CODEWITHSUNDEEP
rtidyversestringi
TinoMass
TinoMass

Reputation: 29

how do I insert a character into a string at a specific location but only counting letters in R

I am trying to create a df where I add a character into a string at a position defined by another column- but I only want the code to count letters and not numbers or other characters while it does this, hopefully the example tables make this clearer

my initial data looks like this.

String Insert_pos
PEPTIDE 3
PE[+10]TIDE 3

i use the following code

library(stringi)
stri_sub(df$String,df$Insert_pos+1, df$insert_pos-1) <- "[+20]"

and it only half works--- I only want it to count the Letters and not the numbers already added when it adds the new characters. as shown below

what I get What I want
PEP[+20]TIDE PEP[+20]TIDE
PE[[+20]+10]TIDE PE[+10]P[+20]TIDE

I think the way to do it would be to specify to only count letters, but I cant find how to specify this in Stringi/ not sure if it is possible.

Any help would be great,

Thanks!

Upvotes: 0

Views: 296

Answers (2)

GKi
GKi

Reputation: 39667

You can use sub:

sub("(([[:alpha:]][^[:alpha:]]*){3})", "\\1[+20]", df$String)
#[1] "PEP[+20]TIDE"     "PE[+10]T[+20]IDE"

In this case using only upper case will also work.

sub("(([A-Z][^A-Z]*){3})", "\\1[+20]", df$String)

sub("(([[:upper:]][^[:upper:]]*){3})", "\\1[+20]", df$String)

Taking the positions from Insert_pos:

mapply(function(s, n) sub(paste0("(([[:alpha:]][^[:alpha:]]*){", n, "})"),
                          "\\1[+20]", s), df$String, df$Insert_pos)

Upvotes: 1

Domingo
Domingo

Reputation: 671

You can change the value of your df$Insert_pos column for the next possible position:

df <- data.frame(
  String = c("PEPTIDE", "PE[+10]TIDE"),
  Insert_pos = c(3,3)
)

df$Insert_pos <- ifelse(
  stri_sub(df$String,df$Insert_pos, df$Insert_pos) %in% c("[", "]", "+", 1:9), 
  df$Insert_pos + 5 , 
  df$Insert_pos)

library(stringi)
stri_sub(df$String,df$Insert_pos+1, df$Insert_pos-1) <- "[+20]"

df

This gives you then:

            String Insert_pos
1     PEP[+20]TIDE          3
2 PE[+10]T[+20]IDE          8

Upvotes: 1

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