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# Detect outliers in a series of numbers

Posted on 2011-03-04
Medium Priority
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Suppose I have a series of numbers like this

-0.0289436618082665
-0.0322635297824615
0.0473380547993016
-0.0483053616147235
0.0561386651052217
-0.0546202231192121
3912478746624.73
-0.0570958411471398
-0.0406567550991673
-0.0191101260081410
-0.0178598058749180
5912378756654.12
-0.00649518615382946
-0.0569007033673227
0.00634860933789683

So you most are within a certain range, say +/- 1.0, but there a couple numbers way out of this range.
Is there an algorithm to determine which numbers are these outliers?
I was thinking if I could detect these, then I could calculate the mean of the non-outliers and replace the outliers with that.
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Question by:allelopath
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Accepted Solution

deighton earned 336 total points
ID: 35038303
yes you calculate the mean and standard deviation of the numbers.

then for each number you calculate the number of standard deviations from the mean

if the number is more than 4 standard deviations from the mean, it can be considred an outsider.

I see your out-lying values are extremely outside the range of the others.
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Assisted Solution

TommySzalapski earned 668 total points
ID: 35038618
if the number is more than 4 standard deviations from the mean, it can be considred an outsider.

This threshold for what makes it an outlier is really application dependent. In fact, for the data you posted, one of the obvious outliers is less than 3 standard deviations from the mean. Many applications would throw out the top and bottom 5-10% of the data before doing any caclulations. If you do that, then your outliers will be over a billion standard deviations from the mean and would be outliers by almost any standard.
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Assisted Solution

aburr earned 332 total points
ID: 35039179
All this outliers business is very fraught with danger.
Physics is full of stories about people ignoring outliers and missing Nobel prizes.
Nevertheless people find it useful to establish algorithms to spot outliers. There is no standard algorithm to which objections cannot be raised.
Several popular ones have been given above.
Obviously if you run your data through whatever algorithm you choose enough times you will end up with one data point. You should not discard any data point without a non-statistical cause. Nevertheless often the problem is not important enough to spend a lot of time on it so one of the algorithms mentioned above will be usually an improvement in the decision making process.
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Assisted Solution

phoffric earned 664 total points
ID: 35041453
I just wanted to remind you that the mean and standard deviation includes the outliers and depending upon the quantity and magnitude of these outliers, these values could be adversely skewed. Depending on your model, the mean and standard deviation may be just what you need.

You may also want to consider determining the median instead. And if you can define from your model what an outlier is, then you might consider a % threshold error (or an absolute threshold error value - depends on your model), so that if the absolute value of the difference between the median and the data point exceeds the threshold, then that point will be considered an outlier.

As others have already alluded, you need to understand your model in order to define what an outlier is.
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Assisted Solution

phoffric earned 664 total points
ID: 35041472
Here are some EE discussions on outliers. Again, make sure that the question in the OP fits your model before applying any points made:

http://rdsrc.us/GyvkW7

http://rdsrc.us/U4sDl9

http://rdsrc.us/VrQc4e

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Assisted Solution

TommySzalapski earned 668 total points
ID: 35044851
Another common thing to do in outlier detection is to consider the mean and standard deviation (sd) of all points but the one in question. This tends to avoid the problem of one massive outlier messing up the statistics.
The best way to do that is to get the mean and sd for the whole set and 'remove' the one in question.
If you have N data points and want to remove x then you just do newmean = (mean*n-x)/(n-1) and that gives you the mean without considering x.
For the sd, remember that the mean of the x^2 minus the mean^2 gives the sd, so if you keep track of the mean of the squares, you can do the same thing for sd.
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