如何在 Pyspark 中加入多个列？

Question

我正在使用 Spark 1.3，并希望使用 python 接口 (SparkSQL) 加入多个列

以下作品：

我首先将它们注册为临时表。

numeric.registerTempTable("numeric")
Ref.registerTempTable("Ref")

test  = numeric.join(Ref, numeric.ID == Ref.ID, joinType='inner')

我现在想根据多个列加入它们。

我得到SyntaxError : invalid syntax ：

test  = numeric.join(Ref,
   numeric.ID == Ref.ID AND numeric.TYPE == Ref.TYPE AND
   numeric.STATUS == Ref.STATUS ,  joinType='inner')

Answer 1

你应该使用& / | 运算符并注意运算符优先级（ ==优先级低于按位AND和OR ）：

df1 = sqlContext.createDataFrame(
    [(1, "a", 2.0), (2, "b", 3.0), (3, "c", 3.0)],
    ("x1", "x2", "x3"))

df2 = sqlContext.createDataFrame(
    [(1, "f", -1.0), (2, "b", 0.0)], ("x1", "x2", "x3"))

df = df1.join(df2, (df1.x1 == df2.x1) & (df1.x2 == df2.x2))
df.show()

## +---+---+---+---+---+---+
## | x1| x2| x3| x1| x2| x3|
## +---+---+---+---+---+---+
## |  2|  b|3.0|  2|  b|0.0|
## +---+---+---+---+---+---+

Answer 2

另一种方法是：

df1 = sqlContext.createDataFrame(
    [(1, "a", 2.0), (2, "b", 3.0), (3, "c", 3.0)],
    ("x1", "x2", "x3"))

df2 = sqlContext.createDataFrame(
    [(1, "f", -1.0), (2, "b", 0.0)], ("x1", "x2", "x4"))

df = df1.join(df2, ['x1','x2'])
df.show()

输出：

+---+---+---+---+
| x1| x2| x3| x4|
+---+---+---+---+
|  2|  b|3.0|0.0|
+---+---+---+---+

主要优点是表所连接的列在输出中不会重复，从而降低了遇到诸如org.apache.spark.sql.AnalysisException: Reference 'x1' is ambiguous, could be: x1#50L, x1#57L.等错误的风险org.apache.spark.sql.AnalysisException: Reference 'x1' is ambiguous, could be: x1#50L, x1#57L.

---

每当两个表中的列具有不同的名称时（假设在上面的示例中， df2具有列y1 、 y2和y4 ），您可以使用以下语法：

df = df1.join(df2.withColumnRenamed('y1','x1').withColumnRenamed('y2','x2'), ['x1','x2'])

Answer 3

test = numeric.join(Ref, 
   on=[
     numeric.ID == Ref.ID, 
     numeric.TYPE == Ref.TYPE,
     numeric.STATUS == Ref.STATUS 
   ], how='inner')

Answer 4

如果列名相同，您还可以提供字符串列表。

df1 = sqlContext.createDataFrame(
    [(1, "a", 2.0), (2, "b", 3.0), (3, "c", 3.0)],
    ("x1", "x2", "x3"))

df2 = sqlContext.createDataFrame(
    [(1, "f", -1.0), (2, "b", 0.0)], ("x1", "x2", "x3"))

df = df1.join(df2, ["x1","x2"])

df.show()
+---+---+---+---+
| x1| x2| x3| x3|
+---+---+---+---+
|  2|  b|3.0|0.0|
+---+---+---+---+

go 关于此的另一种方法是，如果列名不同并且您想依赖列名字符串，则如下所示：

df1 = sqlContext.createDataFrame(
    [(1, "a", 2.0), (2, "b", 3.0), (3, "c", 3.0)],
    ("x1", "x2", "x3"))

df2 = sqlContext.createDataFrame(
    [(1, "f", -1.0), (2, "b", 0.0)], ("y1", "y2", "y3"))

df = df1.join(df2, (col("x1")==col("y1")) & (col("x2")==col("y2")))

df.show()
+---+---+---+---+---+---+
| x1| x2| x3| y1| y2| y3|
+---+---+---+---+---+---+
|  2|  b|3.0|  2|  b|0.0|
+---+---+---+---+---+---+

如果您想动态引用列名，并且在列名中有空格并且您不能使用df.col_name语法的情况下，这很有用。 不过，您应该考虑在这种情况下更改列名。