R: Plotting Multiple Graphs using a "for loop" - r

Question

I am using the R programming language.
Using the following code, I am able to put two plots on the same page:
#load library
library(dbscan)
#specify number of plots per page
par(mfrow = c(1,2))
#load libraries
library(dbscan)
library(dplyr)
#generate data
n <- 100
x <- cbind(
x=runif(10, 0, 5) + rnorm(n, sd=0.4),
y=runif(10, 0, 5) + rnorm(n, sd=0.4)
)
### calculate LOF score
lof <- lof(x, k=3)
### distribution of outlier factors (first plot)
summary(lof)
hist(lof, breaks=10)
### point size is proportional to LOF (second plot)
plot(x, pch = ".", main = "LOF (k=3)")
points(x, cex = (lof-1)*3, pch = 1, col="red")
This produces the following plot:
Now, I am trying to make several plots (e.g. 6 plots, 2 pairs of 3) on the same page. I tried to implement this with a "for loop" (for k = 3, 4, 5):
par(mfrow = c(3,2))
vals <- 3:5
combine <- vector('list', length(vals))
count <- 0
for (i in vals) {
lof_i <- lof(x, k=i)
### distribution of outlier factors
summary(lof_i)
hist(lof_i, breaks=10)
### point size is proportional to LOF
plot(x, pch = ".", main = "LOF (k=i)")
points(x, cex = (lof_i-1)*3, pch = 1, col="red")
}
However, this seems to just repeat the same graph 6 times on the same page:
Can someone please show me how to correct this code?
Is it also possible to save the files "lof_3, lof_4, lof_5"? It seems that none of these files are created, only "lof_i" is created:
> lof_3
Error: object 'lof_3' not found
> head(lof_i)
[1] 1.223307 1.033424 1.077149 1.011407 1.040634 1.431029
Thanks

