问题描述
似乎 StructType
保留顺序,因此两个包含相同 StructField
的 StructType
并不等效.
It seems like StructType
preserves order, so two StructType
containing same StructField
s are not considered equivalent.
例如:
val st1 = StructType(
StructField("ii",StringType,true) ::
StructField("i",StringType,true) :: Nil)
val st2 = StructType(
StructField("i",StringType,true) ::
StructField("ii",StringType,true) :: Nil)
println(st1 == st2)
返回 false
,即使它们都具有 StructField("i",StringType,true)
和 StructField("ii",StringType,true)代码>,只是顺序不同.
returns false
even though they both have StructField("i",StringType,true)
and StructField("ii",StringType,true)
, just in different order.
我需要一个测试,可以说这两个是等效的,因为出于我的目的,这两个没有区别.
I need a test that can say that these two are equivalent because for my purpose, these two are not different.
val schema1 = StructType(StructField("A",ArrayType(st1,true),true) :: Nil)
val schema2 = StructType(StructField("A",ArrayType(st2,true),true) :: Nil)
val final_schema = StructType((schema1 ++ schema2).distinct)
final_schmea 的结果应该只有一个 A
的 StructType
,而不是两个,但是 distinct
会考虑这些两个 StructType
不同,所以我最终得到两个名为 A
的不同 StructField
.所以我的问题是,有没有一种方法可以根据其内容而不是按订单比较两个 StructType
?
The result of final_schmea
should only have one StructType
of A
instead of two, but distinct
considers these two StructType
as different, so I end up getting two different StructField
named A
. So my question is, is there a way to compare two StructType
s based on their contents, not on orders?
经过进一步调查,由于 StructType
基本上是 Seq< StructField>
,因此我可以执行,但是我我正在尝试一种方法,可以最有效地对嵌入式 StructType
进行比较.
After further investigation, since StructType
is basically Seq<StructField>
, I can do content comparison for that works for Seq
, but I am trying to think of a way I can do comparison for embedded StructType
most efficiently.
推荐答案
这可能可以清除,但它可以工作并处理嵌套的StructType:
This can probably be cleaned up, but it works and handles nested StructType:
def isEqual(struct1: StructType, struct2: StructType): Boolean = {
struct1.headOption match {
case Some(field) => {
if(field.dataType.typeName != "struct") {
struct2.find(_ == field) match {
case Some(matchedField) => isEqual(StructType(struct1.filterNot(_ == field)), StructType(struct2.filterNot(_ == field)))
case None => false
}
} else {
val isEqualContents = struct2.find(x => x.name == field.name && x.nullable == field.nullable && x.dataType.typeName == "struct") match {
case Some(matchedField) => isEqual(field.dataType.asInstanceOf[StructType], matchedField.dataType.asInstanceOf[StructType])
case None => false
}
if(isEqualContents) isEqual(StructType(struct1.filterNot(_ == field)), StructType(struct2.filterNot(_ == field))) else false
}
}
case None => struct2.size == 0
}
}
val st1 = StructType(
StructField("ii",StringType,true) ::
StructField("i",StringType,true) ::
StructField("iii", StructType(StructField("iv", StringType, true) :: Nil), true) :: Nil)
val st2 = StructType(
StructField("i",StringType,true) ::
StructField("ii",StringType,true) ::
StructField("iii", StructType(StructField("v", StringType, true) :: Nil), true) :: Nil)
isEqual(st1, st2)
它也可以使用更多的爱来成为尾递归.
It could also use a little more love to become tail recursive, too.
这篇关于如何比较两个共享相同内容的StructType?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!