How to Use Outlier Tests in R Code

How to use Outlier Tests in R Code

I agree with Dirk, It's hard. I would recomend first looking at why you might have outliers. An outlier is just a number that someone thinks is suspicious, it's not a concrete 'bad' value, and unless you can find a reason for it to be an outlier, you may have to live with the uncertainty.

One thing you didn't mention was what kind of outlier you're looking at. Are your data clustered around a mean, do they have a particular distribution or is there some relationship between your data.

Here's some examples

First, we'll create some data, and then taint it with an outlier;

> testout<-data.frame(X1=rnorm(50,mean=50,sd=10),X2=rnorm(50,mean=5,sd=1.5),Y=rnorm(50,mean=200,sd=25))
> #Taint the Data
> testout$X1[10]<-5
> testout$X2[10]<-5
> testout$Y[10]<-530

> testout
         X1         X2        Y
1  44.20043  1.5259458 169.3296
2  40.46721  5.8437076 200.9038
3  48.20571  3.8243373 189.4652
4  60.09808  4.6609190 177.5159
5  50.23627  2.6193455 210.4360
6  43.50972  5.8212863 203.8361
7  44.95626  7.8368405 236.5821
8  66.14391  3.6828843 171.9624
9  45.53040  4.8311616 187.0553
10  5.00000  5.0000000 530.0000
11 64.71719  6.4007245 164.8052
12 54.43665  7.8695891 192.8824
13 45.78278  4.9921489 182.2957
14 49.59998  4.7716099 146.3090
<snip>
48 26.55487  5.8082497 189.7901
49 45.28317  5.0219647 208.1318
50 44.84145  3.6252663 251.5620

It's often most usefull to examine the data graphically (you're brain is much better at spotting outliers than maths is)

> #Use Boxplot to Review the Data
> boxplot(testout$X1, ylab="X1")
> boxplot(testout$X2, ylab="X2")
> boxplot(testout$Y, ylab="Y")

Then you can use a test. If the test returns a cut off value, or the value that might be an outlier, you can use ifelse to remove it

> #Use Outlier test to remove individual values
> testout$newX1<-ifelse(testout$X1==outlier(testout$X1),NA,testout$X1)
> testout
         X1         X2        Y    newX1
1  44.20043  1.5259458 169.3296 44.20043
2  40.46721  5.8437076 200.9038 40.46721
3  48.20571  3.8243373 189.4652 48.20571
4  60.09808  4.6609190 177.5159 60.09808
5  50.23627  2.6193455 210.4360 50.23627
6  43.50972  5.8212863 203.8361 43.50972
7  44.95626  7.8368405 236.5821 44.95626 
8  66.14391  3.6828843 171.9624 66.14391 
9  45.53040  4.8311616 187.0553 45.53040
10  5.00000  5.0000000 530.0000       NA 
11 64.71719  6.4007245 164.8052 64.71719 
12 54.43665  7.8695891 192.8824 54.43665 
13 45.78278  4.9921489 182.2957 45.78278 
14 49.59998  4.7716099 146.3090 49.59998 
15 45.07720  4.2355525 192.9041 45.07720 
16 62.27717  7.1518606 186.6482 62.27717 
17 48.50446  3.0712422 228.3253 48.50446 
18 65.49983  5.4609713 184.8983 65.49983 
19 44.38387  4.9305222 213.9378 44.38387 
20 43.52883  8.3777627 203.5657 43.52883 
<snip>
49 45.28317  5.0219647 208.1318 45.28317 
50 44.84145  3.6252663 251.5620 44.84145

Or for more complicated examples, you can use stats to calculate critical cut off values, here using the Lund Test (See Lund, R. E. 1975, "Tables for An Approximate Test for Outliers in Linear Models", Technometrics, vol. 17, no. 4, pp. 473-476. and Prescott, P. 1975, "An Approximate Test for Outliers in Linear Models", Technometrics, vol. 17, no. 1, pp. 129-132.)

> #Alternative approach using Lund Test
> lundcrit<-function(a, n, q) {
+ # Calculates a Critical value for Outlier Test according to Lund
+ # See Lund, R. E. 1975, "Tables for An Approximate Test for Outliers in Linear Models", Technometrics, vol. 17, no. 4, pp. 473-476.
+ # and Prescott, P. 1975, "An Approximate Test for Outliers in Linear Models", Technometrics, vol. 17, no. 1, pp. 129-132.
+ # a = alpha
+ # n = Number of data elements
+ # q = Number of independent Variables (including intercept)
+ F<-qf(c(1-(a/n)),df1=1,df2=n-q-1,lower.tail=TRUE)
+ crit<-((n-q)*F/(n-q-1+F))^0.5
+ crit
+ }

> testoutlm<-lm(Y~X1+X2,data=testout)

> testout$fitted<-fitted(testoutlm)

> testout$residual<-residuals(testoutlm)

> testout$standardresid<-rstandard(testoutlm)

> n<-nrow(testout)

> q<-length(testoutlm$coefficients)

> crit<-lundcrit(0.1,n,q)

> testout$Ynew<-ifelse(abs(testout$standardresid)>crit,NA,testout$Y)

