How to select certain elements from a list?

1

If I have list like this:

 list_data <- list(c("Red", "Green","red"), 

                   c(21,32,11,2,3,6,5,4), 

                   c(1,2,5,4,TRUE, 51.23, 119.1))

I want to keep the last 2 elements before the last one from the list:
desired output:

  > list_data

  [[1]]

  [1] "Red"   "Green" 



  [[2]]

   [1]  6  5  



  [[3]]

  [1]   1.00  51.23 

Any idea on this?

r

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

add a comment | 

1

If I have list like this:

 list_data <- list(c("Red", "Green","red"), 

                   c(21,32,11,2,3,6,5,4), 

                   c(1,2,5,4,TRUE, 51.23, 119.1))

I want to keep the last 2 elements before the last one from the list:
desired output:

  > list_data

  [[1]]

  [1] "Red"   "Green" 



  [[2]]

   [1]  6  5  



  [[3]]

  [1]   1.00  51.23 

Any idea on this?

r

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

add a comment | 

1

1

1

1

If I have list like this:

 list_data <- list(c("Red", "Green","red"), 

                   c(21,32,11,2,3,6,5,4), 

                   c(1,2,5,4,TRUE, 51.23, 119.1))

I want to keep the last 2 elements before the last one from the list:
desired output:

  > list_data

  [[1]]

  [1] "Red"   "Green" 



  [[2]]

   [1]  6  5  



  [[3]]

  [1]   1.00  51.23 

Any idea on this?

r

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

If I have list like this:

 list_data <- list(c("Red", "Green","red"), 

                   c(21,32,11,2,3,6,5,4), 

                   c(1,2,5,4,TRUE, 51.23, 119.1))

I want to keep the last 2 elements before the last one from the list:
desired output:

  > list_data

  [[1]]

  [1] "Red"   "Green" 



  [[2]]

   [1]  6  5  



  [[3]]

  [1]   1.00  51.23 

Any idea on this?

r

r

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

share|improve this question

share|improve this question

edited Nov 20 '18 at 10:35

Sotos

29k51640

edited Nov 20 '18 at 10:35

Sotos

29k51640

edited Nov 20 '18 at 10:35

Sotos

29k51640

29k51640

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

asked Nov 20 '18 at 10:24

bic tonbic ton

343129

343129

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

2

Another option using tail, i.e.

lapply(list_data, function(i)tail(i, 3)[-3])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

add a comment | 

2

You could do

lapply(list_data, function(x) x[(length(x) - 2:1)])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

add a comment | 

1

Solution with Map (thanks to @Sotos for tail):

Map(function(x)tail(x, 3)[-3],list_data) 

# [[1]]

# [1] "Red"   "Green"

# 

# [[2]]

# [1] 6 5

# 

# [[3]]

# [1]  1.00 51.23

If you name each element, you could get a data.frame by using purrr:

names(list_data) <- letters[1:3]

purrr::map_df(list_data, function(x) x[length(x)-2:1]) # ~ .x[length(.x)-(2:1)]



# # A tibble: 2 x 3

# a         b     c

# <chr> <dbl> <dbl>

# 1 Red       6   1  

# 2 Green     5  51.2

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53390906%2fhow-to-select-certain-elements-from-a-list%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

Another option using tail, i.e.

lapply(list_data, function(i)tail(i, 3)[-3])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

add a comment | 

2

Another option using tail, i.e.

lapply(list_data, function(i)tail(i, 3)[-3])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

add a comment | 

2

2

2

Another option using tail, i.e.

lapply(list_data, function(i)tail(i, 3)[-3])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

Another option using tail, i.e.

lapply(list_data, function(i)tail(i, 3)[-3])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

share|improve this answer

share|improve this answer

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

answered Nov 20 '18 at 10:34

SotosSotos

29k51640

29k51640

add a comment | 

add a comment | 

2

You could do

lapply(list_data, function(x) x[(length(x) - 2:1)])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

add a comment | 

2

You could do

lapply(list_data, function(x) x[(length(x) - 2:1)])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

add a comment | 

2

2

2

You could do

lapply(list_data, function(x) x[(length(x) - 2:1)])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

You could do

lapply(list_data, function(x) x[(length(x) - 2:1)])

#[[1]]

#[1] "Red"   "Green"



#[[2]]

#[1] 6 5



#[[3]]

#[1]  1.00 51.23

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

share|improve this answer

share|improve this answer

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

answered Nov 20 '18 at 10:29

markusmarkus

11.1k1031

11.1k1031

add a comment | 

add a comment | 

1

Solution with Map (thanks to @Sotos for tail):

Map(function(x)tail(x, 3)[-3],list_data) 

# [[1]]

# [1] "Red"   "Green"

# 

# [[2]]

# [1] 6 5

# 

# [[3]]

# [1]  1.00 51.23

If you name each element, you could get a data.frame by using purrr:

names(list_data) <- letters[1:3]

purrr::map_df(list_data, function(x) x[length(x)-2:1]) # ~ .x[length(.x)-(2:1)]



# # A tibble: 2 x 3

# a         b     c

# <chr> <dbl> <dbl>

# 1 Red       6   1  

# 2 Green     5  51.2

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

add a comment | 

1

Solution with Map (thanks to @Sotos for tail):

Map(function(x)tail(x, 3)[-3],list_data) 

# [[1]]

# [1] "Red"   "Green"

# 

# [[2]]

# [1] 6 5

# 

# [[3]]

# [1]  1.00 51.23

If you name each element, you could get a data.frame by using purrr:

names(list_data) <- letters[1:3]

purrr::map_df(list_data, function(x) x[length(x)-2:1]) # ~ .x[length(.x)-(2:1)]



# # A tibble: 2 x 3

# a         b     c

# <chr> <dbl> <dbl>

# 1 Red       6   1  

# 2 Green     5  51.2

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

add a comment | 

1

1

1

Solution with Map (thanks to @Sotos for tail):

Map(function(x)tail(x, 3)[-3],list_data) 

# [[1]]

# [1] "Red"   "Green"

# 

# [[2]]

# [1] 6 5

# 

# [[3]]

# [1]  1.00 51.23

If you name each element, you could get a data.frame by using purrr:

names(list_data) <- letters[1:3]

purrr::map_df(list_data, function(x) x[length(x)-2:1]) # ~ .x[length(.x)-(2:1)]



# # A tibble: 2 x 3

# a         b     c

# <chr> <dbl> <dbl>

# 1 Red       6   1  

# 2 Green     5  51.2

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

Solution with Map (thanks to @Sotos for tail):

Map(function(x)tail(x, 3)[-3],list_data) 

# [[1]]

# [1] "Red"   "Green"

# 

# [[2]]

# [1] 6 5

# 

# [[3]]

# [1]  1.00 51.23

If you name each element, you could get a data.frame by using purrr:

names(list_data) <- letters[1:3]

purrr::map_df(list_data, function(x) x[length(x)-2:1]) # ~ .x[length(.x)-(2:1)]



# # A tibble: 2 x 3

# a         b     c

# <chr> <dbl> <dbl>

# 1 Red       6   1  

# 2 Green     5  51.2

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

share|improve this answer

share|improve this answer

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

answered Nov 20 '18 at 10:37

RLaveRLave

3,9951922

3,9951922

add a comment | 

add a comment | 

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53390906%2fhow-to-select-certain-elements-from-a-list%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here