val descrList = cursorReal.interfaceInfo.interfaces.map {
case values => (values.ifIndex , values.ifName , values.ifType)
}
val ipAddressList = cursorReal.interfaceIpAndIndex.filter(x=> (!x.ifIpAddress.equalsIgnoreCase("0"))).map {
case values => (values.ifIndex,values.ifIpAddress)
}
例如,
val descrList =
List((12,"VoIP-Null0",1), (8,"FastEthernet6",6), (19,"Vlan11",53),
(4,"FastEthernet2",6), (15,"Vlan1",53), (11,"GigabitEthernet0",6),
(9,"FastEthernet7",6), (22,"Vlan20",53), (13,"Wlan-GigabitEthernet0",6),
(16,"Async1",1), (5,"FastEthernet3",6), (10,"FastEthernet8",6),
(21,"Vlan12",53), (6,"FastEthernet4",6), (1,"wlan-ap0",24),
(17,"Virtual-Template1",131), (14,"Null0",1), (20,"Vlan10",53),
(2,"FastEthernet0",6), (18,"NVI0",1), (7,"FastEthernet5",6),
(29,"Virtual-Access7",131), (3,"FastEthernet1",6), (28,"Virtual-Access6",131))
val ipAddressList = List((21,"192.168.12.1"), (19,"192.168.11.1"),
(11,"104.36.252.115"), (20,"192.168.10.1"),
(22,"192.168.20.1"))
在两个列表中,第一个元素是索引,我必须明智地合并这两个列表。它的意思是
(21,"192.168.12.1")
此ipAddress应与(21,"Vlan12",53)
合并,并形成如下(21,"Vlan12",53,"192.168.12.1")
下方的新列表。
答案 0 :(得分:1)
scala> descrList map {case (index, v1, v2) =>
(index, v1, v2, ipAddressList.toMap.getOrElse(index, "empty"))}
res0: List[(Int, String, Int, String)] = List(
(12,VoIP-Null0,1,empty), (8,FastEthernet6,6,empty), (19,Vlan11,53,192.168.11.1),
(4,FastEthernet2,6,empty), (15,Vlan1,53,empty), (11,GigabitEthernet0,6,104.36.252.115),
(9,FastEthernet7,6,empty), (22,Vlan20,53,192.168.20.1), (13,Wlan-GigabitEthernet0,6,empty),
(16,Async1,1,empty), (5,FastEthernet3,6,empty), (10,FastEthernet8,6,empty),
(21,Vlan12,53,192.168.12.1), (6,FastEthernet4,6,empty), (1,wlan-ap0,24,empty), (17,Virtual-
Template1,131,empty), (14,Null0,1,empty), (20,Vlan10,53,192.168.10.1), (2,FastEthernet0,6,empty),
(18,NVI0,1,empty), (7,FastEthernet5,6,empty), (29,Virtual-Access7,131,empty),
(3,FastEthernet1,6,empty), (28,Virtual-Access6,131,empty))
答案 1 :(得分:0)
首先,我建议您生成一个Map而不是List。 Map本质上有一个索引器,在你的情况下,这将是ifIndex值。
一旦你有地图,你可以使用这样的东西(来自其他SO Best way to merge two maps and sum the values of same key?的样本)
来自Rex Kerr: map1 ++ map2.map {case(k,v)=> k - > (v + map1.getOrElse(k,0))}
或者像Matthew Farwell这样: (map1.keySet ++ map2.keySet).map(i =>(i,map1.getOrElse(i,0)+ map2.getOrElse(i,0))}。toMap
如果您出于任何原因无法使用Google地图,请查看您现有的项目库。如果你有Scalaz,那么你已经有了一些工具。
Scalaz:https://github.com/scalaz/scalaz
如果你有Slick,你也有一些很好的工具可以直接使用。
答案 2 :(得分:0)
首先考虑将decrList
转换为Map
,就像这样,
val a = (for ( (k,v1,v2) <- descrList) yield k -> (v1,v2)).toMap
然后我们可以查找ipAddressList
的键并将元素聚合成一个新的元组,如下所示,
for ( (k,ip) <- ipAddressList ; v = a.getOrElse(k,("none","none")) ) yield (k,v._1,v._2,ip)
因此,ipAddressList
,
res: List((21,Vlan12,53,192.168.12.1), (19,Vlan11,53,192.168.11.1),
(11,GigabitEthernet0,6,104.36.252.115), (20,Vlan10,53,192.168.10.1),
(22,Vlan20,53,192.168.20.1))
答案 3 :(得分:0)
鉴于数据:
val descrList =
List((12, "VoIP-Null0", 1), (8, "FastEthernet6", 6), (19, "Vlan11", 53),
(4, "FastEthernet2", 6), (15, "Vlan1", 53), (11, "GigabitEthernet0", 6),
(9, "FastEthernet7", 6), (22, "Vlan20", 53), (13, "Wlan-GigabitEthernet0", 6),
(16, "Async1", 1), (5, "FastEthernet3", 6), (10, "FastEthernet8", 6),
(21, "Vlan12", 53), (6, "FastEthernet4", 6), (1, "wlan-ap0", 24),
(17, "Virtual-Template1", 131), (14, "Null0", 1), (20, "Vlan10", 53),
(2, "FastEthernet0", 6), (18, "NVI0", 1), (7, "FastEthernet5", 6),
(29, "Virtual-Access7", 131), (3, "FastEthernet1", 6), (28, "Virtual-Access6", 131))
val ipAddressList = List((21, "192.168.12.1"), (19, "192.168.11.1"),
(11, "104.36.252.115"), (20, "192.168.10.1"),
(22, "192.168.20.1"))
合并并排序:
val addrMap = ipAddressList.toMap
val output = descrList
.filter(x => addrMap.contains(x._1))
.map(x => x match { case (i, a, b) => (i, a, b, addrMap(i)) })
.sortBy(_._1)
output foreach println
输出:
(11,GigabitEthernet0,6,104.36.252.115)
(19,Vlan11,53,192.168.11.1)
(20,Vlan10,53,192.168.10.1)
(21,Vlan12,53,192.168.12.1)
(22,Vlan20,53,192.168.20.1)