I have a dataframe with yearly energy uses of buildings over 5 years. In order to have a representative yearly energy use for data modelling, I'll have to take the mean of those data. As the data can contain outliers, I want to deal with outliers correctly (but keeping as much proper data as possible). (The df can contain empty cells (and older years are a little more important than recent years if a weighted decision has to be made somewhere).)

What are good methods to deal with outliers when calculating the mean of data?

I thought of: -calculating the mean of the 5 datacolumns (y_2010 - y_2014) and then comparing all 5 datapoints with that mean. If there's a difference of e.g. >20%, this case is deleted and cannot be used for further analysis as there's too much variability in the data for that ID. (Normally the energy data over 5 years should be more or less the same unless retrofits have been done to the building, but most buildings haven't.) - doing something with a rolling mean to come to a proper mean building energy use column - ...

Example of the df:

   ID  y_2010   y_2011   y_2012  y_2013  y_2014  mean
21524   22631    21954    22314   22032   21843   ...
28965   27456    29654    28159   28654   27345   ...
10236   32165      NaN    31678   31895   32459   ...
89754   87621    86542    87542   88456   86961   ...
56457   58951    57486     2000       0       0   ...
25984   24587    25478      NaN   24896   25461   ...
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    $\begingroup$ Why do you think that you should do anything special with those “outlier” points? Do you not expect them to happen again because of some subject matter knowledge, for instance? $\endgroup$
    – Dave
    Commented May 30, 2020 at 20:19
  • 1
    $\begingroup$ Is there a reason not to use the median instead of the mean? For many purposes, the median is considered an outlier-robust version of the mean. $\endgroup$
    – Eike P.
    Commented May 31, 2020 at 11:14

1 Answer 1


At the start, you have a fundamental decision to make: Are you trying to learn something from the data? Or are you trying to teach the data to behave as you suppose they should? This answer is mainly oriented toward the former approach.

Usually, it is a mistake to remove an 'outlier' from a dataset unless you can establish that the observation in question arose from a documentable error (equipment failure, data entry error, etc.) or you know for sure that its value is impossible (person's age above 140, negative height, etc.)

Anecdote: Where I live, the highest heating bill arises in December and January. For personal and family reasons unlikely ever to be repeated, my energy bills for 12/2019 and 1/2020 were extraordinarily high. By your criterion, I might be removed from your list, which I think would be a mistake. I can't foresee the same circumstances ever coming around for me again, but others may well be surprised by similar temporary periods of high energy usage in the future for very similar reasons.

One way to stabilize averages without tampering with data is to use trimmed means. To find a trimmed mean, data are sorted, a certain percentage of the very lowest and of the very highest observations are ignored, and the mean of the more central remaining observations is taken. Depending on circumstances, typical trimming percentages may be 2% to 20% (sometimes higher), leaving the central 96% to 60% (sometimes fewer) to be averaged.

Consider data with $n=1000$ observations from a gamma distribution with shape parameter 10 (perhaps waiting times for finishing multi-phase projects.) Here is a graph of its density curve--made in R.

curve(dgamma(x,10,1), 0, 25, lwd=2, ylab="PDF", 
      main="Density of GAMMA(10, 1)")
abline(v=0, col="green2");  abline(h=0, col="green2")

enter image description here

Almost all such samples have at least one boxplot outlier and the average number of outliers in a sample of 1000 is about 14.

nr.out = replicate(10^5, 
           length(boxplot.stats(rgamma(1000,10,1))$out) )
mean(nr.out);  mean(nr.out>0)
[1] 13.97049
[1] 1

Let's take a look at boxplots of 20 samples of size 1000 from this distribution in order to see the outliers.

m = 20;  n=1000
x = rgamma(m*n,10,1);  g = rep(1:m, n)
boxplot(x~g, col="skyblue2", 
      main="GAMMA(10,1) Population: Boxplots of 20 Samples of 1000")

It seems that 2% trimmed means of the 1000 observations in each sample should allow us to ignore boxplot outliers in finding the means. (But the trimmed values are are not removed, so the ordinary mean and quartiles are not affected.)

enter image description here

My simulated gamma observations have $\mu = 10, \sigma^2 = 10,$ so samples of 1000 have ordinary means average about $10$ with variances about $0.01$ (from theory). By contrast 2% trimmed means of samples average about $9.93$ with variances about $0.01$ (from simulation).

a.02 = replicate(10^5, mean(rgamma(1000,10,1),trim=.02))
mean(a.02);  var(a.02)
[1] 9.932821
[1] 0.009988345

By using trimmed means we have retained all of the data. In a fair and systematic way, we have mainly avoided using boxplot outliers to estimate means. Very roughly speaking we have computed trimmed means by ignoring values that are more than double the ordinary mean. And at the same time we have ignored values that are less than half the ordinary mean. Perhaps we find that temporarily ignoring values that are proportionately far from the ordinary mean (still the best estimate of the population mean), we can make better judgments from our data.

qgamma(c(.02,.98), 10, 1)
[1]  4.618349 17.509813

However, over time we may come to realize that all of the observations have a legitimate role to play in understanding how to use the data to best advantage. In that case, the data are intact and we can do so.

Note: There are distributions with such heavy tails that a trimmed sample mean is a better estimate of the location of the population than an ordinary sample mean. The Cauchy is one such distribution. In that case tails are so heavy that a 38% trimmed mean seems optimal. See a brief discussion here and further information at its links.

  • 2
    $\begingroup$ +1 Related: Should the mean be used when the data skewed. I really like your closing note. :) $\endgroup$
    – Alexis
    Commented May 31, 2020 at 1:51
  • 3
    $\begingroup$ (+1) You addressed the energy usage example very nicely as well as making more general points. A simple but fundamental point arising here, and often elsewhere, but occasionally forgotten, is that a mean remains pertinent -- statistically and practically -- whenever totals have meaning and are important. Here the total energy usage certainly qualifies, not least in implications for what bills have to be paid (unless the consumer can establish a measurement error). $\endgroup$
    – Nick Cox
    Commented May 31, 2020 at 8:05

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