sort two lists by their first element and zip them in scala

Question

 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)
  }

For instance,

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"))

In both the lists first element is index and i have to merge these two list index wise . It means (21,"192.168.12.1") this ipAddress should merge with (21,"Vlan12",53) and form new list like below (21,"Vlan12",53,"192.168.12.1").

score 1 · Accepted Answer · edited Oct 12 '14 at 17:50

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))

score 0 · Answer 2 · edited May 23 '17 at 11:51

First, I would suggest you produce a Map instead of a List. A Map by nature has an indexer, and in your case this would be the ifIndex value.

Once you have Maps in place, you can use something like this (sample from this other SO Best way to merge two maps and sum the values of same key?)

From Rex Kerr: map1 ++ map2.map{ case (k,v) => k -> (v + map1.getOrElse(k,0)) }

Or like this from Matthew Farwell: (map1.keySet ++ map2.keySet).map (i=> (i,map1.getOrElse(i,0) + map2.getOrElse(i,0))}.toMap

If you cannot use Maps for whatever reason, then look into your existing project libraries. If you have Scalaz, then you have some tools already available.

Scalaz: https://github.com/scalaz/scalaz

If you have Slick, you also have some nice tools to directly use.

Slick: http://slick.typesafe.com/docs/

elm · Answer 3 · 2014-10-13T06:06:16.567

Consider first converting decrList to a Map, like this,

val a = (for ( (k,v1,v2) <- descrList) yield k -> (v1,v2)).toMap

Then we can look up keys for ipAddressList and agglomerate elements into a new tuple, as follows,

for ( (k,ip) <- ipAddressList ; v = a.getOrElse(k,("none","none")) ) yield (k,v._1,v._2,ip)

Hence, for 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))

score 0 · Answer 4 · answered Oct 13 '14 at 11:15

Given the data:

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"))

Merge and sort:

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

Output:

(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)

sort two lists by their first element and zip them in scala

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