Creating Dummy Variables from Multiple Character (string) Variables (r)

Question

I need to create a dummy variable (binary) from a character (string) variable The data that I have look like this:

dat <- tribble(
    ~pat_id, ~icd9_1, ~icd9_2,
    1, "414.01", "414.01",
    2, "411.89", NA,
    3, NA, "410.71",
    4, NA, NA,
    5, NA, "410.51",
    6, NA, "272.0, 410.71"
)
dat



# A tibble: 6 x 3
#         pat_id icd9_1        icd9_2
#          <dbl>  <chr>         <chr>
#              1 414.01        414.01
#              2 411.89          <NA>
#              3   <NA>        410.71
#              4   <NA>          <NA>
#              5   <NA>        410.51
#              6   <NA> 272.0, 410.71

I want to create three new binary variables:

icd9_bin_1 == binary (0/1) for icd9_1
icd9_bin_2 == binary (0/1) for icd9_2
icd9_bin == binary for either icd9_1 OR icd9_2

What is the fastest way to create these binary variables?

I've replaced NAs with 0, turned into a factor and then recoded, but that took forever.

# get structure
dat$icd9_1 %>% str()
# get rid of NAs (replace with 0s)
dat$icd9_1[is.na(dat$icd9_1 )] <- 0
# turn into factor
dat$icd9_1 <- factor(dat$icd9_1)
# get levels 
dat$icd9_1 %>% levels()
# use fct_collapse
dat %>%
    mutate(icd9_bin_1 = fct_collapse(
        icd9_1,
        `icd9` = c("411.89","414.01"),
        `no icd9 dx` = c("0")))
# A tibble: 6 x 4
#   pat_id icd9_1        icd9_2 icd9_bin_1
#    <dbl> <fctr>         <chr>     <fctr>
#        1 414.01        414.01       icd9
#        2 411.89          <NA>       icd9
#        3      0        410.71 no icd9 dx
#        4      0          <NA> no icd9 dx
#        5      0        410.51 no icd9 dx
#        6      0 272.0, 410.71 no icd9 dx

I'm looking for a more elegant solution. Ideas?

The first row should be the binary for either since it has both non-na columns. You have labeled it the same as the second row indicating column 9_1 only. — Pierre L
– Pierre L, Commented Jun 14, 2017 at 17:12
Do you just need dat$icd9_bin_1 <- if_else(is.na(dat$icd9_1), "no icd9 dx", "icd9")? I'm tired, so I'm probably missing something... — Phil
– Phil, Commented Jun 14, 2017 at 17:14
@PierreLafortune sorry about that--I was just giving an example of how I was creating the first binary variable, icd9_bin_1. After these two are created, I use mutate and if_else to create the binary for either icd9_1 or icd9_2 — Martin Frigaard
– Martin Frigaard, Commented Jun 14, 2017 at 17:19
Try dat[c('icd9_bin_1', 'icd9_bin_2')] <- paste(c('yes', 'no')[is.na(dat[-1]) + 1L], rep(names(dat[-1]), each=nrow(dat)), sep='-') — Pierre L
– Pierre L, Commented Jun 14, 2017 at 17:23
@Phil, yes that works (and is way fewer lines of code). I guess I was hoping for a dplyr solution that let me create all three variables in one pipe? The actual data has up to 50 different icd9 levels across several variables. — Martin Frigaard
– Martin Frigaard, Commented Jun 14, 2017 at 17:27

TJ Mahr · Accepted Answer · 2017-06-14 19:19:30Z

To create the binary values manually, just apply a function to each column and take the or of the columns to find rows where neither is NA.

is_not_na <- function(...) Negate(is.na)(...)

dat %>%
  mutate(icd9_bin_1 = icd9_1 %>% is_not_na() %>% as.numeric(),
         icd9_bin_2 = icd9_2 %>% is_not_na() %>% as.numeric(),
         icd9_bin = as.numeric(icd9_bin_1 | icd9_bin_2))
#> # A tibble: 6 x 6
#>   pat_id icd9_1        icd9_2 icd9_bin_1 icd9_bin_2 icd9_bin
#>    <dbl>  <chr>         <chr>      <dbl>      <dbl>    <dbl>
#> 1      1 414.01        414.01          1          1        1
#> 2      2 411.89          <NA>          1          0        1
#> 3      3   <NA>        410.71          0          1        1
#> 4      4   <NA>          <NA>          0          0        0
#> 5      5   <NA>        410.51          0          1        1
#> 6      6   <NA> 272.0, 410.71          0          1        1

If you had many, many of these columns, you could use mutate_at().

is_not_na_num <- function(...) as.numeric(Negate(is.na)(...))

# Make up a new column
dat$icd9_3 <- rev(dat$icd9_1)

# To use pattern matching...
data_auto <- dat %>%
  mutate_at(vars(matches("icd9")), funs(bin = is_not_na_num))
data_auto
#> # A tibble: 6 x 7
#>   pat_id icd9_1        icd9_2 icd9_3 icd9_1_bin icd9_2_bin icd9_3_bin
#>    <dbl>  <chr>         <chr>  <chr>      <dbl>      <dbl>      <dbl>
#> 1      1 414.01        414.01   <NA>          1          1          0
#> 2      2 411.89          <NA>   <NA>          1          0          0
#> 3      3   <NA>        410.71   <NA>          0          1          0
#> 4      4   <NA>          <NA>   <NA>          0          0          0
#> 5      5   <NA>        410.51 411.89          0          1          1
#> 6      6   <NA> 272.0, 410.71 414.01          0          1          1

(But to automate that final or you could use reduce()...)

bin_any <- data_auto %>%
  select(matches("_bin")) %>%
  purrr::reduce(~ as.numeric(.x | .y))
data_auto$icd9_bin <- bin_any
data_auto["icd9_bin"]
#> # A tibble: 6 x 1
#>   icd9_bin
#>      <dbl>
#> 1        1
#> 2        1
#> 3        1
#> 4        0
#> 5        1
#> 6        1

thank you! I went with the pattern matching because the actual data is coded according to specific medical conditions (i.e. hypertension is htn_icd9_plst, htn_icd9_enc, etc. ). This function will be very useful! I also wanted to share the dummyVar function from the caret package.

Phil · Accepted Answer · 2017-06-14 17:46:34Z

0

As per your comments, if_else() is a dplyr function that plays well with mutate() if that's what you need:

dat <- dat %>%
  mutate(icd9_bin_1 = if_else(is.na(dat$icd9_1), "no icd9 dx", "icd9"),
         more...)

answered Jun 14, 2017 at 17:46

Phil

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1 Comment

Martin Frigaard Over a year ago

Yes, @Phil --this is similar to what I am currently using. This requires multiple steps (i.e. binary for each variable, then binary for either of the binary variables). I was hoping there was a solution that passed over the first step and looked across a set of variables (because technically they are lists?) and return 1 = at least one icd9 in any of these variables or 0 = NA in all of these variables.

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