Reputation: 15475
hey all, just getting started on hadoop and curious what the best way in mapreduce would be to count unique visitors if your logfiles looked like this...
DATE siteID action username
05-05-2010 siteA pageview jim
05-05-2010 siteB pageview tom
05-05-2010 siteA pageview jim
05-05-2010 siteB pageview bob
05-05-2010 siteA pageview mike
and for each site you wanted to find out the unique visitors for each site?
I was thinking the mapper would emit siteID \t username and the reducer would keep a set() of the unique usersnames per key and then emit the length of that set. However that would be potentially storing millions of usernames in memory which doesn't seem right. Anyone have a better way?
I'm using python streaming by the way
thanks
Upvotes: 9
Views: 6277
Reputation: 19666
Use the secondary sort to sort on user id. That way, you don't need to have anything in memory -- just stream the data through, and increment your distinct counter every time you see the value change for a particular site id.
Here is some documentation.
Upvotes: 1
Reputation: 26689
It is often faster to use HiveQL to sort many simple tasks. Hive will translate your queries into Hadoop MapReduce. In this case you may use
SELECT COUNT(DISTINCT username) FROM logviews
You may find a more advanced example here: http://www.dataminelab.com/blog/calculating-unique-visitors-in-hadoop-and-hive/
Upvotes: 0
Reputation: 10642
My aproach is similar to what tzaman gave with a small twist
Note that the first reduce does not need to go over any of the records is gets presented. You can simply examine the key and produce the output.
HTH
Upvotes: 1
Reputation: 47790
You could do it as a 2-stage operation:
First step, emit (username => siteID)
, and have the reducer just collapse multiple occurrences of siteID using a set
- since you'd typically have far less sites than users, this should be fine.
Then in the second step, you can emit (siteID => username)
and do a simple count, since the duplicates have been removed.
Upvotes: 3