How to naming list elements using only lapply in R
1
This is my list:
l <- vector("list", 4)
l[[1]][1] <- list(c(1,2,3))
l[[1]][2] <- list(c(1,2,3))
l[[2]][1] <- list(c(1,2,3))
l[[2]][2] <- list(c(1,2,3))
l[[3]][1] <- list(c(1,2,3))
l[[3]][2] <- list(c(1,2,3))
l[[4]][1] <- list(c(1,2,3))
l[[4]][2] <- list(c(1,2,3))
I have this names vectors: mynames <- c("number one","number two")
How can I name the list elements using the lapply function with the mynames vector?
I tried this, but didn't work:
lapply(l, names(x) <- mynames)
Any help?
r apply lapply
edited Nov 12 '18 at 23:10
Florian
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
add a comment |
1
This is my list:
l <- vector("list", 4)
l[[1]][1] <- list(c(1,2,3))
l[[1]][2] <- list(c(1,2,3))
l[[2]][1] <- list(c(1,2,3))
l[[2]][2] <- list(c(1,2,3))
l[[3]][1] <- list(c(1,2,3))
l[[3]][2] <- list(c(1,2,3))
l[[4]][1] <- list(c(1,2,3))
l[[4]][2] <- list(c(1,2,3))
I have this names vectors: mynames <- c("number one","number two")
How can I name the list elements using the lapply function with the mynames vector?
I tried this, but didn't work:
lapply(l, names(x) <- mynames)
Any help?
r apply lapply
edited Nov 12 '18 at 23:10
Florian
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
1
You can use the functional form of a "special" function, something likelapply(l, `names<-`, mynames)(which is equivalent tolapply(l, function(a) `names<-`(a, mynames))which is equivalent tolapply(l, function(a) { names(a) <- mynames; a; })).
– r2evans
Nov 12 '18 at 23:05
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read?
– Laura
Nov 12 '18 at 23:11
1
When you do something likenames(x) <- c('a','b','c'), there is a function callednameswhich only returns the names of the objectx, and there is a function callednames<-(no space) that is called whennames(x)is on the left side of an assignment operator (<-or=). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to usesetNamesis simpler and just as good.
– r2evans
Nov 12 '18 at 23:29
add a comment |
1
1
1
1
This is my list:
l <- vector("list", 4)
l[[1]][1] <- list(c(1,2,3))
l[[1]][2] <- list(c(1,2,3))
l[[2]][1] <- list(c(1,2,3))
l[[2]][2] <- list(c(1,2,3))
l[[3]][1] <- list(c(1,2,3))
l[[3]][2] <- list(c(1,2,3))
l[[4]][1] <- list(c(1,2,3))
l[[4]][2] <- list(c(1,2,3))
I have this names vectors: mynames <- c("number one","number two")
How can I name the list elements using the lapply function with the mynames vector?
I tried this, but didn't work:
lapply(l, names(x) <- mynames)
Any help?
r apply lapply
edited Nov 12 '18 at 23:10
Florian
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
This is my list:
l <- vector("list", 4)
l[[1]][1] <- list(c(1,2,3))
l[[1]][2] <- list(c(1,2,3))
l[[2]][1] <- list(c(1,2,3))
l[[2]][2] <- list(c(1,2,3))
l[[3]][1] <- list(c(1,2,3))
l[[3]][2] <- list(c(1,2,3))
l[[4]][1] <- list(c(1,2,3))
l[[4]][2] <- list(c(1,2,3))
I have this names vectors: mynames <- c("number one","number two")
How can I name the list elements using the lapply function with the mynames vector?
I tried this, but didn't work:
lapply(l, names(x) <- mynames)
Any help?
r apply lapply
r apply lapply
edited Nov 12 '18 at 23:10
Florian
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
edited Nov 12 '18 at 23:10
Florian
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
edited Nov 12 '18 at 23:10
Florian
1,072817
edited Nov 12 '18 at 23:10
Florian
1,072817
edited Nov 12 '18 at 23:10
Florian
1,072817
1,072817
asked Nov 12 '18 at 22:56
LauraLaura
35319
asked Nov 12 '18 at 22:56
LauraLaura
35319
asked Nov 12 '18 at 22:56
LauraLaura
35319
35319
1
You can use the functional form of a "special" function, something likelapply(l, `names<-`, mynames)(which is equivalent tolapply(l, function(a) `names<-`(a, mynames))which is equivalent tolapply(l, function(a) { names(a) <- mynames; a; })).
– r2evans
Nov 12 '18 at 23:05
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read?
– Laura
Nov 12 '18 at 23:11
1
When you do something likenames(x) <- c('a','b','c'), there is a function callednameswhich only returns the names of the objectx, and there is a function callednames<-(no space) that is called whennames(x)is on the left side of an assignment operator (<-or=). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to usesetNamesis simpler and just as good.
