如何根据条件（组中的值）更新列？

Question

我有以下df：

+---+----+-----+
|sno|dept|color|
+---+----+-----+
|  1|  fn|  red|
|  2|  fn| blue|
|  3|  fn|green|
+---+----+-----+

如果任何颜色列值为red，那么我应将颜色列的所有值更新为red，如下所示：

+---+----+-----+
|sno|dept|color|
+---+----+-----+
|  1|  fn|  red|
|  2|  fn|  red|
|  3|  fn|  red|
+---+----+-----+

我无法理解。请帮忙;我试过以下代码：

val gp=jdbcDF.filter($"dept".contains("fn"))
     //.withColumn("newone",when($"dept"==="fn","RED").otherwise("NULL"))
    gp.show()
gp.map(
  row=>{
    val row1=row.getAs[String](1)
    var row2=row.getAs[String](2)
    val make=if(row1 =="fn") row2="red"
    Row(row(0),row(1),make)
  }
).collect().foreach(println)

Answer 1

假设：

val df = Seq(
  (1, "fn", "red"),
  (2, "fn", "blue"),
  (3, "fn", "green"),
  (4, "aa", "blue"),
  (5, "aa", "green"),
  (6, "bb", "red"),
  (7, "bb", "red"),
  (8, "aa", "blue")
).toDF("id", "fn", "color")

进行计算：

val redOrNot = df.groupBy("fn")
  .agg(collect_set('color) as "values")
  .withColumn("hasRed", array_contains('values, "red"))

// gives null for no option
val colorPicker = when('hasRed, "red")
val result = df.join(redOrNot, "fn")
  .withColumn("resultColor", colorPicker) 
  .withColumn("color", coalesce('resultColor, 'color)) // skips nulls that leads to the answer
  .select('id, 'fn, 'color)

result看起来如下（这似乎是一个答案）：

scala> result.show
+---+---+-----+
| id| fn|color|
+---+---+-----+
|  1| fn|  red|
|  2| fn|  red|
|  3| fn|  red|
|  4| aa| blue|
|  5| aa|green|
|  6| bb|  red|
|  7| bb|  red|
|  8| aa| blue|
+---+---+-----+

您可以链接when运算符，并使用otherwise的默认值。咨询scaladoc of when operator。

我认为你可以使用窗口操作符或用户定义的聚合函数（UDAF）做一些非常相似（也许更有效）的事情，但是......好吧......目前还不知道怎么做。将评论留在这里以激励他人; - ）

P.S。学到了很多！谢谢你的想法！

Answer 2

高效的解决方案，不需要昂贵的分组：

// All groups with `red`
df.where($"color" === "red").select($"fn".alias("fn_")).distinct
  // Join with input
  .join(df.as("df"), $"fn" === $"fn_", "rightouter")
  // Replace `color`
  .withColumn("color", when($"fn_"isNull, $"color").otherwise(lit("red")))
  .drop("fn_")

Answer 3

如果DataFrame满足某个属性，则会有条件地更新它。在这种情况下，属性为＆＃34;颜色列包含红色＆＃39;＆＃34;。表达这一点的惯用方法是使用所需的谓词进行过滤，然后确定是否有任何行满足它。没有必要加入。

import org.apache.spark.sql.functions.lit
import org.apache.spark.sql.DataFrame

def makeAllRedIfAnyAreRed(df: DataFrame) = {
    val containsRed = df.filter(df("color") === "red").count() > 0
    if (containsRed) df.withColumn("color", lit("red")) else df
}

Answer 4

由于过滤后的数据框中可能只有很少的行，因此我添加了isin()和.withColumn()组合的解决方案。

示例DataFrame

val df = Seq(
  (1, "fn", "red"),
  (2, "fn", "blue"),
  (3, "fn", "green"),
  (4, "aa", "blue"),
  (5, "aa", "green"),
  (6, "bb", "red"),
  (7, "bb", "red"),
  (8, "aa", "blue")
).toDF("id", "dept", "color")

现在，我们只选择至少有一个红色 dept行的color，并将其放在broadcast变量中，如下所示。< / p>

val depts = sc.broadcast(df.filter($"color" === "red").select(collect_set("dept")).first.getSeq[String](0)))

更新已过滤的depts条记录的红色颜色。

  

isin()采用vararg转换列表为vararg（depts.value:_*）

//creating new column by giving diff name (clr) to see the diff
val result = df.withColumn("clr", when($"dept".isin(depts.value:_*),lit("red"))
                    .otherwise($"color"))

result.show()

+---+----+-----+-----+
| id|dept|color|  clr|
+---+----+-----+-----+
|  1|  fn|  red|  red|
|  2|  fn| blue|  red|
|  3|  fn|green|  red|
|  4|  aa| blue| blue|
|  5|  aa|green|green|
|  6|  bb|  red|  red|
|  7|  bb|  red|  red|
|  8|  aa| blue| blue|
+---+----+-----+-----+

Answer 5

Spark 2.2.0： 样本数据帧（取自上面的例子）

    val df = Seq(
  (1, "fn", "red"),
  (2, "fn", "blue"),
  (3, "fn", "green"),
  (4, "aa", "blue"),
  (5, "aa", "green"),
  (6, "bb", "red"),
  (7, "bb", "red"),
  (8, "aa", "blue")
).toDF("id", "dept", "color")

通过检查条件创建了一个UDF来执行更新。

val replace_val = udf((x: String,y:String) => if (Option(x).getOrElse("").equalsIgnoreCase("fn")&&(!y.equalsIgnoreCase("red"))) "red" else y)

val final_df = df.withColumn("color", replace_val($"dept",$"color"))
final_df.show()

输出：

火花1.6：

val conf = new SparkConf().setMaster("local").setAppName("My app")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)

import sqlContext.implicits._
// For implicit conversions like converting RDDs to DataFrames
val df = sc.parallelize(Seq(
  (1, "fn", "red"),
  (2, "fn", "blue"),
  (3, "fn", "green"),
  (4, "aa", "blue"),
  (5, "aa", "green"),
  (6, "bb", "red"),
  (7, "bb", "red"),
  (8, "aa", "blue")
) ).toDF("id","dept","color")


val replace_val = udf((x: String,y:String) => if (Option(x).getOrElse("").equalsIgnoreCase("fn")&&(!y.equalsIgnoreCase("red"))) "red" else y)
val final_df = df.withColumn("color", replace_val($"dept",$"color"))

final_df.show()