2
与OpenFlow
工作,我想计算两个所有组合比较多少双流满足以下条件(给定任意两个流,例如f1
和f2
):有效地Scala的集合
f1
“ sSrc ip
必须等于f2
的Dst ip
。f1
的Dst ip
必须等于f2
的Src ip
。f1
和f2
必须具有相同的协议。
这个我认为应该是两个元素(NFlows Choose 2)
的组合。由于this question我使用的方法combinations
象下面这样:
val pairFlows = matchs.combinations(2).filter{ list =>
val f1 = list.head
val f2 = list.tail.head
f1.dl_type == f2.dl_type &&
f1.nw_dst == f2.nw_src &&
f1.nw_src == f2.nw_dst
}
我的问题是,这是执行这种计算的最有效的方法是什么?或者为了跟踪每个src
,dst
和protocol type
,创建一个HashMap
会更好吗?
这里是数据的一个示例:
{
1: [
{
"actions": [
"OUTPUT:2"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 18,
"hard_timeout": 0,
"byte_count": 1764,
"duration_sec": 6884,
"duration_nsec": 5000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:02",
"nw_src": "10.0.0.1",
"in_port": 1,
"nw_dst": "10.0.0.2"
}
},
{
"actions": [
"OUTPUT:2"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 18,
"hard_timeout": 0,
"byte_count": 1764,
"duration_sec": 1123,
"duration_nsec": 181000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:02",
"nw_src": "10.0.0.3",
"in_port": 3,
"nw_dst": "10.0.0.2"
}
},
{
"actions": [
"OUTPUT:1"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 10,
"hard_timeout": 0,
"byte_count": 980,
"duration_sec": 1132,
"duration_nsec": 253000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:01",
"nw_src": "10.0.0.3",
"in_port": 3,
"nw_dst": "10.0.0.1"
}
},
{
"actions": [
"OUTPUT:1"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 16,
"hard_timeout": 0,
"byte_count": 1568,
"duration_sec": 1141,
"duration_nsec": 652000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:01",
"nw_src": "10.0.0.2",
"in_port": 2,
"nw_dst": "10.0.0.1"
}
},
{
"actions": [
"OUTPUT:3"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 20,
"hard_timeout": 0,
"byte_count": 1960,
"duration_sec": 6883,
"duration_nsec": 961000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:03",
"nw_src": "10.0.0.2",
"in_port": 2,
"nw_dst": "10.0.0.3"
}
},
{
"actions": [
"OUTPUT:3"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 12,
"hard_timeout": 0,
"byte_count": 1176,
"duration_sec": 6883,
"duration_nsec": 984000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:03",
"nw_src": "10.0.0.1",
"in_port": 1,
"nw_dst": "10.0.0.3"
}
},
{
"actions": [
"OUTPUT:CONTROLLER"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 0,
"hard_timeout": 0,
"byte_count": 0,
"duration_sec": 9156,
"duration_nsec": 252000000,
"priority": 65535,
"length": 96,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 35020,
"dl_dst": "01:80:c2:00:00:0e"
}
},
{
"actions": [
"OUTPUT:CONTROLLER"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 22,
"hard_timeout": 0,
"byte_count": 1260,
"duration_sec": 9156,
"duration_nsec": 268000000,
"priority": 0,
"length": 80,
"flags": 0,
"table_id": 0,
"match": {}
}
],
}
在这里有8间流动,其中3个是对。
与GROUPBY(_。dl_type)开始会给你几个较小的列表,而不是一个巨大的,检查。根据您的数据的外观,这可以帮助很多,一点或根本没有。另外,看着通过h来自n个元素列表的2个元素的所有组合是O(n^2)。如果您保留两个列表,一个按dst排序,另一个按src排序(这是在O(nlogn)中完成的),您可以遍历它们一次并收集所有匹配。这会给你比上面的例子更好的时间复杂度。 – Buhb
@Buhb谢谢,我已经设法在'O(n) – elbaulp