作另一个已验证的调用。
左连接：保留所有客户，缺失匹配用 null 表示
左连接保留左表的每一行，并附上右表中匹配的数据。Charlie 找不到服务商，仍然出现在结果里，但右侧的 id 与 name 为空。这里的空值是 null，不是数字 0，也不是空字符串。
left.join(right, left_on="customer_id", right_on="id", how="left")| customer_id | customer | id | name |
|---|---|---|---|
| 1 | Alice | 1 | CityCabs |
| 2 | Bob | 2 | MetroRide |
| 3 | Charlie | null | null |
官方页的支持列表也包含右连接。其保留方向与左连接相反：以右表为完整保留的一侧。原页没有单独给出右连接的代码与结果，本文也不额外伪造一组官方输出。
全连接：两侧未匹配的数据都留下
全连接把两张表中的行都纳入结果。有匹配的行合并成一行；只存在于一侧的行，另一侧的列填 null。它适合同时检查“客户没有服务商”和“服务商没有客户”这两类缺口。
left.join(right, left_on="customer_id", right_on="id", how="full")| customer_id | customer | id | name |
|---|---|---|---|
| 1 | Alice | 1 | CityCabs |
| 2 | Bob | 2 | MetroRide |
| null | null | 5 | UrbanGo |
| 3 | Charlie | null | null |
左半连接：只问是否匹配，不带入右表列
左半连接留下左表中至少能在右表找到一次匹配的行，输出仍然只有左表的列。右表即便有多个匹配，也不会因为这些右侧匹配而重复扩展同一条左侧记录。这不等于对左表本身做去重；它解决的是“存在匹配就保留”的筛选需求。
left.join(right, left_on="customer_id", right_on="id", how="semi")| customer_id | customer |
|---|---|
| 1 | Alice |
| 2 | Bob |
左反连接：找出完全没有匹配的左表行
左反连接保留左表里找不到任何右侧匹配的行。一旦某条左表记录至少有一个匹配，它就会被排除。示例中只有 Charlie 符合条件，输出也只包含左表列。
left.join(right, left_on="customer_id", right_on="id", how="anti")| customer_id | customer |
|---|---|
| 3 | Charlie |
由此可以把选择标准说得很具体：需要两边的字段并只看匹配项，用内连接；要完整保留一侧，用对应方向的外连接；要看双方的缺口，用全连接；只根据右表是否存在匹配来筛左表，用半连接或反连接。
同名连接键：51.0.0 起默认合并
接下来将左表的连接键也改为 id。这里讨论的“重复键”是连接结果中两个键列名字相同，并不是说输入中出现了重复键值。从 datafusion-python 51.0.0 开始，默认会合并两侧同名的连接键，减少后续选列的歧义。
left = ctx.from_pydict(
{
"id": [1, 2, 3],
"customer": ["Alice", "Bob", "Charlie"],
}
)
right = ctx.from_pylist([
{"id": 1, "name": "CityCabs"},
{"id": 2, "name": "MetroRide"},
{"id": 5, "name": "UrbanGo"},
])
left.join(right, "id", how="inner")| id | customer | name |
|---|---|---|
| 1 | Alice | CityCabs |
| 2 | Bob | MetroRide |
如果确实需要保留两侧的键列，将 coalesce_duplicate_keys 显式设为 False：
left.join(right, "id", how="inner", coalesce_duplicate_keys=False)| id（左） | customer | id（右） | name |
|---|---|---|---|
| 1 | Alice | 1 | CityCabs |
| 2 | Bob | 2 | MetroRide |
上表中的“左”“右”是为了阅读而加的说明；原始结果的两个列名都叫 id，不会因为关闭合并而自动获得本文的标签。升级旧项目时，应检查后续选择列的逻辑是否依赖过去存在两个同名键列的行为。
同名的普通列：通过 DataFrame.col() 指明来源
连接键之外，两张表也可能各自有一列 val。这时仅凭列名无法表达想要哪一侧的数据。应在连接前，通过各自 DataFrame 的 col() 取得带来源限定的列引用，再把这些引用用于连接条件与结果投影。接受表达式条件的 join_on() 很适合这种写法。
left = ctx.from_pydict(
{
"id": [1, 2, 3],
"val": [10, 20, 30],
}
)
right = ctx.from_pydict(
{
"id": [1, 2, 3],
"val": [40, 50, 60],
}
)
joined = left.join_on(
right, left.col("id") == right.col("id"), how="inner"
)
joined.select(left.col("id"), left.col("val"), right.col("val"))| id | val（左） | val（右） |
|---|---|---|
| 1 | 10 | 40 |
| 2 | 20 | 50 |
| 3 | 30 | 60 |
这里限定的是表达式指向的来源。原文结果仍然显示两个 val 列，并未自动重命名。上表再次用括号标出来源，仅为避免读者混淆。实际整理下游数据时，可以另行制定清晰的字段命名，但不要把同名键默认合并误解为所有同名普通列都会自动消歧。
本文只静态检查了代码的参数、字段对应关系和结果语义，没有安装或运行 DataFusion。示例只使用内存中的常量数据，没有凭据、拼接 SQL 或破坏性操作；这项检查也不意味着已完成安全审计。将示例迁移到真实数据时，还应单独核对输入中重复键值、空连接键、数据类型和排序要求，它们并未由这组小表覆盖。
原始版权与许可
Apache DataFusion Python 文档及示例采用 Apache License 2.0；中文翻译与技术整理已在正文标示。 官方许可来源
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