> testout
         X1         X2        Y    newX1   fitted    residual standardresid
1  44.20043  1.5259458 169.3296 44.20043 209.8467 -40.5171222  -1.009507695
2  40.46721  5.8437076 200.9038 40.46721 231.9221 -31.0183107  -0.747624895
3  48.20571  3.8243373 189.4652 48.20571 203.4786 -14.0134646  -0.335955648
4  60.09808  4.6609190 177.5159 60.09808 169.6108   7.9050960   0.190908291
5  50.23627  2.6193455 210.4360 50.23627 194.3285  16.1075799   0.391537883
6  43.50972  5.8212863 203.8361 43.50972 222.6667 -18.8306252  -0.452070155
7  44.95626  7.8368405 236.5821 44.95626 223.3287  13.2534226   0.326339981
8  66.14391  3.6828843 171.9624 66.14391 148.8870  23.0754677   0.568829360
9  45.53040  4.8311616 187.0553 45.53040 214.0832 -27.0279262  -0.646090667
10  5.00000  5.0000000 530.0000       NA 337.0535 192.9465135   5.714275585
11 64.71719  6.4007245 164.8052 64.71719 159.9911   4.8141018   0.118618011
12 54.43665  7.8695891 192.8824 54.43665 194.7454  -1.8630426  -0.046004311
13 45.78278  4.9921489 182.2957 45.78278 213.7223 -31.4266180  -0.751115595
14 49.59998  4.7716099 146.3090 49.59998 201.6296 -55.3205552  -1.321042392
15 45.07720  4.2355525 192.9041 45.07720 213.9655 -21.0613819  -0.504406009
16 62.27717  7.1518606 186.6482 62.27717 169.2455  17.4027250   0.430262983
17 48.50446  3.0712422 228.3253 48.50446 200.6938  27.6314695   0.667366651
18 65.49983  5.4609713 184.8983 65.49983 155.2768  29.6214506   0.726319931
19 44.38387  4.9305222 213.9378 44.38387 217.7981  -3.8603382  -0.092354925
20 43.52883  8.3777627 203.5657 43.52883 228.9961 -25.4303732  -0.634725264
<snip>
49 45.28317  5.0219647 208.1318 45.28317 215.3075  -7.1756966  -0.171560291
50 44.84145  3.6252663 251.5620 44.84145 213.1535  38.4084869   0.923804784
       Ynew
1  169.3296
2  200.9038
3  189.4652
4  177.5159
5  210.4360
6  203.8361
7  236.5821
8  171.9624
9  187.0553
10       NA
11 164.8052
12 192.8824
13 182.2957
14 146.3090
15 192.9041
16 186.6482
17 228.3253
18 184.8983
19 213.9378
20 203.5657
<snip>
49 208.1318
50 251.5620

Edit: I've just noticed an issue in my code. The Lund test produces a critical value that should be compared to the absolute value of the studantized residual (i.e. without sign)

Several Grubbs tests simultaneously in R

You can use lapply:

library(outliers)

df = data.frame(a=runif(20),b=runif(20),c=runif(20))
tests = lapply(df,grubbs.test) 
# or with parameters:
tests = lapply(df,grubbs.test,opposite=T)

Results:

> tests
$a

    Grubbs test for one outlier

data:  X[[i]]
G = 1.80680, U = 0.81914, p-value = 0.6158
alternative hypothesis: highest value 0.963759744539857 is an outlier

$b

    Grubbs test for one outlier

data:  X[[i]]
G = 1.53140, U = 0.87008, p-value = 1
alternative hypothesis: highest value 0.975481075001881 is an outlier

$c

    Grubbs test for one outlier

data:  X[[i]]
G = 1.57910, U = 0.86186, p-value = 1
alternative hypothesis: lowest value 0.0136249314527959 is an outlier

You can access the results as follows:

> tests$a$statistic
        G         U 
1.8067906 0.8191417

Hope this helps.

Outlier detection of time series data in R

You may have a look at the following packages available in R.

The R package forecast uses loess decomposition of time series to identify and replace outliers.

The R package tsoutliers implements the Chen and Liu procedure for detection of outliers in time series. A description of the procedure and the implementation is given in the documentation attached to the package. You may also see this post.

R: iterative outliers detection

I think this may work. The dropout function will do iterative looping to test for outliers. For each element you pass in, it will return 1 if the element is not an outliers, otherwise it will return the p-value < .05 for the outlier test.

library(outliers)
dropout<-function(x) {
    if(length(x)<2) return (1)
    vals <- rep.int(1, length(x))
    r <- chisq.out.test(x)
    while (r$p.value<.05 & sum(vals==1)>2) {
        if (grepl("lowest", r$alternative)) {
            d <- which.min(ifelse(vals==1,x, NA))
        } else {
            d <- which.max(ifelse(vals==1, x, NA))
        }
        vals[d] <- r$p.value
        r <- chisq.out.test(x[vals==1])
    }
    vals
}

With that helper function in place, we can now apply it to each of the sub-groups defined by V, V1. To do that, we use the ave function.

with(dd, ave(V3, V1, V2, FUN = dropout))

It appears your sample data has no outliers in any of the sub-groups given chisq.out.test definition of outliers.

And surely this iterative process is not statistically meaningful given the problem of resting for outliers in general and certainly with the multiple testing problem. Nevertheless, that discussion is for https://stats.stackexchange.com/, here we just focus on the code.

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