– r2evans
Nov 12 '18 at 23:29
add a comment |
1
You can use the functional form of a "special" function, something likelapply(l, `names<-`, mynames)(which is equivalent tolapply(l, function(a) `names<-`(a, mynames))which is equivalent tolapply(l, function(a) { names(a) <- mynames; a; })).
– r2evans
Nov 12 '18 at 23:05
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read?
– Laura
Nov 12 '18 at 23:11
1
When you do something likenames(x) <- c('a','b','c'), there is a function callednameswhich only returns the names of the objectx, and there is a function callednames<-(no space) that is called whennames(x)is on the left side of an assignment operator (<-or=). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to usesetNamesis simpler and just as good.
– r2evans
Nov 12 '18 at 23:29
1
1
You can use the functional form of a "special" function, something like
lapply(l, `names<-`, mynames) (which is equivalent to lapply(l, function(a) `names<-`(a, mynames)) which is equivalent to lapply(l, function(a) { names(a) <- mynames; a; })).– r2evans
Nov 12 '18 at 23:05
You can use the functional form of a "special" function, something like
lapply(l, `names<-`, mynames) (which is equivalent to lapply(l, function(a) `names<-`(a, mynames)) which is equivalent to lapply(l, function(a) { names(a) <- mynames; a; })).– r2evans
Nov 12 '18 at 23:05
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such as
names<-? I had no idea that I could use this in lapply. Any suggestion to read?– Laura
Nov 12 '18 at 23:11
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such as
names<-? I had no idea that I could use this in lapply. Any suggestion to read?– Laura
Nov 12 '18 at 23:11
1
1
When you do something like
names(x) <- c('a','b','c'), there is a function called names which only returns the names of the object x, and there is a function called names<- (no space) that is called when names(x) is on the left side of an assignment operator (<- or =). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to use setNames is simpler and just as good.– r2evans
Nov 12 '18 at 23:29
When you do something like
names(x) <- c('a','b','c'), there is a function called names which only returns the names of the object x, and there is a function called names<- (no space) that is called when names(x) is on the left side of an assignment operator (<- or =). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to use setNames is simpler and just as good.– r2evans
Nov 12 '18 at 23:29
add a comment |
1 Answer
1
active
oldest
votes
2
The second argument of lappyl() has to be a function. One can use setNames():
named_list <- lapply(l, setNames, nm=mynames)
named_list[1:2]
[[1]]
[[1]]$`number one`
[1] 1 2 3
[[1]]$`number two`
[1] 1 2 3
[[2]]
[[2]]$`number one`
[1] 1 2 3
[[2]]$`number two`
[1] 1 2 3
An alternative version based on the replacement function `names<-` is:
named_list2 <- lapply(l, function(x, names) { names(x) <- names; x },
names=mynames)
identical(named_list, named_list2)
[1] TRUE
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps returnafrom the inner function?
– r2evans
Nov 12 '18 at 23:06
1
Nice preservation of scope by adding thenames=argument.
– r2evans
Nov 12 '18 at 23:07
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintaxnames<-has any name?
– Laura
Nov 12 '18 at 23:15
2
Consider even the left hand function,setNames:lapply(l, function(x) setNames(x, mynames))
– Parfait
Nov 12 '18 at 23:20
|
show 2 more comments
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
2
The second argument of lappyl() has to be a function. One can use setNames():
named_list <- lapply(l, setNames, nm=mynames)
named_list[1:2]
[[1]]
[[1]]$`number one`
[1] 1 2 3
[[1]]$`number two`
[1] 1 2 3
[[2]]
[[2]]$`number one`
[1] 1 2 3
[[2]]$`number two`
[1] 1 2 3
An alternative version based on the replacement function `names<-` is:
named_list2 <- lapply(l, function(x, names) { names(x) <- names; x },
names=mynames)
identical(named_list, named_list2)
[1] TRUE
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps returnafrom the inner function?
– r2evans
Nov 12 '18 at 23:06
1
Nice preservation of scope by adding thenames=argument.
– r2evans
Nov 12 '18 at 23:07
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintaxnames<-has any name?
– Laura
Nov 12 '18 at 23:15
2
Consider even the left hand function,setNames:lapply(l, function(x) setNames(x, mynames))
– Parfait
Nov 12 '18 at 23:20
|
show 2 more comments
2
The second argument of lappyl() has to be a function. One can use setNames():
named_list <- lapply(l, setNames, nm=mynames)
named_list[1:2]
[[1]]
[[1]]$`number one`
[1] 1 2 3
[[1]]$`number two`
[1] 1 2 3
[[2]]
[[2]]$`number one`
[1] 1 2 3
[[2]]$`number two`
[1] 1 2 3
An alternative version based on the replacement function `names<-` is:
named_list2 <- lapply(l, function(x, names) { names(x) <- names; x },
names=mynames)
identical(named_list, named_list2)
[1] TRUE
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps returnafrom the inner function?
