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How to Order/Sort the structType based on the field names in Java Spark 2.1.1

-1

I have a structType schema, i need to order / sort the schema based on the field name, below i am sharing the StructType.

StructType schema = StructType(StructField(zzz,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(aaa,StringType,true),
 StructField(lll,StringType,true))

i would like to get the above structType, as below

StructType schema = StructType(StructField(aaa,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(lll,StringType,true),
 StructField(zzz,StringType,true))

apache-spark apache-spark-sql apache-spark-dataset

share|improve this question

asked Mar 7 at 4:50

svk 041994svk 041994

487

add a comment | 

-1

I have a structType schema, i need to order / sort the schema based on the field name, below i am sharing the StructType.

StructType schema = StructType(StructField(zzz,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(aaa,StringType,true),
 StructField(lll,StringType,true))

i would like to get the above structType, as below

StructType schema = StructType(StructField(aaa,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(lll,StringType,true),
 StructField(zzz,StringType,true))

apache-spark apache-spark-sql apache-spark-dataset

share|improve this question

asked Mar 7 at 4:50

svk 041994svk 041994

487

add a comment | 

I have a structType schema, i need to order / sort the schema based on the field name, below i am sharing the StructType.

StructType schema = StructType(StructField(zzz,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(aaa,StringType,true),
 StructField(lll,StringType,true))

i would like to get the above structType, as below

StructType schema = StructType(StructField(aaa,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(lll,StringType,true),
 StructField(zzz,StringType,true))

apache-spark apache-spark-sql apache-spark-dataset

share|improve this question

asked Mar 7 at 4:50

svk 041994svk 041994

487

StructType schema = StructType(StructField(zzz,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(aaa,StringType,true),
 StructField(lll,StringType,true))

StructType schema = StructType(StructField(aaa,StringType,true), 
 StructField(kkk,StringType,true), 
 StructField(lll,StringType,true),
 StructField(zzz,StringType,true))

share|improve this question

asked Mar 7 at 4:50

svk 041994svk 041994

487

share|improve this question

asked Mar 7 at 4:50

svk 041994svk 041994

487

asked Mar 7 at 4:50

svk 041994svk 041994

487

asked Mar 7 at 4:50

svk 041994svk 041994

487

1 Answer
1

active

oldest

votes

1

In scala

var schema = StructType(Seq(StructField("zzz",StringType,true), 
 StructField("kkk",StringType,true), 
 StructField("aaa",StringType,true),
 StructField("lll",StringType,true)))
def reorderSchema: StructType => StructType = schema => StructType(schema.sortBy(_.name))

var newSchema = reorderSchema(schema)

In java,
You can do something like,

 new StructType(Stream.of(schema.fields())
.sorted(Comparator.comparing(StructField::name))
.collect(Collectors.toList()).toArray(new StructField[schema.fields().length]))

Please accept the answer if it solves.

share|improve this answer

edited Mar 7 at 7:00

answered Mar 7 at 6:09

deodeo

8615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As i am using java, the mentioned answer is not working. we should pass the two parameters for sortBy(function f, order)
  
  – svk 041994
  Mar 7 at 6:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated the answer to work with java
  
  – deo
  Mar 7 at 7:01
  

  
  
  
  

add a comment | 

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1

In scala

var schema = StructType(Seq(StructField("zzz",StringType,true), 
 StructField("kkk",StringType,true), 
 StructField("aaa",StringType,true),
 StructField("lll",StringType,true)))
def reorderSchema: StructType => StructType = schema => StructType(schema.sortBy(_.name))

var newSchema = reorderSchema(schema)

In java,
You can do something like,

 new StructType(Stream.of(schema.fields())
.sorted(Comparator.comparing(StructField::name))
.collect(Collectors.toList()).toArray(new StructField[schema.fields().length]))

Please accept the answer if it solves.

share|improve this answer

edited Mar 7 at 7:00

answered Mar 7 at 6:09

deodeo

8615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As i am using java, the mentioned answer is not working. we should pass the two parameters for sortBy(function f, order)
  
  – svk 041994
  Mar 7 at 6:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated the answer to work with java
  
  – deo
  Mar 7 at 7:01
  

  
  
  
  

add a comment | 

1

In scala

var schema = StructType(Seq(StructField("zzz",StringType,true), 
 StructField("kkk",StringType,true), 
 StructField("aaa",StringType,true),
 StructField("lll",StringType,true)))
def reorderSchema: StructType => StructType = schema => StructType(schema.sortBy(_.name))

var newSchema = reorderSchema(schema)

In java,
You can do something like,

 new StructType(Stream.of(schema.fields())
.sorted(Comparator.comparing(StructField::name))
.collect(Collectors.toList()).toArray(new StructField[schema.fields().length]))

Please accept the answer if it solves.

share|improve this answer

edited Mar 7 at 7:00

answered Mar 7 at 6:09

deodeo

8615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As i am using java, the mentioned answer is not working. we should pass the two parameters for sortBy(function f, order)
  
  – svk 041994
  Mar 7 at 6:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated the answer to work with java
  
  – deo
  Mar 7 at 7:01
  

  
  
  
  

add a comment | 

In scala

var schema = StructType(Seq(StructField("zzz",StringType,true), 
 StructField("kkk",StringType,true), 
 StructField("aaa",StringType,true),
 StructField("lll",StringType,true)))
def reorderSchema: StructType => StructType = schema => StructType(schema.sortBy(_.name))

var newSchema = reorderSchema(schema)

In java,
You can do something like,

 new StructType(Stream.of(schema.fields())
.sorted(Comparator.comparing(StructField::name))
.collect(Collectors.toList()).toArray(new StructField[schema.fields().length]))

Please accept the answer if it solves.

share|improve this answer

edited Mar 7 at 7:00

answered Mar 7 at 6:09

deodeo

8615

var schema = StructType(Seq(StructField("zzz",StringType,true), 
 StructField("kkk",StringType,true), 
 StructField("aaa",StringType,true),
 StructField("lll",StringType,true)))
def reorderSchema: StructType => StructType = schema => StructType(schema.sortBy(_.name))

var newSchema = reorderSchema(schema)

 new StructType(Stream.of(schema.fields())
.sorted(Comparator.comparing(StructField::name))
.collect(Collectors.toList()).toArray(new StructField[schema.fields().length]))

share|improve this answer

edited Mar 7 at 7:00

answered Mar 7 at 6:09

deodeo

8615

answered Mar 7 at 6:09

deodeo

8615

answered Mar 7 at 6:09

deodeo

8615