问题
How I can calculate the arithmetic mean of a large vector(series) in distributed computing where I partition the data on multiple nodes. I do not want to use map reduce paradigm. Is there any distributed algorithm to efficiently compute the mean besides the trivial computation of individual sum on each node and then bringing the result at master node and dividing with the size of the vector(series).
回答1:
distributed average consensus is an alternative.
The problem with the trivial approach of map-reduce with a master is that if you have a vast set of data, in essence to make everything dependent on each other, it could take a very long time to calculate the data, by which time the information is very out of date, and therefore wrong, unless you lock the entire dataset - impractical for a massive set of distributed data. Using distributed average consensus (the same methods work for alternative algorithms to Mean), you get a more up to date, better guess at the current value of the Mean without locking the data, and in real time. Here is a link to a paper on it, but it's math heavy : http://web.stanford.edu/~boyd/papers/pdf/lms_consensus.pdf You can google for many papers on it.
The general concept is like this: say on each node you have a socket listener. You evaluate your local sum and average, then publish it to the other nodes. Each node listens for the other nodes, and receives their sum and averages on a timescale that makes sense. You can then evaluate a good guess at the total average by (sumForAllNodes(storedAverage[node] * storedCount[node]) / (sumForAllNodes(storedCount[node])). If you have a truly large dataset, you could just listen for new values as they are stored in the node, and amend the local count and average, then publish them.
If even this is taking too long, you could average over a random subset of the data in each node.
Here is some c# code that gives you an idea (uses fleck to run on more versions of windows than windows-10-only microsoft websockets implementation). Run this on two nodes, one with
<appSettings>
<add key="thisNodeName" value="UK" />
</appSettings>
in the app.config, and use "EU-North" in the other. Here is some sample code. The two instances exchange means using websockets. You just need to add your back end enumeration of the database.
using Fleck;
namespace WebSocketServer
{
class Program
{
static List<IWebSocketConnection> _allSockets;
static Dictionary<string,decimal> _allMeans;
static Dictionary<string,decimal> _allCounts;
private static decimal _localMean;
private static decimal _localCount;
private static decimal _localAggregate_count;
private static decimal _localAggregate_average;
static void Main(string[] args)
{
_allSockets = new List<IWebSocketConnection>();
_allMeans = new Dictionary<string, decimal>();
_allCounts = new Dictionary<string, decimal>();
var serverAddresses = new Dictionary<string,string>();
//serverAddresses.Add("USA-WestCoast", "ws://127.0.0.1:58951");
//serverAddresses.Add("USA-EastCoast", "ws://127.0.0.1:58952");
serverAddresses.Add("UK", "ws://127.0.0.1:58953");
serverAddresses.Add("EU-North", "ws://127.0.0.1:58954");
//serverAddresses.Add("EU-South", "ws://127.0.0.1:58955");
foreach (var serverAddress in serverAddresses)
{
_allMeans.Add(serverAddress.Key, 0m);
_allCounts.Add(serverAddress.Key, 0m);
}
var thisNodeName = ConfigurationSettings.AppSettings["thisNodeName"]; //for example "UK"
var serverSocketAddress = serverAddresses.First(x=>x.Key==thisNodeName);
serverAddresses.Remove(thisNodeName);
var websocketServer = new Fleck.WebSocketServer(serverSocketAddress.Value);
websocketServer.Start(socket =>
{
socket.OnOpen = () =>
{
Console.WriteLine("Open!");
_allSockets.Add(socket);
};
socket.OnClose = () =>
{
Console.WriteLine("Close!");
_allSockets.Remove(socket);
};
socket.OnMessage = message =>
{
Console.WriteLine(message + " received");
var parameters = message.Split('~');
var remoteHost = parameters[0];
var remoteMean = decimal.Parse(parameters[1]);
var remoteCount = decimal.Parse(parameters[2]);
_allMeans[remoteHost] = remoteMean;
_allCounts[remoteHost] = remoteCount;
};
});
while (true)
{
//evaluate my local average and count
Random rand = new Random(DateTime.Now.Millisecond);
_localMean = 234.00m + (rand.Next(0, 100) - 50)/10.0m;
_localCount = 222m + rand.Next(0, 100);
//evaluate my local aggregate average using means and counts sent from all other nodes
//could publish aggregate averages to other nodes, if you wanted to monitor disagreement between nodes
var total_mean_times_count = 0m;
var total_count = 0m;
foreach (var server in serverAddresses)
{
total_mean_times_count += _allCounts[server.Key]*_allMeans[server.Key];
total_count += _allCounts[server.Key];
}
//add on local mean and count which were removed from the server list earlier, so won't be processed
total_mean_times_count += (_localMean * _localCount);
total_count = total_count + _localCount;
_localAggregate_average = (total_mean_times_count/total_count);
_localAggregate_count = total_count;
Console.WriteLine("local aggregate average = {0}", _localAggregate_average);
System.Threading.Thread.Sleep(10000);
foreach (var serverAddress in serverAddresses)
{
using (var wscli = new ClientWebSocket())
{
var tokSrc = new CancellationTokenSource();
using (var task = wscli.ConnectAsync(new Uri(serverAddress.Value), tokSrc.Token))
{
task.Wait();
}
using (var task = wscli.SendAsync(new ArraySegment<byte>(Encoding.UTF8.GetBytes(thisNodeName+"~"+_localMean + "~"+_localCount)),
WebSocketMessageType.Text,
false,
tokSrc.Token
))
{
task.Wait();
}
}
}
}
}
}
}
Don't forget to add static lock or separate activity by synchronising at given times. (not shown for simplicity)
回答2:
There are two simple approaches you can use.
One is, as you correctly noted, to calculate the sum on every node and then combine the sums and divide by the total amount of data:
avg = (sum1+sum2+sum3)/(cnt1+cnt2+cnt3)
Another possibility is to calculate the average on every node and then use weighted average:
avg = (avg1*cnt1 + avg2*cnt2 + avg3*cnt3) / (cnt1+cnt2+cnt3)
= avg1*cnt1/(cnt1+cnt2+cnt3) + avg2*cnt2/(cnt1+cnt2+cnt3) + avg3*cnt3/(cnt1+cnt2+cnt3)
I don't see anything wrong with these trivial ways and am wondering why you would want to use a different approach.
来源：https://stackoverflow.com/questions/42428424/how-to-calculate-mean-of-distributed-data