How to naming list elements using only lapply in R

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?

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

35319

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

@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 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 |

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?

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

35319

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

@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 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 |

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?

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

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

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

35319

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

35319

edited Nov 12 '18 at 23:10

Florian

1,092817

edited Nov 12 '18 at 23:10

Florian

1,092817

edited Nov 12 '18 at 23:10

Florian

1,092817

asked Nov 12 '18 at 22:56

Laura

35319

asked Nov 12 '18 at 22:56

Laura

35319

asked Nov 12 '18 at 22:56

Laura

35319

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

@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 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

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

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

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

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

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

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

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

1

Nice preservation of scope by adding the names= 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 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

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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1 Answer
1

active

oldest

votes

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

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

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

1

Nice preservation of scope by adding the names= 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 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

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

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

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

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

1

Nice preservation of scope by adding the names= 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 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

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

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

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

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

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

edited Nov 13 '18 at 3:31

answered Nov 12 '18 at 23:05

Florian

1,092817

answered Nov 12 '18 at 23:05

Florian

1,092817

answered Nov 12 '18 at 23:05

Florian

1,092817

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

1

Nice preservation of scope by adding the names= 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 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

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 return a from the inner function?

– r2evans
Nov 12 '18 at 23:06

1

Nice preservation of scope by adding the names= 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 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

Consider even the left hand function, setNames: lapply(l, function(x) setNames(x, mynames))

– Parfait
Nov 12 '18 at 23:20

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

Nice preservation of scope by adding the names= argument.

– r2evans
Nov 12 '18 at 23:07

@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

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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