This faster than your version, and doesn't return NA:
df$htn <- apply(df[,2:4], 1, function(x)any(grepl("^410", x)))
> df
ID DX1 DX2 DX3 htn
1 1 4109 4280 7102 TRUE
2 2 734 311 490 FALSE
3 3 4011 42822 4101 TRUE
> system.time({
+ for(j in 1:10000) {
+ for (i in 1:nrow(df)) {
+ df[i,"htn"] <- any(sapply('410', function(x) which( grepl(x,
df[i, 2:4], fixed = TRUE) )))
+ }
+ }
+ })
user system elapsed
6.648 0.008 6.657
There were 50 or more warnings (use warnings() to see the first 50)
>
>
>
> system.time({
+ for(j in 1:10000) {
+ df$htn <- apply(df[,2:4], 1, function(x)any(grepl("^410", x)))
+ }
+ })
user system elapsed
1.826 0.000 1.826
On Mon, Jun 15, 2015 at 4:12 PM, Federman, Douglas
<douglas.feder...@utoledo.edu> wrote:
> I'm trying to do the following: search each patient's list of diagnoses for a
> specific code then create a new column based upon the the presence of the
> specific code.
> Simplified data follows:
>
> con <- textConnection("
> ID DX1 DX2 DX3
> 1 4109 4280 7102
> 2 734 311 490
> 3 4011 42822 4101
> ")
> df <- read.table(con, header = TRUE, strip.white = TRUE,
> colClasses="character")
> #
> # I would like to add a column such the result of searching for 410 would
> give: The search string would always be at the start of a word and doesn't
> need regex.
> #
> # ID DX1 DX2 DX3 htn
> # 1 4109 4280 7102 1
> # 2 734 311 490 0
> # 3 4011 42822 4101 1
> #
> # The following works but is slow and returns NA if the search string is not
> found:
>
> for (i in 1:nrow(df)) {
> df[i,"htn"] <- any(sapply('410', function(x) which( grepl(x, df[i, 2:4],
> fixed = TRUE) )))
> }
>
> Thanks in advance. I never fail to learn new things from this list.
>
--
Sarah Goslee
http://www.functionaldiversity.org
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