Question

我想根据多个条件使用另外两列更新一列中的值。例如-流就像：

    +---+---+----+---+
    | A | B | C  | D |
    +---+---+----+---+
    | a | T | 10 | 0 |
    | a | T | 100| 0 |
    | a | L | 0  | 0 |
    | a | L | 1  | 0 |
    +---+---+----+---+

我有多种情况-

（B =“ T” && C> 20）或（B =“ L” && C = 0）

值"T"，20，"L"和0是动态的。 AND/OR个运算符也在运行时提供。只要条件成立，我就想制作D = 1，否则应该保留D = 0。条件的数量也是动态的。

我尝试通过UPDATE（即spark-sql）中的UPDATE df SET D = '1' WHERE CONDITIONS命令使用它。但是它说尚不支持该更新。结果数据框应为-

+---+---+----+---+
| A | B | C  | D |
+---+---+----+---+
| a | T | 10 | 0 |
| a | T | 100| 1 |
| a | L | 0  | 1 |
| a | L | 1  | 0 |
+---+---+----+---+

有什么办法可以做到这一点？

Answer 1

我希望您正在使用Python。同样也将为Scala发布！使用udf

PYTHON

>>> df.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  0|
|  a|  L|  0|  0|
|  a|  L|  1|  0|
+---+---+---+---+

>>> def get_column(B, C):
...     return int((B == "T" and C > 20) or (B == "L" and C == 0))
...
>>> fun = udf(get_column)
>>> res = df.withColumn("D", fun(df['B'], df['C']))>>> res.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  1|
|  a|  L|  0|  1|
|  a|  L|  1|  0|
+---+---+---+---+

SCALA

scala> import org.apache.spark.sql.functions._
import org.apache.spark.sql.functions._

scala> df.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  0|
|  a|  L|  0|  0|
|  a|  L|  1|  0|
+---+---+---+---+


scala> def get_column(B : String, C : Int) : Int = {     
     |     if((B == "T" && C > 20) || (B == "L" && C == 0))
     |         1     
     |     else
     |         0
     | }
get_column: (B: String, C: Int)Int

scala> val fun = udf(get_column _)
fun: org.apache.spark.sql.expressions.UserDefinedFunction = UserDefinedFunction(<function2>,IntegerType,Some(List(StringType, IntegerType)
))

scala> val res = df.withColumn("D", fun(df("B"), df("C")))
res: org.apache.spark.sql.DataFrame = [A: string, B: string ... 2 more fields]

scala> res.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  1|
|  a|  L|  0|  1|
|  a|  L|  1|  0|
+---+---+---+---+

您还可以像这样使用case when和otherwise：

PYTHON

>>> df.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  0|
|  a|  L|  0|  0|
|  a|  L|  1|  0|
+---+---+---+---+

>>> new_column = when(
        (col("B") == "T") & (col("C") > 20), 1
    ).when((col("B") == "L") & (col("C") == 0), 1).otherwise(0)

>>> res = df.withColumn("D", new_column)
>>> res.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  1|
|  a|  L|  0|  1|
|  a|  L|  1|  0|
+---+---+---+---+

SCALA

scala> df.show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  0|
|  a|  L|  0|  0|
|  a|  L|  1|  0|
+---+---+---+---+

scala> val new_column = when(
     |     col("B") === "T" && col("C") > 20, 1
     | ).when(col("B") === "L" && col("C") === 0, 1 ).otherwise(0)

new_column: org.apache.spark.sql.Column = CASE WHEN ((B = T) AND (C > 20)) THEN 1 WHEN ((B = L) AND (C = 0)) THEN 1 ELSE 0 END

scala> df.withColumn("D", new_column).show()
+---+---+---+---+
|  A|  B|  C|  D|
+---+---+---+---+
|  a|  T| 10|  0|
|  a|  T|100|  1|
|  a|  L|  0|  1|
|  a|  L|  1|  0|
+---+---+---+---+

根据Spark结构化流中的多个条件从其他列更新列值

1 个答案: