Parsing Delimited Data In a DataFrame Into Separate Columns in R - r

Question

I have a data frame which looks as such
A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14
I would like to parse the C column into separate columns as such...
A B X1 X2 X3
1 3 7 8 9
2 4 10 11 12
5 6 13 14 NA
How would one go about doing this in R?

Answer 1

First, here's the sample data in data.frame form
dd<-data.frame(
A = c(1L, 2L, 5L),
B = c(3L, 4L, 6L),
C = c("X1=7;X2=8;X3=9",
"X1=10;X2=11;X3=12", "X1=13;X2=14"),
stringsAsFactors=F
)
Now I define a small helper function to take vectors like c("A=1","B=2") and changed them into named vectors like c(A="1", B="2").
namev<-function(x) {
a<-strsplit(x,"=")
setNames(sapply(a,'[',2), sapply(a,'[',1))
}
and now I perform the transformations
#turn each row into a named vector
vv<-lapply(strsplit(dd$C,";"), namev)
#find list of all column names
nm<-unique(unlist(sapply(vv, names)))
#extract data from all rows for every column
nv<-do.call(rbind, lapply(vv, '[', nm))
#convert everything to numeric (optional)
class(nv)<-"numeric"
#rejoin with original data
cbind(dd[,-3], nv)
and that gives you
A B X1 X2 X3
1 1 3 7 8 9
2 2 4 10 11 12
3 5 6 13 14 NA

Answer 2

My cSplit function makes solving problems like these fun. Here it is in action:
## Load some packages
library(data.table)
library(devtools) ## Just for source_gist, really
library(reshape2)
## Load `cSplit`
source_gist("https://gist.github.com/mrdwab/11380733")
First, split your values up and create a "long" dataset:
ddL <- cSplit(cSplit(dd, "C", ";", "long"), "C", "=")
ddL
# A B C_1 C_2
# 1: 1 3 X1 7
# 2: 1 3 X2 8
# 3: 1 3 X3 9
# 4: 2 4 X1 10
# 5: 2 4 X2 11
# 6: 2 4 X3 12
# 7: 5 6 X1 13
# 8: 5 6 X2 14
Next, use dcast.data.table (or just dcast) to go from "long" to "wide":
dcast.data.table(ddL, A + B ~ C_1, value.var="C_2")
# A B X1 X2 X3
# 1: 1 3 7 8 9
# 2: 2 4 10 11 12
# 3: 5 6 13 14 NA

Answer 3

Here's one possible approach:
dat <- read.table(text="A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14", header=TRUE, stringsAsFactors = FALSE)
library(qdapTools)
dat_C <- strsplit(dat$C, ";")
dat_C2 <- sapply(dat_C, function(x) {
y <- strsplit(x, "=")
rep(sapply(y, "[", 1), as.numeric(sapply(y, "[", 2)))
})
data.frame(dat[, -3], mtabulate(dat_C2))
## A B X1 X2 X3
## 1 1 3 7 8 9
## 2 2 4 10 11 12
## 3 5 6 13 14 0
EDIT To obtain the NA values
m <- mtabulate(dat_C2)
m[m==0] <- NA
data.frame(dat[, -3], m)

Answer 4

Here's a nice, somewhat hacky way to get you there.
## read your data
> dat <- read.table(h=T, text = "A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14", stringsAsFactors = FALSE)
## ---
> s <- strsplit(dat$C, ";|=")
> xx <- unique(unlist(s)[grepl('[A-Z]', unlist(s))])
> sap <- t(sapply(seq(s), function(i){
wh <- which(!xx %in% s[[i]]); n <- suppressWarnings(as.numeric(s[[i]]))
nn <- n[!is.na(n)]; if(length(wh)){ append(nn, NA, wh-1) } else { nn }
})) ## see below for explanation
> data.frame(dat[1:2], sap)
# A B X1 X2 X3
# 1 1 3 7 8 9
# 2 2 4 10 11 12
# 3 5 6 13 14 NA
Basically what's happening in sap is
check which values are missing
change each list element of s to numeric
remove the NA values from (2)
insert NA into the correct position with append
transpose the result