R Split a column based on pattern

Question

I have a long list of vectors which I got using the str_extract_all().

The head(list) output is as following:

[1] "ARGENTINA"

[[2]]
[1] "BUENOS "   "AIRES"     "BUENOS "   "AIRES"     "ARGENTINA"

[[3]]
[1] "ARGENTINA" "ARGENTINA"

[[4]]
[1] "ARGENTINA" "ARGENTINA"

[[5]]
[1] "ARGENTINA"

[[6]]
[1] "ARGENTINA"

I now want to move the output to excel with each element occupying a different column within the same row. example:

p1          p2         p3       p4     p5
ARGENTINA   NA         NA       NA     NA
BUENOS      AIRES      BUENOS   AIRES  ARGENTINA
ARGENTINA   ARGENTINA  ARGENTINA NA    NA

But I get the following error while trying to do it:

Error in data.frame("ARGENTINA", c("BUENOS ", "AIRES", "BUENOS ", "AIRES", : arguments imply differing number of rows: 1, 5, 2, 3, 6, 4, 0, 9, 8, 7, 38,

Any help will be appreciated.

Answer 1

Since the result is from str_extract_all, just use simplify = T argument which will return a matrix instead of a list. The matrix will consist of empty strings and not NA. You can replace them and then turn it into a dataframe.

mat <- str_extract_all(your_string, pattern, simplify = TRUE)

mat[mat == ""] <- NA_character_

df <- as.data.frame(mat)

You can rename the column names now

Answer 2

As the lengths of the list elements are different, we may need to pad NA at the end before rbinding by assinging length<- to the maximum length of the elements in the list.

lst <- lapply(lst, trimws)
d1 <- as.data.frame(do.call(rbind, lapply(lst, `length<-`, 
                  max(lengths(lst)))), stringsAsFactors=FALSE)
names(d1) <- paste0("p", seq_along(d1))

---

If we are using packages, stri_list2matrix is a convenient function from stringi

library(stringi)
stri_list2matrix(lst, byrow=TRUE)

data

lst <- list("ARGENTINA", c("BUENOS ", "AIRES", "BUENOS ", "AIRES", 
   "ARGENTINA"
 ), c("ARGENTINA", "ARGENTINA"), c("ARGENTINA", "ARGENTINA"), 
    "ARGENTINA", "ARGENTINA")