以下是兩個具有相同列名的Dataset按照同名列相等進行join操作，join結果中同名列只會出現一個：

val df1 = Seq((1, 2, 3),(1, 1, 1)).toDF("a", "b", "c")
val df2 = Seq((1, 2, 4),(2, 2, 2)).toDF("a", "b", "c")

df1.show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  2|  3|
|  1|  1|  1|
+---+---+---+

df2.show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1
 
|  2|  4|
|  2|  2|  2|
+---+---+---+

df1.join(df2, Seq("a","b"), "inner").show
+---+---+---+---+
|  a|  b|  c|  c|
+---+---+---+---+
|  1|  2|  3|  4|
+---+---+---+---+

df1.join(df2, Seq("a","b"), "outer").show
+---+---+----+----+
|  a|  b|   c|   c|
+---+---+----+----+
|  2|  2|null|   2|
|  1|  2
 
|   3|   4|
|  1|  1|   1|null|
+---+---+----+----+

df1.join(df2, Seq("a","b"), "right").show
+---+---+----+---+
|  a|  b|   c|  c|
+---+---+----+---+
|  1|  2|   3|  4|
|  2|  2|null|  2|
+---+---+----+---+

df1.join(df2, Seq("a","b"), "left").show
+---+---+---+----+
|  a|  b|  c|   c|
+---+---+---+----+
|  1
 
|  2|  3|   4|
|  1|  1|  1|null|
+---+---+---+----+

df1.join(df2, Seq("a","b"), "leftsemi").show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  2|  3|
+---+---+---+

df1.join(df2, Seq("a","b"), "leftanti").show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  1|  1|
+---+---+---+

以下是具有不同列名的Dataset按照不同名列相等進行join操作，join結果中兩個不同名列都會出現（因為列名不同不能只保留某一列）：

val df1 = Seq((1, 2, 3),(1, 1, 1)).toDF("a", "b", "c")
val df2 = Seq((1, 2, 4),(2, 2, 2)).toDF("a1", "b1", "c1")

df1.show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  2|  3|
|  1|  1|  1|
+---+---+---+

df2.show
+---+---+---+
| a1| b1| c1|
+---+---+---+
|  1|  2|  4|
|  2|  2|  2|
+---+---+---+

df1.join(df2, col("a") === col("a1") && col("b") === col("b1"), "outer").show
+----+----+----+----+----+----+                                                 
|   a|   b|   c|  a1|  b1|  c1|
+----+----+----+----+----+----+
|null|null|null|   2|   2|   2|
|   1|   2|   3|   1|   2|   4|
|   1|   1|   1|null|null|null|
+----+----+----+----+----+----+

若是具有部分相同、部分不同列名的兩個Dataset按照部分相同、部分不同列相等進行join操作，有以下幾種方式：

val df1 = Seq((1, 2, 3),(1, 1, 1)).toDF("a", "b", "c")
val df2 = Seq((1, 2, 4),(2, 2, 2)).toDF("a", "b1", "d")

df1.show
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  2|  3|
|  1|  1|  1|
+---+---+---+
df2.show
+---+---+---+
|  a| b1|  d|
+---+---+---+
|  1|  2|  4|
|  2|  2|  2|
+---+---+---+
//join條件：df1("a") == df2("a") && df1("b") == df2("b1") 

//若是直接join會報錯：org.apache.spark.sql.AnalysisException: Reference 'a' is ambiguous, could be:...
df1.join(df2, col("a") === col("a") && col("b") === col("b1"), "outer").show
//可以改為這樣：
df1.join(df2, df1("a") === df2("a") && col("b") === col("b1"), "outer").show
+----+----+----+----+----+----+
|   a|   b|   c|   a|  b1|   d|
+----+----+----+----+----+----+
|null|null|null|   2|   2|   2|
|   1|   2|   3|   1|   2|   4|
|   1|   1|   1|null|null|null|
+----+----+----+----+----+----+

//當然也可以將其中一個Dataset的列改名，改為都相同或都不同，再用上面的方法join
df1.join(df2.withColumnRenamed("b1", "b"), Seq("a", "b"), "outer").show
+---+---+----+----+
|  a|  b|   c|   d|
+---+---+----+----+
|  2|  2|null|   2|
|  1|  2|   3|   4|
|  1|  1|   1|null|
+---+---+----+----+

//還可以用Dataset的as方法（與alias方法等效），給Dataset命名，然後消除歧義。（Dataset的別名類似SQL中表的別名）
df1.alias("df1")
    .join(df2.as("df2"), col("df1.a") === col("df2.a") && col("b") === col("b1"), "outer")
    .show
+----+----+----+----+----+----+
|   a|   b|   c|   a|  b1|   d|
+----+----+----+----+----+----+
|null|null|null|   2|   2|   2|
|   1|   2|   3|   1|   2|   4|
|   1|   1|   1|null|null|null|
+----+----+----+----+----+----+
//如果只想保留df2的a列：
val t = df1.alias("df1")
    .join(df2.as("df2"), col("df1.a") === col("df2.a") && col("b") === col("b1"), "outer")
    .drop(col("df1.a")).show
+----+----+----+----+----+
|   b|   c|   a|  b1|   d|
+----+----+----+----+----+
|null|null|   2|   2|   2|
|   2|   3|   1|   2|   4|
|   1|   1|null|null|null|
+----+----+----+----+----+

補充：
Dataset的as方法（與alias方法等效）：為Dataset物件起別名，Dataset的別名類似SQL中表的別名。

val df = Seq((1, 2),(1, 1)).toDF("a", "b")
df.select("a").show
+---+
|  a|
+---+
|  1|
|  1|
+---+

df.select("df.a").show
//報錯：org.apache.spark.sql.AnalysisException: cannot resolve '`df.a`' given input columns: [a, b];

df.as("df").select("df.a").show
+---+
|  a|
+---+
|  1|
|  1|
+---+

注意：

以上Dataset的as方法是指的：def as(alias: Symbol): Dataset[T]。
另外Dataset還有一個as方法：def as[U](implicit arg0: Encoder[U]): Dataset[U]，這篇文章和此方法無關。