Computing median in map reduce

Question

Can someone example the computation of median/quantiles in map reduce?

My understanding of Datafu's median is that the 'n' mappers sort the data and send the data to "1" reducer which is responsible for sorting all the data from n mappers and finding the median(middle value) Is my understanding correct?,

if so, does this approach scale for massive amounts of data as i can clearly see the one single reducer struggling to do the final task. Thanks

Answer 1

In many real-world scenarios, the cardinality of values in a dataset will be relatively small. In such cases, the problem can be efficiently solved with two MapReduce jobs:

1. Calculate frequencies of values in your dataset (Word Count job, basically)
2. Identity mapper + a reducer which calculates median based on < value - frequency> pairs

Job 1. will drastically reduce the amount of data and can be executed fully in parallel. Reducer of job 2. will only have to process n (n = cardinality of your value set) items instead of all values, as with the naive approach.

Below, an example reducer of the job 2. It's is python script that could be used directly in Hadoop streaming. Assumes values in your dataset are ints, but can be easily adopted for doubles

import sys

item_to_index_range = []
total_count = 0

# Store in memory a mapping of a value to the range of indexes it has in a sorted list of all values
for line in sys.stdin:
    item, count = line.strip().split("\t", 1)
    new_total_count = total_count + int(count)
    item_to_index_range.append((item, (total_count + 1,   new_total_count + 1)))
    total_count = new_total_count

# Calculate index(es) of middle items
middle_items_indexes = [(total_count / 2) + 1]
if total_count % 2 == 0:
    middle_items_indexes += [total_count / 2]

# Retrieve middle item(s) 
middle_items = []
for i in middle_items_indexes:
    for item, index_range in item_to_index_range:
        if i in range(*index_range):
            middle_items.append(item)
            continue

print sum(middle_items) / float(len(middle_items))

This answer builds up on top of a suggestion initially coming from the answer of Chris White. The answer suggests using a combiner as a mean to calculate frequencies of values. However, in MapReduce, combiners are not guaranteed to be always executed. This has some side effects:

• reducer will first have to compute final < value - frequency > pairs and then calculate median.
• In the worst case scenario, combiners will never be executed and the reducer will still have to struggle with processing all individual values

Answer 2

Do you really need the exact median and quantiles?

A lot of the time, you are better off with just getting approximate values, and working with them, in particular if you use this for e.g. data partitioning.

In fact, you can use the approximate quantiles to speed up finding the exact quantiles (actually in O(n/p) time), here is a rough outline of the strategy:

1. Have a mapper for each partition compute the desired quantiles, and output them to a new data set. This data set should be several order of magnitues smaller (unless you ask for too many quantiles!)
2. Within this data set, compute the quantiles again, similar to "median of medians". These are your initial estimates.
3. Repartition the data according to these quantiles (or even additional partitions obtained this way). The goal is that in the end, the true quantile is guaranteed to be in one partition, and there should be at most one of the desired quantiles in each partition
4. Within each of the partitions, perform a QuickSelect (in O(n)) to find the true quantile.

Each of the steps is in linear time. The most costly step is part 3, as it will require the whole data set to be redistributed, so it generates O(n) network traffic. You can probably optimize the process by choosing "alternate" quantiles for the first iteration. Say, you want to find the global median. You can't find it in a linear process easily, but you can probably narrow it down to 1/kth of the data set, when it is split into k partitions. So instead of having each node report its median, have each node additionally report the objects at (k-1)/(2k) and (k+1)/(2k). This should allow you to narrow down the range of values where the true median must lie signficantly. So in the next step, you can each node send those objects that are within the desired range to a single master node, and choose the median within this range only.

Answer 3

O((n log n)/p) to sort it then O(1) to get the median.

Yes... you can get O(n/p) but you can't use the out of the box sort functionality in Hadoop. I would just sort and get the center item unless you can justify the 2-20 hours of development time to code the parallel kth largest algorithm.

Answer 4

Trying to find the median (middle number) in a series is going to require that 1 reducer is passed the entire range of numbers to determine which is the 'middle' value.

Depending on the range and uniqueness of values in your input set, you could introduce a combiner to output the frequency of each value - reducing the number of map outputs sent to your single reducer. Your reducer can then consume the sort value / frequency pairs to identify the median.

Another way you could scale this (again if you know the range and rough distribution of values) is to use a custom partitioner that distributes the keys by range buckets (0-99 go to reducer 0, 100-199 to reducer 2, and so on). This will however require some secondary job to examine the reducer outputs and perform the final median calculation (knowing for example the number of keys in each reducer, you can calculate which reducer output will contain the median, and at which offset)