R: How to automatically create flag variables for sequences of values? - r

Question

Suppose, you're given the following dataframe:
a <- data.frame(var = c(",1,2,3,", ",2,3,5,", ",1,3,5,5,"))
What I am looking for is to create the variables flag_1, ..., flag_7 in a containing the information of how many times the respective values occur. For a, I would expect the following result:
var flag_1 flag_2 flag_3 flag_4 flag_5
",1,2,3," 1. 1. 1. 0. 0.
",2,3,5," 0. 1. 1. 0. 1.
",1,3,5,5," 1. 0. 1. 0. 2.
I managed to get the result using a nested for-loop and an if-condition but there must be a nicer (more aesthetic and better performing) solution.

Answer 1

One option would be to do strsplit, get the table and then cbind with original data
cbind(a, do.call(rbind, lapply(strsplit(as.character(a$var), ","),
function(x) table(factor(x[nzchar(x)], levels = 1:5, labels = paste0("flag_", 1:5))))))
# var flag_1 flag_2 flag_3 flag_4 flag_5
#1 ,1,2,3, 1 1 1 0 0
#2 ,2,3,5, 0 1 1 0 1
#3 ,1,3,5,5, 1 0 1 0 2
Another option is with tidyverse
library(tidyverse)
str_extract_all(a$var, "[0-9]") %>%
map(~ as.integer(.x) %>%
as_tibble) %>%
bind_rows(.id = 'grp') %>%
count(grp, value = factor(value, levels = min(value):max(value))) %>%
spread(value, n, drop = FALSE, fill = 0) %>%
select(-grp) %>%
bind_cols(a, .) %>%
rename_at(vars(matches("^[0-9]+$")), ~ paste0("flag_", .))
# var flag_1 flag_2 flag_3 flag_4 flag_5
#1 ,1,2,3, 1 1 1 0 0
#2 ,2,3,5, 0 1 1 0 1
#3 ,1,3,5,5, 1 0 1 0 2

Answer 2

First, don't make the strings into factors. Nothing good comes from that.
a <- data.frame(var = c(",1,2,3,", ",2,3,5,", ",1,3,5,5,"),
stringsAsFactors = FALSE)
To get from strings to your table is simple enough if we take it in small steps. Here, I've written (or renamed) a function per step and then gone through the steps using lapply one at a time. You can string it all together in a pipeline if like, but it would be roughly these steps.
First, I extract the numbers from the strings. That involves splitting on commas, getting rid of empty strings, you have those because you can begin and end a string with a comma, but otherwise, that step wouldn't be necessary. Then we need to translate the strings into numbers, count how often we see each (we can do that with the as.numeric and table functions, respectively), and then it is just a question of mapping the observed counts into a table that also includes those we haven't observed.
pick_indices <- function(str) unlist(strsplit(str, split = ","))
remove_empty <- function(chrs) chrs[nchar(chrs) > 0]
get_indices <- as.numeric
to_counts <- table
to_flag_vect <- function(counts, len) {
vec <- rep(0, len)
names(vec) <- 1:len
vec[names(counts)] <- counts
vec
}
strings <- lapply(a$var, pick_indices)
cleaned <- lapply(strings, remove_empty)
indices <- lapply(cleaned, get_indices)
counts <- lapply(indices, to_counts)
flags <- lapply(counts, to_flag_vect, len = 5)
We now have the flag-counts in a list, so to make it into the table you want, with the column names you want, we simply do this:
tbl <- do.call(rbind, flags)
colnames(tbl) <- paste0("flag_", 1:5)
tbl
Done.

Answer 3

Split and unlist the values into a factor with appropriate levels
x = strsplit(a$var, ",")
xp = factor(unlist(x), levels = seq_len(5))
Create an index that maps the values of xp to the rows they came from
i = rep(seq_along(x), lengths(x))
use xtabs() to cross-tabulate the entries by row
xt = xtabs(~ i + xp)
and cbind() the matrix representation of the result to the original
> cbind(a, unclass(xt))
var 1 2 3 4 5
1 ,1,2,3, 1 1 1 0 0
2 ,2,3,5, 0 1 1 0 1
3 ,1,3,5,5, 1 0 1 0 2