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How to use transform higher-order function?


How to use transform higher-order function?

By : nhocyeuhoc
Date : November 22 2020, 04:01 AM
To fix this issue
Is there anyway to use it as a standard function located in package org.apache.spark.sql.functions._ ?
code :
org.apache.spark.sql.functions._

expr("transform(i, x -> x + 1)"): Column
spark.udf.register("f", (x: Int) => x + 1)

Seq((1, Seq(1, 2, 3))).toDF("id", "xs")
  .withColumn("xsinc", expr("transform(xs, x -> f(x))"))
  .show
+---+---------+---------+
| id|       xs|    xsinc|
+---+---------+---------+
|  1|[1, 2, 3]|[2, 3, 4]|
+---+---------+---------+
from typing import Optional
from pyspark.sql.functions import expr

sc.version
'2.4.0'
def f(x: Optional[int]) -> Optional[int]:
    return x + 1 if x is not None else None

spark.udf.register('f', f, "integer")

df = (spark
    .createDataFrame([(1, [1, 2, 3])], ("id", "xs"))
    .withColumn("xsinc", expr("transform(xs, x -> f(x))")))

df.printSchema()
root
 |-- id: long (nullable = true)
 |-- xs: array (nullable = true)
 |    |-- element: long (containsNull = true)
 |-- xsinc: array (nullable = true)
 |    |-- element: integer (containsNull = true)
df.show()
+---+---------+-----+
| id|       xs|xsinc|
+---+---------+-----+
|  1|[1, 2, 3]| [,,]|
+---+---------+-----+


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Find the priority function / alphabet order for extreme higher order elements relation

Find the priority function / alphabet order for extreme higher order elements relation


By : SHAILESH DEV SAHU
Date : March 29 2020, 07:55 AM
hop of those help? My intuition is that all variations on this problem will prove to be NP-hard. So I'd begin looking for heuristics that produce reasonable answers. This may involve some trial and error.
A simplistic approach is to write down a possible permutation. And then try possible swaps until you've arrived at a local minimum. Try several times, and pick the best answer.
C++11 How to use lambda and higher order functions to transform a vector to a vector of different type

C++11 How to use lambda and higher order functions to transform a vector to a vector of different type


By : byrneout64
Date : March 29 2020, 07:55 AM
hope this fix your issue std::transform does not return a vector, it applies the specified transformation and stores the results in the destination range you pass it
code :
std::transform(m_Foo_strings.begin(),
               m_Foo_strings.end(),
               std::back_inserter(m_Foos),
               [this](std::string &s){ return toFoo(s); } );
NodeJS: Higher Order Function to Create Higher Order Function with Callback

NodeJS: Higher Order Function to Create Higher Order Function with Callback


By : egil137
Date : March 29 2020, 07:55 AM
I hope this helps you . All my functions work. I just spend many hours testing with the wrong input. An input that works would be the following:
code :
const inputFunc1 = function (func2) {func2()};
const inputFunc2 = function (callback) { callback()};
putFunctionInFunction(inputFunc1,inputFunc2)(() => {console.log("hello")});
Higher order function sort() used it with other higher order function its working with let content also

Higher order function sort() used it with other higher order function its working with let content also


By : Elle
Date : March 29 2020, 07:55 AM
like below fixes the issue The sort func you are using is the mutating func, that means sort function internally sort the array in the same variable.
Mutating Func : To modify the properties of your structure or enumeration within a particular method, you can opt in to mutating behavior for that method. The method can then mutate (that is, change) its properties from within the method, and any changes that it makes are written back to the original structure when the method ends.
spark higher order function transform output struct

spark higher order function transform output struct


By : PearlChen
Date : March 29 2020, 07:55 AM
I think the issue was by ths following ,
In this way, you can achieve your desired output. you cannot directly access other value bcoz foo and other are sharing the same hierarchy. so you need to access other separately.
code :
scala>  df.withColumn("cleaned", expr("transform(bazes, x -> struct(concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3),cast(x.other as string)))")).show(false)
+---+----------------------------------------------------+------------------------------------------------+
|id |bazes                                               |cleaned                                         |
+---+----------------------------------------------------+------------------------------------------------+
scala>  df.withColumn("cleaned", expr("transform(bazes, x -> struct(concat(x.foo.thing1, '::', x.foo.thing2, '::', x.foo.thing3),cast(x.other as string)))")).printSchema
root
 |-- id: integer (nullable = false)
 |-- bazes: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- foo: struct (nullable = true)
 |    |    |    |-- thing1: string (nullable = true)
 |    |    |    |-- thing2: string (nullable = true)
 |    |    |    |-- thing3: string (nullable = true)
 |    |    |-- other: string (nullable = true)
 |-- cleaned: array (nullable = true)
 |    |-- element: struct (containsNull = false)
 |    |    |-- col1: string (nullable = true)
 |    |    |-- col2: string (nullable = true)
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