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March 18, 2018 20:56
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Find most popular AS in BGP table (empty and 0 first results should be ignored)
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| 'use strict'; | |
| const fs = require('fs'); | |
| const split = require('split'); | |
| const through = require('through'); | |
| const regex = /\*\>?\s+[\d+|\.]+\/\d+\s+[\d+|\.]+\s+([\d+\s]+)/; | |
| const filter = through(function(buffer) { | |
| const line = buffer.toString(); | |
| if (line.match(regex)) { | |
| this.queue(line); | |
| } | |
| }); | |
| const popularity = new Map(); | |
| const longest = through(function(buffer) { | |
| const line = buffer.toString(); | |
| const results = line.match(regex); | |
| const current = new Set(results[1].split(' ')); | |
| current.forEach((id) => { | |
| if (!popularity.has(id)) { | |
| popularity.set(id, 1); | |
| return; | |
| } | |
| const count = popularity.get(id); | |
| popularity.set(id, count + 1); | |
| }); | |
| }); | |
| longest.on('close', () => { | |
| const values = [0, 0, 0, 0, 0]; | |
| const keys = [-1, -1, -1, -1, -1]; | |
| popularity.forEach((value, key) => { | |
| for(let i = 0; i < values.length; i++) { | |
| if (values[i] <= value) { | |
| console.log(`Popular AS: ${key} and it's popularity is: ${value}`); | |
| for(let j = values.length - 2; j >= i; j--) { | |
| values[j + 1] = values[j]; | |
| keys[j + 1] = keys[j]; | |
| } | |
| values[i] = value; | |
| keys[i] = key; | |
| break; | |
| } | |
| } | |
| }); | |
| console.log(keys); | |
| console.log(values); | |
| }); | |
| const fileStream = fs.createReadStream('ipv4bgp2018apnic.txt'); | |
| fileStream | |
| .pipe(split()) | |
| .pipe(filter) | |
| .pipe(longest) |
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