# Tutorial :Incremental median computation with max memory efficiency ### Question:

I have a process that generates values and that I observe. When the process terminates, I want to compute the median of those values.

If I had to compute the mean, I could just store the sum and the number of generated values and thus have O(1) memory requirement. How about the median? Is there a way to save on the obvious O(n) coming from storing all the values?

Edit: Interested in 2 cases: 1) the stream length is known, 2) it's not.

### Solution:1

You are going to need to store at least ceil(n/2) points, because any one of the first n/2 points could be the median. It is probably simplest to just store the points and find the median. If saving ceil(n/2) points is of value, then read in the first n/2 points into a sorted list (a binary tree is probably best), then as new points are added throw out the low or high points and keep track of the number of points on either end thrown out.

Edit:

If the stream length is unknown, then obviously, as Stephen observed in the comments, then we have no choice but to remember everything. If duplicate items are likely, we could possibly save a bit of memory using Dolphins idea of storing values and counts.

### Solution:2

You can

• Use statistics, if that's acceptable - for example, you could use sampling.
• using a counting sort like approach: `k` distinct values means storing `O(k)` memory)
• or toss out known outliers and keep a (high,low) counter.
• If you know you have no duplicates, you could use a bitmap... but that's just a smaller constant for `O(n)`.

### Solution:3

If you have discrete values and lots of repetition you could store the values and counts, which would save a bit of space.

Possibly at stages through the computation you could discard the top 'n' and bottom 'n' values, as long as you are sure that the median is not in that top or bottom range.
e.g. Let's say you are expecting 100,000 values. Every time your stored number gets to (say) 12,000 you could discard the highest 1000 and lowest 1000, dropping storage back to 10,000.

If the distribution of values is fairly consistent, this would work well. However if there is a possibility that you will receive a large number of very high or very low values near the end, that might distort your computation. Basically if you discard a "high" value that is less than the (eventual) median or a "low" value that is equal or greater than the (eventual) median then your calculation is off.

Update
Bit of an example
Let's say that the data set is the numbers 1,2,3,4,5,6,7,8,9.
By inspection the median is 5.

Let's say that the first 5 numbers you get are 1,3,5,7,9.
To save space we discard the highest and lowest, leaving 3,5,7
Now get two more, 2,6 so our storage is 2,3,5,6,7
Discard the highest and lowest, leaving 3,5,6
Get the last two 4,8 and we have 3,4,5,6,8
Median is still 5 and the world is a good place.

However, lets say that the first five numbers we get are 1,2,3,4,5
Discard top and bottom leaving 2,3,4
Get two more 6,7 and we have 2,3,4,6,7
Discard top and bottom leaving 3,4,6
Get last two 8,9 and we have 3,4,6,8,9
With a median of 6 which is incorrect.

If our numbers are well distributed, we can keep trimming the extremities. If they might be bunched in lots of large or lots of small numbers, then discarding is risky.

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