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I want to make $5^{th}$,$10^{th}$,$15^{th}$,$20^{th}$ and $25^{th}$ values of vector an outlier in all xs by using x1 [5]+OT1,x1 [10]+OT1 and so on. For this purpose I have made this R code,

n=25

x1<-runif(n,0,1)
x2<-runif(n,0,1)
x3<-runif(n,0,1)
x4<-runif(n,0,1)
x<-data.frame(x1,x2,x3,x4)

OT1<-mean(x1)+100
OT2<-mean(x2)+100
OT3<-mean(x3)+100
OT4<-mean(x4)+100

I have tried command replace() and also modify(), but none of them replace them at once at least in one vector. Kindly help me in this manner.

Edit

by using comment of @user2974951 I tried this

x1[seq(5,25,5)]=x1[seq(5,25,5)]+100
Nx1<-replace(x1,x1==x1[5],x1 [5]+OT1)

x2[seq(5,25,5)]=x2[seq(5,25,5)]+100
Nx2<-replace(x2,x2==x2[5],x1 [5]+OT2)

x3[seq(5,25,5)]=x3[seq(5,25,5)]+100
Nx3<-replace(x3,x3==x3[5],x3 [5]+OT3)

x4[seq(5,25,5)]=x4[seq(5,25,5)]+100
Nx4<-replace(x4,x4==x4[5],x4 [5]+OT4)
Nx<-data.frame(Nx1,Nx2,Nx3,Nx4)

but its not working well

Results

         Nx1          Nx2         Nx3          Nx4
1    0.46815292   0.08606537   0.3307899 4.362630e-01
2    0.59723633   0.12122892   0.4819987 7.753236e-01
3    0.56219881   0.25936144   0.4990369 8.125097e-03
4    0.58366209   0.90552595   0.7368288 9.701722e-01
5  201.53593455 201.43976570 201.2130687 2.014071e+02
6    0.05521220   0.61975750   0.8296397 9.942981e-02
7    0.99058967   0.59373303   0.1156678 2.632295e-01
8    0.96428154   0.41710719   0.2547667 4.605275e-01
9    0.49978441   0.98922281   0.7526796 6.978671e-01
10 100.63831600 100.24166490 100.1790951 1.009707e+02
11   0.42694764   0.67506156   0.3142930 8.022078e-02
12   0.76015772   0.93265460   0.5734483 2.417875e-01
13   0.92832414   0.95247906   0.2578651 2.536677e-01
14   0.38818813   0.47634761   0.7163780 4.091937e-01
15 100.95118175 100.35951345 100.5519005 1.007286e+02
16   0.34262275   0.42573721   0.7594048 2.707246e-01
17   0.91930401   0.33828510   0.2679736 7.299545e-01
18   0.45901144   0.95876530   0.6419959 9.764771e-01
19   0.08840004   0.34092442   0.7492228 5.148988e-01
20 100.63958996 100.25792655 100.1351512 1.007377e+02
21   0.81191203   0.88845305   0.6504586 6.138992e-01
22   0.05737578   0.27700759   0.1193294 9.060633e-01
23   0.72447661   0.41372855   0.3055600 5.396204e-01
24   0.47942584   0.71890752   0.4814340 3.924752e-01
25 100.68191347 100.11710451 100.0443692 1.002326e+02

Why observation $5^{th}$ is 200+ and no outlier in $x4$???

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  • 1
    $\begingroup$ Something like x[seq(5,25,5)]=x[seq(5,25,5)]+100? $\endgroup$ – user2974951 Oct 5 '18 at 10:22
  • $\begingroup$ @user2974951 Kindly see my edit. $\endgroup$ – Angel Oct 6 '18 at 3:23
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You have a data frame with 4 columns, each columns has 25 rows. You want to replace every 5th element with an outlying value (the mean of the column).

apply(x,2,function(i){
    i[seq(5,25,5)]=i[seq(5,25,5)]+mean(i)
})
| improve this answer | |
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  • $\begingroup$ For my further work, i have to increase my $n$ and replacement is till 25th observation, but in that replacement is done till end. How to control this? $\endgroup$ – Angel Oct 24 '18 at 3:49
  • $\begingroup$ Your code is different from mine, have you tried using my code to get the desired result? It should work. As for increasing $n$, all you have to do is change $25$ to the new value, or for ex. nrow(x). $\endgroup$ – user2974951 Oct 24 '18 at 6:05
  • $\begingroup$ For Increasing n, suppose $n=50$ only want to change in 1st 25 values... $\endgroup$ – Angel Oct 26 '18 at 4:36
  • $\begingroup$ OK then it's the same code. $\endgroup$ – user2974951 Oct 26 '18 at 5:00
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You added 100 twice. And Nx4 is the same -- except that it's in scientific notations.

Nx1 = x1
Nx1[seq(5, 25, 5)] = mean(x1) + 100

will do.

| improve this answer | |
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