我有以下两个列表-
list1 = List(("jira","ABC"),
("collApp","ABC"),
("amarColl","ABC"),
("Ubuntu1404","PQR"),
("relColAdmin","XYZ"))
list2 = List(("Linux_server","Hyper-V","suseLinux"),
("Linux_server","VmWare","collApp"),
("Linux_server","VmWare","relColAdmin"),
("Linux_server","VmWare","amarColl"),
("Linux_server","Hyper-V","ubuntu1310"))从上面的两个列表中,我想检查第二个列表是否包含list1的每个元素(元组)的第一个值,如果匹配,则在list2中插入第二个值,否则在list2中插入空值。
以下是示例输出-
output = List(("Linux_server","Hyper-V","suseLinux",""),
("Linux_server","VmWare","collApp","ABC"),
("Linux_server","VmWare","relColAdmin","XYZ"),
("Linux_server","VmWare","amarColl","ABC"),
("Linux_server","Hyper-V","ubuntu1310",""))发布于 2015-02-06 16:11:15
list1已经处于将其转变为地图的完美状态。我们将把它转换成一个默认值为"“的地图。因此,如果我们从map中请求一个不存在的键,我们会得到"“
scala> val map1 = list1.toMap.withDefaultValue("")
map1: scala.collection.immutable.Map[String,String] =
Map(amarColl -> ABC, Ubuntu1404 -> PQR, relColAdmin -> XYZ,
jira -> ABC, collApp -> ABC)现在我们要做的就是将3元组映射到4元组:
scala> list2.map { case (a,b,c) => (a,b,c,map1(c)) }
res0: List[(String, String, String, String)] =
List((Linux_server,Hyper-V,suseLinux,""),
(Linux_server,VmWare,collApp,ABC),
(Linux_server,VmWare,relColAdmin,XYZ),
(Linux_server,VmWare,amarColl,ABC),
(Linux_server,Hyper-V,ubuntu1310,""))发布于 2015-02-06 15:59:07
如果list1的元组的第一个元素可以放在list2的元组的任意位置,请使用以下代码:
list2.map {
case (a, b, c) =>
(a, b, c,
list1.find(e =>
Seq(a, b, c).contains(e._1)
).map(_._2).getOrElse(""))
}如果list1的元组的第一个元素只能放在list2的元组的第三个位置:
list2.map {
case (a, b, c) =>
(a, b, c,
list1.find(_._1 == c)
.map(_._2).getOrElse(""))
}发布于 2015-02-06 16:06:39
首先,把它们拉在一起
val zipped = list1 zip list2 然后,映射到拉链对上并构建结果
val output = zipped.map(case ((key, value), targetTuple) => buildResultTuple(key, value, targetTuple))
def buildResultTuple(key:String, value: String, t:Tuple3):Tuple4 = {
val l = t.productIterator.toList
if (l contains key) (t._1, t._2, t._3, value)
else (t._1, t._2, t._3, "")
} https://stackoverflow.com/questions/28360808
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