Replace last comma in character with “ &”
12
I have many different characters which have the following structure:
# Example
x <- "char1, char2, char3"
I want to remove the last comma of this character with " &", i.e. the desired output should look as follows:
# Desired output
"char1, char2 & char3"
How could I replace the last comma of a character with " &"?
r replace character
asked Feb 7 at 16:11
JSPJSP
75611020
add a comment |
12
I have many different characters which have the following structure:
# Example
x <- "char1, char2, char3"
I want to remove the last comma of this character with " &", i.e. the desired output should look as follows:
# Desired output
"char1, char2 & char3"
How could I replace the last comma of a character with " &"?
r replace character
asked Feb 7 at 16:11
JSPJSP
75611020
1
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
2
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
1
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32
add a comment |
12
12
12
I have many different characters which have the following structure:
# Example
x <- "char1, char2, char3"
I want to remove the last comma of this character with " &", i.e. the desired output should look as follows:
# Desired output
"char1, char2 & char3"
How could I replace the last comma of a character with " &"?
r replace character
asked Feb 7 at 16:11
JSPJSP
75611020
I have many different characters which have the following structure:
# Example
x <- "char1, char2, char3"
I want to remove the last comma of this character with " &", i.e. the desired output should look as follows:
# Desired output
"char1, char2 & char3"
How could I replace the last comma of a character with " &"?
r replace character
r replace character
asked Feb 7 at 16:11
JSPJSP
75611020
asked Feb 7 at 16:11
JSPJSP
75611020
asked Feb 7 at 16:11
JSPJSP
75611020
asked Feb 7 at 16:11
JSPJSP
75611020
asked Feb 7 at 16:11
JSPJSP
75611020
75611020
1
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
2
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
1
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32
add a comment |
1
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
2
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
1
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32
1
1
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
2
2
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
1
1
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32
add a comment |
4 Answers
4
active
oldest
votes
13
One option is stri_replace_last from stringi
library(stringi)
stri_replace_last(x, fixed = ',', ' &')
#[1] "char1, char2 & char3"
answered Feb 7 at 16:13
akrunakrun
418k13206281
add a comment |
8
You can use sub :
sub(",([^,]*)$"," &\1", x)
# [1] "char1, char2 & char3"
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
add a comment |
3
You could split and unsplit it.
u <- unlist(strsplit(x, ""))
u[tail(grep(",", u), 1)] <- " &"
paste0(u, collapse="")
# [1] "char1, char2 & char3"
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
add a comment |
3
We can also use str_locate_all with str_sub:
library(stringr)
pos <- str_locate_all(x, ',')[[1]][2, ]
str_sub(x, pos[1], pos[2]) <- " &"
# [1] "char1, char2 & char3"
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
13
One option is stri_replace_last from stringi
library(stringi)
stri_replace_last(x, fixed = ',', ' &')
#[1] "char1, char2 & char3"
answered Feb 7 at 16:13
akrunakrun
418k13206281
add a comment |
13
One option is stri_replace_last from stringi
library(stringi)
stri_replace_last(x, fixed = ',', ' &')
#[1] "char1, char2 & char3"
answered Feb 7 at 16:13
akrunakrun
418k13206281
add a comment |
13
13
13
One option is stri_replace_last from stringi
library(stringi)
stri_replace_last(x, fixed = ',', ' &')
#[1] "char1, char2 & char3"
answered Feb 7 at 16:13
akrunakrun
418k13206281
One option is stri_replace_last from stringi
library(stringi)
stri_replace_last(x, fixed = ',', ' &')
#[1] "char1, char2 & char3"
answered Feb 7 at 16:13
akrunakrun
418k13206281
answered Feb 7 at 16:13
akrunakrun
418k13206281
answered Feb 7 at 16:13
akrunakrun
418k13206281
answered Feb 7 at 16:13
akrunakrun
418k13206281
418k13206281
add a comment |
add a comment |
8
You can use sub :
sub(",([^,]*)$"," &\1", x)
# [1] "char1, char2 & char3"
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
add a comment |
8
You can use sub :
sub(",([^,]*)$"," &\1", x)
# [1] "char1, char2 & char3"
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
add a comment |
8
8
8
You can use sub :
sub(",([^,]*)$"," &\1", x)
# [1] "char1, char2 & char3"
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
You can use sub :
sub(",([^,]*)$"," &\1", x)
# [1] "char1, char2 & char3"
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
answered Feb 7 at 16:29
Moody_MudskipperMoody_Mudskipper
24.4k33568
24.4k33568
add a comment |
add a comment |
3
You could split and unsplit it.
u <- unlist(strsplit(x, ""))
u[tail(grep(",", u), 1)] <- " &"
paste0(u, collapse="")
# [1] "char1, char2 & char3"
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
add a comment |
3
You could split and unsplit it.
u <- unlist(strsplit(x, ""))
u[tail(grep(",", u), 1)] <- " &"
paste0(u, collapse="")
# [1] "char1, char2 & char3"
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
add a comment |
3
3
3
You could split and unsplit it.
u <- unlist(strsplit(x, ""))
u[tail(grep(",", u), 1)] <- " &"
paste0(u, collapse="")
# [1] "char1, char2 & char3"
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
You could split and unsplit it.
u <- unlist(strsplit(x, ""))
u[tail(grep(",", u), 1)] <- " &"
paste0(u, collapse="")
# [1] "char1, char2 & char3"
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
answered Feb 7 at 16:22
jay.sfjay.sf
6,56231841
6,56231841
add a comment |
add a comment |
3
We can also use str_locate_all with str_sub:
library(stringr)
pos <- str_locate_all(x, ',')[[1]][2, ]
str_sub(x, pos[1], pos[2]) <- " &"
# [1] "char1, char2 & char3"
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
add a comment |
3
We can also use str_locate_all with str_sub:
library(stringr)
pos <- str_locate_all(x, ',')[[1]][2, ]
str_sub(x, pos[1], pos[2]) <- " &"
# [1] "char1, char2 & char3"
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
add a comment |
3
3
3
We can also use str_locate_all with str_sub:
library(stringr)
pos <- str_locate_all(x, ',')[[1]][2, ]
str_sub(x, pos[1], pos[2]) <- " &"
# [1] "char1, char2 & char3"
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
We can also use str_locate_all with str_sub:
library(stringr)
pos <- str_locate_all(x, ',')[[1]][2, ]
str_sub(x, pos[1], pos[2]) <- " &"
# [1] "char1, char2 & char3"
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
answered Feb 7 at 16:32
avid_useRavid_useR
13.3k41933
13.3k41933
add a comment |
add a comment |
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1
It seems like there is much interest in this topic. Therefore, I have written an article which summarizes the responses of this thread: statistical-programming.com/… Thanks again for all the great responses!
– JSP
Feb 7 at 21:28
2
this is a string not a character, right?
– Will Barnwell
Feb 7 at 23:06
1
Ah, no, r calls strings characters. Confusing
– Will Barnwell
Feb 7 at 23:32