– r2evans
Nov 12 '18 at 23:06
1
Nice preservation of scope by adding thenames=argument.
– r2evans
Nov 12 '18 at 23:07
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintaxnames<-has any name?
– Laura
Nov 12 '18 at 23:15
2
Consider even the left hand function,setNames:lapply(l, function(x) setNames(x, mynames))
– Parfait
Nov 12 '18 at 23:20
|
show 2 more comments
2
2
2
The second argument of lappyl() has to be a function. One can use setNames():
named_list <- lapply(l, setNames, nm=mynames)
named_list[1:2]
[[1]]
[[1]]$`number one`
[1] 1 2 3
[[1]]$`number two`
[1] 1 2 3
[[2]]
[[2]]$`number one`
[1] 1 2 3
[[2]]$`number two`
[1] 1 2 3
An alternative version based on the replacement function `names<-` is:
named_list2 <- lapply(l, function(x, names) { names(x) <- names; x },
names=mynames)
identical(named_list, named_list2)
[1] TRUE
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
The second argument of lappyl() has to be a function. One can use setNames():
named_list <- lapply(l, setNames, nm=mynames)
named_list[1:2]
[[1]]
[[1]]$`number one`
[1] 1 2 3
[[1]]$`number two`
[1] 1 2 3
[[2]]
[[2]]$`number one`
[1] 1 2 3
[[2]]$`number two`
[1] 1 2 3
An alternative version based on the replacement function `names<-` is:
named_list2 <- lapply(l, function(x, names) { names(x) <- names; x },
names=mynames)
identical(named_list, named_list2)
[1] TRUE
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
edited Nov 13 '18 at 3:31
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
answered Nov 12 '18 at 23:05
FlorianFlorian
1,072817
1,072817
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps returnafrom the inner function?
– r2evans
Nov 12 '18 at 23:06
1
Nice preservation of scope by adding thenames=argument.
– r2evans
Nov 12 '18 at 23:07
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintaxnames<-has any name?
– Laura
Nov 12 '18 at 23:15
2
Consider even the left hand function,setNames:lapply(l, function(x) setNames(x, mynames))
– Parfait
Nov 12 '18 at 23:20
|
show 2 more comments
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps returnafrom the inner function?
– r2evans
Nov 12 '18 at 23:06
1
Nice preservation of scope by adding thenames=argument.
– r2evans
Nov 12 '18 at 23:07
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such asnames<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintaxnames<-has any name?
– Laura
Nov 12 '18 at 23:15
2
Consider even the left hand function,setNames:lapply(l, function(x) setNames(x, mynames))
– Parfait
Nov 12 '18 at 23:20
1
1
I assumed that Laura would prefer to not discard the data when naming it. Perhaps return
a from the inner function?– r2evans
Nov 12 '18 at 23:06
I assumed that Laura would prefer to not discard the data when naming it. Perhaps return
a from the inner function?– r2evans
Nov 12 '18 at 23:06
1
1
Nice preservation of scope by adding the
names= argument.– r2evans
Nov 12 '18 at 23:07
Nice preservation of scope by adding the
names= argument.– r2evans
Nov 12 '18 at 23:07
1
1
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@r2evans thanks for the comment. I updated the answer.
– Florian
Nov 12 '18 at 23:14
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such as
names<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintax names<- has any name?– Laura
Nov 12 '18 at 23:15
@Florian many thanks. @r2evans many thanks. As I said, I am a begginer in lapply. How can I understand more this sintax such as
names<-? I had no idea that I could use this in lapply. Any suggestion to read? This sintax names<- has any name?– Laura
Nov 12 '18 at 23:15
2
2
Consider even the left hand function,
setNames: lapply(l, function(x) setNames(x, mynames))– Parfait
Nov 12 '18 at 23:20
Consider even the left hand function,
setNames: lapply(l, function(x) setNames(x, mynames))– Parfait
Nov 12 '18 at 23:20
|
show 2 more comments
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1
You can use the functional form of a "special" function, something like
lapply(l, `names<-`, mynames)(which is equivalent tolapply(l, function(a) `names<-`(a, mynames))which is equivalent tolapply(l, function(a) { names(a) <- mynames; a; })).– r2evans
Nov 12 '18 at 23:05
@r2evans many thanks. This is amazing. I am a begginer in lapply. How can I understand more this sintax such as
names<-? I had no idea that I could use this in lapply. Any suggestion to read?– Laura
Nov 12 '18 at 23:11
1
When you do something like
names(x) <- c('a','b','c'), there is a function callednameswhich only returns the names of the objectx, and there is a function callednames<-(no space) that is called whennames(x)is on the left side of an assignment operator (<-or=). Most functions do not have this assignment equivalent; more the point, it does not make sense for most functions to do something on the assignment-side (left-hand side). There are some other minor differences, not really notable here. BTW: I think @Parfait's suggestion to usesetNamesis simpler and just as good.– r2evans
Nov 12 '18 at 23:29