How to divide the column values in to ranges - r

Question

I would like to know how to divide the column values in to three different ranges based on scores.
Here's the following data I have
Name V1.1 V1.2 V2.1 V2.2 V3.1 V3.2
John French 86 Math 78 English 56
Sam Math 97 French 86 English 79
Viru English 93 Math 44 French 34
If I consider three ranges. First rangewith 0-60, Second rangewith 61-90 and third range with 91-100.
I would like to the subject name across all the skills.
Expected result would be
Name Level1 Level2 Level3
John English Math,French Null
Sam Null French,Eng Math
Viru Math,Fren Null English

Answer 1

First you need to convert the data to long form, one row per observation (where an observation is a single score. You need to do a melt, but it is complicated because your wide form consists of not only observations but observation classes. One way to do it is to use melt.data.table twice, but you may be more comfortable with dplyr, which has more accessible syntax.
# first you need to convert to long form
library("data.table")
setDT(df)
lhs <- melt.data.table(df, id = "Name", measure = patterns("\\.2"),
variable.name = "obs", value.name = "score")
lhs[, obs := gsub("(V\\d+)\\.\\d+","\\1",obs)]
lhs
rhs <- melt.data.table(df, id = "Name", measure = patterns("V\\d\\.1"),
variable.name = "obs", value.name = "subject")
rhs[, obs := gsub("(V\\d+)\\.\\d+","\\1",obs)]
rhs
df2 <- merge(lhs, rhs, by = c("Name","obs"))
# Name obs score1 subject1
# 1: John V1 86 French
# 2: John V2 78 Math
# 3: John V3 56 English
# 4: Sam V1 97 Math
# 5: Sam V2 86 French
# 6: Sam V3 79 English
# 7: Viru V1 93 English
# 8: Viru V2 44 Math
# 9: Viru V3 34 French
Then you need to use cut or some other function to create the three levels based on score1.
Then you should group by these levels and apply concatenation to the subjects, such as paste(..., collapse = ",").
Then you need to use cast or spread to return it to wide form.
Do give it some effort, and edit your question with what you've tried, and try to come up with a more specific question, not just "please do this for me".

Answer 2

Another option using splitstackshape and nested ifelse
library(splitstackshape)
library(tidyr)
# prepare data to convert in long format
data$subjects = do.call(paste, c(data[,grep("\\.1", colnames(data))], sep = ','))
data$marks = do.call(paste, c(data[,grep("\\.2", colnames(data))], sep = ','))
data = data[,-grep("V", colnames(data))]
# use cSplit to convert wide to long
out = cSplit(setDT(data), sep = ",", c("subjects", "marks"), "long")
# nested ifelse to assign level based on the score range
out[, level := ifelse(marks <= 60, "level1",
ifelse(marks <= 90, "level2", "level3"))]
req = out[, toString(subjects), by= c("Name","level")]
this will give
#> req
# Name level V1
#1: John level2 French, Math
#2: John level1 English
#3: Sam level3 Math
#4: Sam level2 French, English
#5: Viru level3 English
#6: Viru level1 Math, French
you can reshape either using dcast or spread from tidyr
spread(req, level, V1)
# Name level1 level2 level3
#1: John English French, Math NA
#2: Sam NA French, English Math
#3: Viru Math, French NA English
data
data = structure(list(Name = structure(1:3, .Label = c("John", "Sam",
"Viru"), class = "factor"), V1.1 = structure(c(2L, 3L, 1L), .Label = c("English",
"French", "Math"), class = "factor"), V1.2 = c(86L, 97L, 93L),
V2.1 = structure(c(2L, 1L, 2L), .Label = c("French", "Math"
), class = "factor"), V2.2 = c(78L, 86L, 44L), V3.1 = structure(c(1L,
1L, 2L), .Label = c("English", "French"), class = "factor"),
V3.2 = c(56L, 79L, 34L)), .Names = c("Name", "V1.1", "V1.2",
"V2.1", "V2.2", "V3.1", "V3.2"), class = "data.frame", row.names = c(NA,
-3L))

Answer 3

Not very intuitive, but leads to the requested output. Requires the package sjmisc!
library(sjmisc)
mydat <- data.frame(Name = c("John", "Sam", "Viru"),
V1.1 = c("French", "Math", "English"),
V1.2 = c(86, 97, 93),
V2.1 = c("Math", "French", "Math"),
V2.2 = c(78, 86, 44),
V3.1 = c("English", "English", "French"),
V3.2 = c(56, 79, 34))
# recode into groups
rec(mydat[, c(3,5,7)]) <- "min:60=1; 61:90=2; 91:max=3"
# convert to long format
newdf <- to_long(mydat, "no_use",
c("subject", "score"),
c("V1.1", "V2.1", "V3.1"),
c("V1.2", "V2.2", "V3.2")) %>%
select(-no_use) %>%
arrange(Name, score)
# at this point we are at a similar stage as described in the
# other answers, so we have our data in a long format
newdf
fdf <- list()
# iterate all unique names
for (i in unique(newdf$Name)) {
dummy <- c()
# iterare all three scores
for (s in 1:3) {
# find subjects related to the score
dat <- newdf$subject[newdf$Name == i & newdf$score == s]
if (length(dat) == 0) dat <- ""
dat <- paste0(dat, collapse = ",")
dummy <- c(dummy, dat)
}
# add character vector with sorted subjects to list
fdf[[length(fdf) + 1]] <- dummy
}
# list to data frame
finaldf <- as.data.frame(t(as.data.frame(fdf)))
finaldf <- cbind(unique(newdf$Name), finaldf)
# proper row/col names
colnames(finaldf) <- c("Names", "Level1", "Level2", "Level3")
rownames(finaldf) <- 1:nrow(finaldf)
finaldf