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Remove specific characters from column names in r

Have data set with several hundreds of columns, the column names look like this "drop.loc1.genom1.tret1.gwas2.a", I need to remove everything except loc1 and tret1 -- so it will look like this "loc1.trt1" ---- any hint or help will be highly appreciated
thanks

Do you really want to drop the e in tret? can anything follow loc and tret other than numbers?
– G5W
3 hours ago

3 Answers
3

Another option is to use strsplit:

strsplit

sapply(strsplit(strings, "\."), function(x)
paste0(x[c(2, 4)], collapse = "."))
[1] "loc1.tret1" "loc2.tret2" "loc100.tret100"


(From ManuelBickel's answer)

strings = c("drop.loc1.genom1.tret1.gwas2.a",
"drop.loc2.genom1.tret2.gwas2.a",
"drop.loc100.genom1.tret100.gwas2.a")


Amazing way of tackling this. Although all the functions you are using are vectorized. So no need of sapply eg you could do do.call(paste,c(sep=".",do.call(rbind.data.frame,strsplit(strings, "\."))[c(2,4)])) i guess.. But still you have my +1
– Onyambu
2 hours ago

sapply

do.call(paste,c(sep=".",do.call(rbind.data.frame,strsplit(strings, "\."))[c(2,4)]))

thank you it worked fine
– hema
2 hours ago

@Onyambu Thanks for your example
– Maurits Evers
1 hour ago

@hema You're very welcome; I've changed my solution slightly to make it more succinct (and used the fact that strsplit is vectorised in x as Onyambu pointed out).
– Maurits Evers
1 hour ago

strsplit

x

You might try something like..

UPDATE: Have updated the code with benchmark of all version proposed so far.
In case @Onyambu posts an answer you should accept that one, since the approach is the fastest.

strings = c("drop.loc1.genom1.tret1.gwas2.a",
"drop.loc2.genom1.tret2.gwas2.a",
"drop.loc100.genom1.tret100.gwas2.a")
gsub("(^.*\.)(loc\d+)(\..*\.)(tret\d+)(\..*$)", "\2.\4", strings, perl = T)
[1] "loc1.tret1" "loc2.tret2" "loc100.tret100"

f1 = function(strings)
unname(sapply(strings, function(x)
paste0(unlist(strsplit(x, "\."))[c(2, 4)], collapse = ".")))


f2 = function(strings)
gsub("(^.*\.)(loc\d+)(\..*\.)(tret\d+)(\..*$)", "\2.\4", strings, perl = T)


f2b = function(strings)
sub(".*(loc\d+).*(tret\d+).*","\1.\2",strings)


microbenchmark::microbenchmark(
f1(strings),
f2(strings),
f2b(strings)
)

# Unit: microseconds
# expr min lq mean median uq max neval
# f1(strings) 58.818 64.1475 136.31964 68.687 76.1880 5691.106 100
# f2(strings) 78.161 79.9380 106.08183 83.293 88.6215 2110.333 100
# f2b(strings) 27.238 29.6070 53.29592 32.765 35.1330 1872.299 100


Nice way of tackling this. Although you can shorten it as sub(".*(loc\d+).*(tret\d+).*","\1.\2",strings). still got my +1
– Onyambu
2 hours ago

sub(".*(loc\d+).*(tret\d+).*","\1.\2",strings)

Thank you for the hint and the improvements, you are absolutely right. You should post a separate answer to get the credits.
– Manuel Bickel
2 hours ago

You could used dplyr::rename_all() or dplyr::select_all() and gsub() using Onyambu's regex pattern from the comment to Manuel Bickel's answer:

dplyr::rename_all()

dplyr::select_all()

gsub()

library(dplyr)

# sample data
df <- data_frame(drop.loc1.genom1.tret1.gwas2.a = 1:2,
drop.loc23.genom2.tret2.gwas2.a = 3:4,
drop.loc3.genom3.tret34.gwas3.a = 5:6)

# both rename_all and select_all give the same results:
df %>%
rename_all(~gsub(".*(loc\d+).*(tret\d+).*","\1.\2", .))

df %>%
select_all(~gsub(".*(loc\d+).*(tret\d+).*","\1.\2", .))

# A tibble: 2 x 3
loc1.tret1 loc23.tret2 loc3.tret34
<int> <int> <int>
1 1 3 5
2 2 4 6


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