Find unique set of strings in vector where vector elements can be multiple strings - r

Question

I have a series of batch records that are labeled sequentially. Sometimes batches overlap.
x <- c("1","1","1/2","2","3","4","5/4","5")
> data.frame(x)
x
1 1
2 1
3 1/2
4 2
5 3
6 4
7 5/4
8 5
I want to find the set of batches that are not overlapping and label those periods. Batch "1/2" includes both "1" and "2" so it is not unique. When batch = "3" that is not contained in any previous batches, so it starts a new period. I'm having difficulty dealing with the combined batches, otherwise this would be straightforward. The result of this would be:
x period
1 1 1
2 1 1
3 1/2 1
4 2 1
5 3 2
6 4 3
7 5/4 3
8 5 3
My experience is in more functional programming paradigms, so I know the way I did this is very un-R. I'm looking for the way to do this in R that is clean and simple. Any help is appreciated.
Here's my un-R code that works, but is super clunky and not extensible.
x <- c("1","1","1/2","2","3","4","5/4","5")
p <- 1 #period number
temp <- NULL #temp variable for storing cases of x (batches)
temp[1] <- x[1]
period <- NULL
rl <- 0 #length to repeat period
for (i in 1:length(x)){
#check for "/", split and add to temp
if (grepl("/", x[i])){
z <- strsplit(x[i], "/") #split character
z <- unlist(z) #convert to vector
temp <- c(temp, z, x[i]) #add to temp vector for comparison
}
#check if x in temp
if(x[i] %in% temp){
temp <- append(temp, x[i]) #add to search vector
rl <- rl + 1 #increase length
} else {
period <- append(period, rep(p, rl)) #add to period vector
p <- p + 1 #increase period count
temp <- NULL #reset
rl <- 1 #reset
}
}
#add last batch
rl <- length(x) - length(period)
period <- append(period, rep(p,rl))
df <- data.frame(x,period)
> df
x period
1 1 1
2 1 1
3 1/2 1
4 2 1
5 3 2
6 4 3
7 5/4 3
8 5 3

Answer 1

R has functional paradigm influences, so you can solve this with Map and Reduce. Note that this solution follows your approach in unioning seen values. A simpler approach is possible if you assume batch numbers are consecutive, as they are in your example.
x <- c("1","1","1/2","2","3","4","5/4","5")
s<-strsplit(x,"/")
r<-Reduce(union,s,init=list(),acc=TRUE)
p<-cumsum(Map(function(x,y) length(intersect(x,y))==0,s,r[-length(r)]))
data.frame(x,period=p)
x period
1 1 1
2 1 1
3 1/2 1
4 2 1
5 3 2
6 4 3
7 5/4 3
8 5 3
What this does is first calculate a cumulative union of seen values. Then, it maps across this to determine the places where none of the current values have been seen before. (Alternatively, this second step could be included within the reduce, but this would be wordier without support for destructuring.) The cumulative sum provides the "period" numbers based on the number of times the intersections have come up empty.
If you do make the assumption that the batch numbers are consecutive then you can do the following instead
x <- c("1","1","1/2","2","3","4","5/4","5")
s<-strsplit(x,"/")
n<-mapply(function(x) range(as.numeric(x)),s)
p<-cumsum(c(1,n[1,-1]>n[2,-ncol(n)]))
data.frame(x,period=p)
For the same result (not repeated here).

Answer 2

A little bit shorter:
x <- c("1","1","1/2","2","3","4","5/4","5")
x<-data.frame(x=x, period=-1, stringsAsFactors = F)
period=0
prevBatch=-1
for (i in 1:nrow(x))
{
spl=unlist(strsplit(x$x[i], "/"))
currentBatch=min(spl)
if (currentBatch<prevBatch) { stop("Error in sequence") }
if (currentBatch>prevBatch)
period=period+1;
x$period[i]=period;
prevBatch=max(spl)
}
x

Answer 3

Here's a twist on the original that uses tidyr to split the data into two columns so it's easier to use:
# sample data
x <- c("1","1","1/2","2","3","4","5/4","5")
df <- data.frame(x)
library(tidyr)
# separate x into two columns, with second NA if only one number
df <- separate(df, x, c('x1', 'x2'), sep = '/', remove = FALSE, convert = TRUE)
Now df looks like:
> df
x x1 x2
1 1 1 NA
2 1 1 NA
3 1/2 1 2
4 2 2 NA
5 3 3 NA
6 4 4 NA
7 5/4 5 4
8 5 5 NA
Now the loop can be a lot simpler:
period <- 1
for(i in 1:nrow(df)){
period <- c(period,
# test if either x1 or x2 of row i are in any x1 or x2 above it
ifelse(any(df[i, 2:3] %in% unlist(df[1:(i-1),2:3])),
period[i], # if so, repeat the terminal value
period[i] + 1)) # else append the terminal value + 1
}
# rebuild df with x and period, which loses its extra initializing value here
df <- data.frame(x = df$x, period = period[2:length(period)])
The resulting df:
> df
x period
1 1 1
2 1 1
3 1/2 1
4 2 1
5 3 2
6 4 3
7 5/4 3
8 5 3