R How to loop over paired columns to create new columns - r

Question

I am trying to loop over specific pair columns (they have similar names) and create columns based on a conditional statement.
Example dataset:
set.seed(2)
df <- data.frame (id=rep(1:5),
s1=rnorm(5, 0, 3),
s2=rnorm(5, 0, 3),
s2a=rnorm(5, 0, 3),
st1=rnorm(5, 3, 3),
st2=rnorm(5, 3, 3),
st2a=rnorm(5, 3, 3))
> df
id s1 s2 s2a st1 st2 st2a
1 1 -2.6907436 0.3972609 1.252952 -3.933207 9.2724576 -4.355119
2 2 0.5545476 2.1238642 2.945258 5.635814 -0.5997775 4.431712
3 3 4.7635360 -0.7190941 -1.178086 3.107420 7.7689146 1.210325
4 4 -3.3911270 5.9534218 -3.119007 6.038486 8.8639549 5.376610
5 5 -0.2407553 -0.4163610 5.346687 4.296795 3.0148133 3.868910
Column s1 is paired with column st1 etc. I want to indicate 1/0 if the equality between these columns is -3 to 0. E.g. df$ys1<-ifelse(df$s1<=-3 & df$st1>=0, 1, 0). The ultimate aim is to create the final variable yes_no (1/0) to indicate if any of the differences between the pairs of columns are 1 e.g. df$yes_no<-ifelse(df$ys1==1 | df$ys2==1 | df$ys2a==1, 1, 0)
The new dataset should look like this:
> df
id s1 s2 s2a st1 st2 st2a ys1 ys2 ys2a yes_no
1 1 -2.6907436 0.3972609 1.252952 -3.933207 9.2724576 -4.355119 0 0 0 0
2 2 0.5545476 2.1238642 2.945258 5.635814 -0.5997775 4.431712 0 0 0 0
3 3 4.7635360 -0.7190941 -1.178086 3.107420 7.7689146 1.210325 0 0 0 0
4 4 -3.3911270 5.9534218 -3.119007 6.038486 8.8639549 5.376610 1 0 1 1
5 5 -0.2407553 -0.4163610 5.346687 4.296795 3.0148133 3.868910 0 0 0 0
I'm sure there is a way of doing a loop without actually creating all additional columns (i.e. just create the final column, yes_no ) but I would be interested in how to create these just to know how to do it, in addition to a neater method.
I think a way of doing it would be to break up the dataset into two sets based on the pairs and then use in a loop:
firstt<-(df[,c(2:4)])
final<-(df[,c(5:7)])
or skip that and try directly in a loop
for(i in names(df[,c(2:4)])){
r<-(df[,c(5:7)])
df[i] <-ifelse(df$[i]<=-3 & df$[r]>=0, 1, 0)
}
Obviously that wont work but that is the idea of what I was trying.
Any help would be appreciated.

Answer 1

Here a solution in base R:
df$yes_no <-
rowSums(mapply(function(i,r)
ifelse(df[[r]]<=-3 & df[[i]]>=0, 1, 0)
, grep("st",names(df),value=TRUE),
gsub("t","",grep("st",names(df),value=TRUE)))) >0
1- I amusing regex to extract names. You can use indices here also. $X
X = "st1" "st2" "st2a"
Y = "s1" "s2" "s2a"
2- I am using mapply to apply to the paired elemend ( first elt of X and first element of Y and so ..)
3- rowSums to aggregate the 3 columns in one , and >0 to convert it to a logical vector

Answer 2

Here is another solution with for loops
a <- names(df[,c(2:4)])
b <- names(df[,c(5:7)])
for(i in seq_along(a)){
df$temp<-ifelse(df[,names(df)[names(df)==a[i]]]<=-3 & df[,names(df)[names(df)==b[i]]]>=0, 1, 0)
names(df)[names(df)=="temp"] <- paste0("ys", i)
}
df$yes_no <- apply(df[grep("ys", names(df))]==1,1, function(k) ifelse(TRUE %in% k, 1, 0) )
print(df)
id s1 s2 s2a st1 st2 st2a ys1 ys2 ys3 yes_no
1 1 -2.6907436 0.3972609 1.252952 -3.933207 9.2724576 -4.355119 0 0 0 0
2 2 0.5545476 2.1238642 2.945258 5.635814 -0.5997775 4.431712 0 0 0 0
3 3 4.7635360 -0.7190941 -1.178086 3.107420 7.7689146 1.210325 0 0 0 0
4 4 -3.3911270 5.9534218 -3.119007 6.038486 8.8639549 5.376610 1 0 1 1
5 5 -0.2407553 -0.4163610 5.346687 4.296795 3.0148133 3.868910 0 0 0 0