Answer 1

Looking at your plots you seem to have generated and plotted different plots, but to have the labels correct you need to pass a variable and not a fixed character to your title (e.g. using the paste command).
To get the calculated values out of your loop you could either generate an empty list and assign the results in the loop to individual list elements, or use something like lapply that will automatically return the results in a list form.
To simplify things a bit you could define a function that either plots or returns the calculated values, e.g. like this:
library(dbscan)
#generate data
set.seed(123)
n <- 100
x <- cbind(
x=runif(10, 0, 5) + rnorm(n, sd=0.4),
y=runif(10, 0, 5) + rnorm(n, sd=0.4)
)
plotLOF <- function(i, plot=TRUE){
lof <- lof(x, k=i)
if (plot){
hist(lof, breaks=10)
plot(x, pch = ".", main = paste0("LOF (k=", i, ")"))
points(x, cex = (lof-1)*3, pch = 1, col="red")
} else return(lof)
}
par(mfrow = c(3,2))
invisible(lapply(3:5, plotLOF))
lapply(3:5, plotLOF, plot=FALSE)
#> [[1]]
#> [1] 1.1419243 0.9551471 1.0777472 1.1224447 0.8799095 1.0377858 0.8416306
#> [8] 1.0487133 1.0250496 1.3183819 0.9896833 1.0353398 1.3088266 1.0123238
#> [15] 1.1233530 0.9685039 1.0589151 1.3147785 1.0488644 0.9212146 1.2568698
#> [22] 1.0086274 1.0454450 0.9661698 1.0644528 1.1107202 1.0942201 1.5147076
#> [29] 1.0321698 1.0553455 1.1149748 0.9341090 1.2352716 0.9478602 1.4096464
#> [36] 1.0519127 1.0507267 1.3199825 1.2525485 0.9361488 1.0958563 1.2131615
#> [43] 0.9943090 1.0123238 1.1060491 1.0377766 0.9803135 0.9627699 1.1165421
#> [50] 0.9796819 0.9946925 2.1576989 1.6015310 1.5670315 0.9343637 1.0033725
#> [57] 0.8769431 0.9783065 1.0800050 1.2768800 0.9735274 1.0377472 1.0743988
#> [64] 1.7583562 1.2662485 0.9685039 1.1662145 1.2491499 1.1131718 1.0085023
#> [71] 0.9636864 1.1538360 1.2126138 1.0609829 1.0679010 1.0490234 1.1403292
#> [78] 0.9638900 1.1863703 0.9651060 0.9503445 1.0098536 0.8440855 0.9052420
#> [85] 1.2662485 1.4447713 1.0845415 1.0661381 0.9282678 0.9380078 1.1414628
#> [92] 1.0407138 1.0942201 1.0589805 1.0370938 1.0147094 1.1067291 0.8834466
#> [99] 1.7027132 1.1766560
#>
#> [[2]]
#> [1] 1.1667311 1.0409009 1.0920953 1.0068953 0.9894195 1.1332413 0.9764505
#> [8] 1.0228796 1.0446905 1.0893386 1.1211637 1.1029415 1.3453498 0.9712910
#> [15] 1.1635936 1.0265746 0.9480282 1.2144437 1.0570346 0.9314618 1.3345561
#> [22] 0.9816097 0.9929112 1.0322014 1.2739621 1.2947553 1.0202948 1.6153264
#> [29] 1.0790922 0.9987830 1.0378609 0.9622779 1.2974938 0.9129639 1.2601398
#> [36] 1.0265746 1.0241622 1.2420568 1.2204376 0.9297345 1.1148404 1.2546361
#> [43] 1.0059582 0.9819820 1.0342491 0.9452673 1.0369500 0.9791091 1.2000825
#> [50] 0.9878844 1.0205586 2.0057587 1.2757014 1.5347815 0.9622614 1.0692613
#> [57] 1.0026404 0.9408510 1.0280687 1.3534531 0.9669894 0.9300601 0.9929112
#> [64] 1.7567871 1.3861828 1.0265746 1.1120151 1.3542396 1.1562077 0.9842179
#> [71] 1.0301098 1.2326327 1.1866352 1.0403814 1.0577086 0.8745912 1.0017905
#> [78] 0.9904356 1.0602487 0.9501681 1.0176457 1.0405430 0.9718224 1.0046821
#> [85] 1.1909982 1.6151918 0.9640852 1.0141963 1.0270237 0.9867738 1.1474414
#> [92] 1.1293307 1.0323945 1.0859417 0.9622614 1.0290635 1.0186381 0.9225209
#> [99] 1.6456612 1.1366753
#>
#> [[3]]
#> [1] 1.1299335 1.0122028 1.2077092 0.9485150 1.0115694 1.1190314 0.9989174
#> [8] 1.0145663 1.0357546 0.9783702 1.1050504 1.0661798 1.3571416 1.0024603
#> [15] 1.1484745 1.0162149 0.9601474 1.1310442 1.0957731 1.0065501 1.2687934
#> [22] 0.9297323 0.9725355 0.9876444 1.2314822 1.2209304 0.9906446 1.4249452
#> [29] 1.2156607 0.9959685 1.0304305 0.9976110 1.1711354 1.0048161 0.9813000
#> [36] 1.0128909 0.9730295 1.1741982 1.3317209 0.9708714 1.0994309 1.1900047
#> [43] 0.9960765 0.9659553 0.9744357 0.9556112 1.0508484 0.9669406 1.3919743
#> [50] 0.9467537 1.0596883 1.7396644 1.1323109 1.6516971 0.9922995 1.0223594
#> [57] 0.9917594 0.9542419 1.0672565 1.2274498 1.0589385 0.9649404 0.9953886
#> [64] 1.7666795 1.3111620 0.9860706 1.0576620 1.2547512 1.0038281 0.9825967
#> [71] 1.0104708 1.1739417 1.1884817 1.0199412 0.9956941 0.9720389 0.9601474
#> [78] 0.9898781 1.1025485 0.9797453 1.0086780 1.0556471 1.0150204 1.0339022
#> [85] 1.1174116 1.5252177 0.9721734 0.9486663 1.0161640 0.9903872 1.2339874
#> [92] 1.0753099 0.9819882 1.0439012 1.0016272 1.0122706 1.0536213 0.9948601
#> [99] 1.4693656 1.0274264
Created on 2021-02-22 by the reprex package (v1.0.0)

Answer 2

for i in vector
eval(parse(text = sprintf("plot(df$%s)",i)))
This is very powerful line of code...can be very handy to plot graphs with loops.
{
eval(parse(text= sprintf('lof_%s <- lof(x, k=%s)',i,i)))
### distribution of outlier factors
eval(parse(text=sprintf('summary(lof_%s)',i)))
eval(parse(text=sprintf('hist(lof_%s, breaks=10)',i)))
### point size is proportional to LOF
eval(parse(text=sprintf("plot(x, pch = '.', main = 'LOF (k=%s)')",i)))
eval(parse(text=sprintf("points(x, cex = (lof_%s-1)*3, pch = 1, col='red')",i)))
}```
Exaplaination-
eval() - it evaluates the expression
parse() - it parse the text for evaluation
sprintf() - it creates a string(text) by concatenating with the parameter parsed.
Your code is not working because inside the loop i is being treated as character. It is not holding the values from the iterator.In case you need to understand above function then i would suggest you to just run this function and see the output sprintf('lof_%s <- lof(x, k=%s)',i,i).