Build a vector/frame by combining regmatches results - r

Question

I have a regular expression that parses a bunch of text, an when doing regmatches(myText,myRegex) it returns a list which looks like:
[[1]]
[1] "a=1" "b=3" "a=9" "c=2" "b=4"
...
I'd like to build a data.frame or table - whatever suits best - to finally have something like:
a b c
1 3 2
9 4 ...
Is it possible to make this in a simple fashion? What are your suggestions?
Thanks in advance.

Answer 1

Its not entirely clear what the general case is here but this works on the data provided.
Assuming this input:
x <- c("a=1", "b=3", "a=9", "c=2", "b=4")
split the values by the names producing s and massage into a data.frame:
s <- split(as.numeric(sub(".*=", "", x)), sub("=.*", "", x))
as.data.frame(do.call(cbind, lapply(s, ts)))
giving:
a b c
1 1 3 2
2 9 4 NA
No packages needed.

Answer 2

You can either use base R methods
d1 <- read.table(text=gsub("[[:punct:]]", " " , unlist(lst)))
d2 <- transform(d1, indx=ave(seq_along(V1), V1, FUN=seq_along))
res <- reshape(d2, timevar='V1', idvar='indx', direction='wide')[,-1]
colnames(res) <- gsub(".*\\.", "", colnames(res))
res
# a b c
#1 1 3 2
#3 9 4 2
#6 4 5 NA
#9 9 NA NA
Or using dcast from reshape2 on d2
library(reshape2)
dcast(d2,indx~V1, value.var='V2')[,-1]
# a b c
#1 1 3 2
#2 9 4 2
#3 4 5 NA
#4 9 NA NA
data
lst <- list(c('a=1', 'b=3', 'a=9', 'c=2', 'b=4'),
c('a=4', 'c=2', 'b=5', 'a=9'))

Answer 3

Using rex may make this type of extraction task a little simpler.
x <- c("a=1", "b=3", "a=9", "c=2", "b=4", "a=2")
First extract the names and values from the strings.
library(rex)
matches <- re_matches(x,
rex(
capture(name="name", letter),
"=",
capture(name="value", digit)
))
#> name value
#>1 a 1
#>2 b 3
#>3 a 9
#>4 c 2
#>5 b 4
#>6 a 2
Then tally the groups using split().
groups <- split(as.numeric(matches$value), matches$name)
#>$a
#>[1] 1 9 2
#>
#>$b
#>[1] 3 4
#>
#>$c
#>[1] 2
If we try to convert directly to a data.frame from split() the groups with fewer members will have their members recycled rather than NA, so instead explicitly fill with NA.
largest_group <- max(sapply(groups, length))
#>[1] 3
groups <- lapply(groups, function(group) {
if (length(group) < largest_group) {
group[largest_group] <- NA
}
group
})
#>$a
#>[1] 1 9 2
#>
#>$b
#>[1] 3 4 NA
#>
#>$c
#>[1] 2 NA NA
Finally we can create the data.frame
do.call('data.frame', groups)
#> a b c
#>1 1 3 2
#>2 9 4 NA
#>3 2 NA NA

Answer 4

Here's an approach using tools from my "splitstackshape" package:
library(splitstackshape)
dcast.data.table( ## Makes the long data wide
getanID( ## Adds an ID variable for dcast
## create a single column data.table and split it by the "="
cSplit(as.data.table(unlist(lst)), "V1", "="), "V1_1"),
.id ~ V1_1, value.var = "V1_2")
# .id a b c
# 1: 1 1 3 2
# 2: 2 9 4 2
# 3: 3 4 5 NA
# 4: 4 9 NA NA
This uses #akrun's sample data:
lst <- list(c('a=1', 'b=3', 'a=9', 'c=2', 'b=4'),
c('a=4', 'c=2', 'b=5', 'a=9'))