How to select certain elements from a list?

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?

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

add a comment |

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?

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

add a comment |

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?

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

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?

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

edited Nov 20 '18 at 10:35

Sotos

30k51640

edited Nov 20 '18 at 10:35

Sotos

30k51640

edited Nov 20 '18 at 10:35

Sotos

30k51640

asked Nov 20 '18 at 10:24

bic ton

353129

asked Nov 20 '18 at 10:24

bic ton

353129

asked Nov 20 '18 at 10:24

bic ton

353129

add a comment |

3 Answers
3

active

oldest

votes

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

answered Nov 20 '18 at 10:34

Sotos

30k51640

add a comment |

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

answered Nov 20 '18 at 10:29

markus

13.1k1234

add a comment |

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

answered Nov 20 '18 at 10:37

RLave

4,75711124

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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

answered Nov 20 '18 at 10:34

Sotos

30k51640

add a comment |

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

answered Nov 20 '18 at 10:34

Sotos

30k51640

add a comment |

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

answered Nov 20 '18 at 10:34

Sotos

30k51640

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

answered Nov 20 '18 at 10:34

Sotos

30k51640

answered Nov 20 '18 at 10:34

Sotos

30k51640

answered Nov 20 '18 at 10:34

Sotos

30k51640

answered Nov 20 '18 at 10:34

Sotos

30k51640

add a comment |

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

answered Nov 20 '18 at 10:29

markus

13.1k1234

add a comment |

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

answered Nov 20 '18 at 10:29

markus

13.1k1234

add a comment |

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

answered Nov 20 '18 at 10:29

markus

13.1k1234

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

answered Nov 20 '18 at 10:29

markus

13.1k1234

answered Nov 20 '18 at 10:29

markus

13.1k1234

answered Nov 20 '18 at 10:29

markus

13.1k1234

answered Nov 20 '18 at 10:29

markus

13.1k1234

add a comment |

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

answered Nov 20 '18 at 10:37

RLave

4,75711124

add a comment |

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

answered Nov 20 '18 at 10:37

RLave

4,75711124

add a comment |

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

answered Nov 20 '18 at 10:37

RLave

4,75711124

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

answered Nov 20 '18 at 10:37

RLave

4,75711124

answered Nov 20 '18 at 10:37

RLave

4,75711124

answered Nov 20 '18 at 10:37

RLave

4,75711124

answered Nov 20 '18 at 10:37

RLave

4,75711124

add a comment |

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