R creating new column without for loop

Question

Suppose I have a data frame with a few numbers in the first column. I want to take these numbers, use them as locations in a string, and take a substring that includes 2 characters before and after that location. To clarify,

aggSN <- data.frame(V1=c(5,6,7,8),V2="blah")
gen <- "AJSDAFKSDAFJKLASDFKJKA"  # <- take this string
aggSN                            # <- take the numbers in the first column
# V1    V2
#  5  blah
#  6  blah
#  7  blah
#  8  blah

and create a new column V3 that looks like

aggSN                           
# V1    V2    V3
#  5  blah SDAFK   # <- took the two characters before and after the 5th character
#  6  blah DAFKS   # <- took the two characters before and after the 6th character 
#  7  blah AFKSD   # <- took the two characters before and after the 7th character 
# 10  blah SDAFJ   # <- took the two characters before and after the 10th character 
#  2  blah AJSD   # <- here you can see that it the substring cuts off

Currently I am using a for loop, which works, but takes a lot of time on very large data frames and large strings. Are there any alternatives to this? Thank you.

fillvector <- ""
for(j in 1:nrow(aggSN)){fillvector[j] <- substr(gen,aggSN[j,V1]-2,aggSN[j,V1]+2)}
aggSN$V9 <- fillvector

What happens if boundaries are out of range 1 .. length(gen)? — zero323
– zero323, Commented Aug 3, 2015 at 2:24
I will edit my post shortly after this comment, but I would want the substring to just cut off. — alki
– alki, Commented Aug 3, 2015 at 2:26

Rich Scriven · Accepted Answer · 2015-08-03 02:52:20Z

4

You can use substring() without writing a loop

aggSN <- data.frame(V1=c(5,6,7,8,2),V2="blah")
gen <- "AJSDAFKSDAFJKLASDFKJKA" 

with(aggSN, substring(gen, V1-2, V1+2))
# [1] "SDAFK" "DAFKS" "AFKSD" "FKSDA" "AJSD"

So to add the new column,

aggSN$V3 <- with(aggSN, substring(gen, V1-2, V1+2))
aggSN
#   V1   V2    V3
# 1  5 blah SDAFK
# 2  6 blah DAFKS
# 3  7 blah AFKSD
# 4  8 blah FKSDA
# 5  2 blah  AJSD

If you are after something a bit faster, I would go with stringi::stri_sub in place of substring().

edited Aug 3, 2015 at 2:52

answered Aug 3, 2015 at 2:28

Rich Scriven

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4 Comments

MichaelChirico Over a year ago

guess your answer didn't load. this is the way to go. one of these days I'll learn how to use with like a pro.

Ricky Over a year ago

interesting, I didn't know the difference between substr and substring. My initial attempt was with with and substr which produced a constant vector, before I switched to sapply. This is definitely better.

Rich Scriven Over a year ago

@MichaelChirico - I had deleted it for a few minutes then realized I had it right. Sorry about that.

MichaelChirico Over a year ago

@Ricky I had to ?substr to remind myself. I initially thought it was substr that accepts vector indices. See the examples in ?substr

Ricky · Accepted Answer · 2015-08-03 02:34:54Z

2

aggSN$V3 <- sapply(aggSN$V1, function(x) substr(gen, x-2, x+2))

should do the trick.

> aggSN
  V1   V2    V3
1  5 blah SDAFK
2  6 blah DAFKS
3  7 blah AFKSD
4  8 blah FKSDA

With your different example

> aggSN
  V1   V2    V3
1  5 blah SDAFK
2  6 blah DAFKS
3  7 blah AFKSD
4 10 blah SDAFJ
5  2 blah  AJSD

answered Aug 3, 2015 at 2:34

Ricky

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1 Comment

Ricky Over a year ago

the reproducible example differs from the example where the V3 is shown. The latter one is what I refer to as "different example".

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