How to splitting a list of vectors to small lists in decreasing order in r











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Suppose I have a function which returns me a list of several vectors. Assume that I would like to split this list to small lists of different numbers of vectors. The number of the vectors in each list is different from one list to another. I need to create these lists in a decreasing order.



Genearal example:



Suppose I have n variables. Then, k = n:2. My function will return me a list of n(n-1)/2 vectors. Then, the number of the new sub-lists is n - 1. Hence, I need to have n-1 different lists. The number of the vectors in each list is J = 1:k-1.



Numerical example:



If n=4, then, k=4, 3, 2, then, J=3,2,1. Hence, my function will return me 6 vectors. These vectors should be stored into different lists. The number of the vectors in each list is based on J. Hence, I will have 3 different lists as follows:




  • 3 vectors in the first list.

  • 2 vectors in the second list.

  • one vector in the last list.


In other words, the returned list (the output of my function) should be split into sub-lists in a decreasing order based on J.



My function is very complicated. Hence, I will provide a list of 6 vectors as the output of my function.



Suppose my function return me the following list:



x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6), x4=c(4,8,4), x5=c(4,33,4), x6=c(9,6,7))


How I can split it into sub-lists as described above? The excepted output is:



x_sub1 <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6))
x_sub2 <- list(x4=c(4,8,4), x5=c(4,33,4))
x_sub3 <- list(x6=c(9,6,7))


I tried this:



x_sub <- list()
for(j in 1:(k-1)){
x_sub[[j]] <- x[[i]]
}


and of course, it is not what I expected.



Any idea, please? How I generate it for an arbitrary number of vectors? for example, how I can apply the split idea over n vectors?



Many thanks for all helps.










share|improve this question
























  • Sorry, it is not clear to me
    – akrun
    Nov 11 at 5:45










  • @MrFlick, thank you so much. This is a typo. I will edit it.
    – Maryam
    Nov 11 at 5:48










  • @akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
    – Maryam
    Nov 12 at 4:48






  • 1




    @akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
    – Maryam
    Nov 13 at 9:28















up vote
2
down vote

favorite












Suppose I have a function which returns me a list of several vectors. Assume that I would like to split this list to small lists of different numbers of vectors. The number of the vectors in each list is different from one list to another. I need to create these lists in a decreasing order.



Genearal example:



Suppose I have n variables. Then, k = n:2. My function will return me a list of n(n-1)/2 vectors. Then, the number of the new sub-lists is n - 1. Hence, I need to have n-1 different lists. The number of the vectors in each list is J = 1:k-1.



Numerical example:



If n=4, then, k=4, 3, 2, then, J=3,2,1. Hence, my function will return me 6 vectors. These vectors should be stored into different lists. The number of the vectors in each list is based on J. Hence, I will have 3 different lists as follows:




  • 3 vectors in the first list.

  • 2 vectors in the second list.

  • one vector in the last list.


In other words, the returned list (the output of my function) should be split into sub-lists in a decreasing order based on J.



My function is very complicated. Hence, I will provide a list of 6 vectors as the output of my function.



Suppose my function return me the following list:



x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6), x4=c(4,8,4), x5=c(4,33,4), x6=c(9,6,7))


How I can split it into sub-lists as described above? The excepted output is:



x_sub1 <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6))
x_sub2 <- list(x4=c(4,8,4), x5=c(4,33,4))
x_sub3 <- list(x6=c(9,6,7))


I tried this:



x_sub <- list()
for(j in 1:(k-1)){
x_sub[[j]] <- x[[i]]
}


and of course, it is not what I expected.



Any idea, please? How I generate it for an arbitrary number of vectors? for example, how I can apply the split idea over n vectors?



Many thanks for all helps.










share|improve this question
























  • Sorry, it is not clear to me
    – akrun
    Nov 11 at 5:45










  • @MrFlick, thank you so much. This is a typo. I will edit it.
    – Maryam
    Nov 11 at 5:48










  • @akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
    – Maryam
    Nov 12 at 4:48






  • 1




    @akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
    – Maryam
    Nov 13 at 9:28













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Suppose I have a function which returns me a list of several vectors. Assume that I would like to split this list to small lists of different numbers of vectors. The number of the vectors in each list is different from one list to another. I need to create these lists in a decreasing order.



Genearal example:



Suppose I have n variables. Then, k = n:2. My function will return me a list of n(n-1)/2 vectors. Then, the number of the new sub-lists is n - 1. Hence, I need to have n-1 different lists. The number of the vectors in each list is J = 1:k-1.



Numerical example:



If n=4, then, k=4, 3, 2, then, J=3,2,1. Hence, my function will return me 6 vectors. These vectors should be stored into different lists. The number of the vectors in each list is based on J. Hence, I will have 3 different lists as follows:




  • 3 vectors in the first list.

  • 2 vectors in the second list.

  • one vector in the last list.


In other words, the returned list (the output of my function) should be split into sub-lists in a decreasing order based on J.



My function is very complicated. Hence, I will provide a list of 6 vectors as the output of my function.



Suppose my function return me the following list:



x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6), x4=c(4,8,4), x5=c(4,33,4), x6=c(9,6,7))


How I can split it into sub-lists as described above? The excepted output is:



x_sub1 <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6))
x_sub2 <- list(x4=c(4,8,4), x5=c(4,33,4))
x_sub3 <- list(x6=c(9,6,7))


I tried this:



x_sub <- list()
for(j in 1:(k-1)){
x_sub[[j]] <- x[[i]]
}


and of course, it is not what I expected.



Any idea, please? How I generate it for an arbitrary number of vectors? for example, how I can apply the split idea over n vectors?



Many thanks for all helps.










share|improve this question















Suppose I have a function which returns me a list of several vectors. Assume that I would like to split this list to small lists of different numbers of vectors. The number of the vectors in each list is different from one list to another. I need to create these lists in a decreasing order.



Genearal example:



Suppose I have n variables. Then, k = n:2. My function will return me a list of n(n-1)/2 vectors. Then, the number of the new sub-lists is n - 1. Hence, I need to have n-1 different lists. The number of the vectors in each list is J = 1:k-1.



Numerical example:



If n=4, then, k=4, 3, 2, then, J=3,2,1. Hence, my function will return me 6 vectors. These vectors should be stored into different lists. The number of the vectors in each list is based on J. Hence, I will have 3 different lists as follows:




  • 3 vectors in the first list.

  • 2 vectors in the second list.

  • one vector in the last list.


In other words, the returned list (the output of my function) should be split into sub-lists in a decreasing order based on J.



My function is very complicated. Hence, I will provide a list of 6 vectors as the output of my function.



Suppose my function return me the following list:



x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6), x4=c(4,8,4), x5=c(4,33,4), x6=c(9,6,7))


How I can split it into sub-lists as described above? The excepted output is:



x_sub1 <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6))
x_sub2 <- list(x4=c(4,8,4), x5=c(4,33,4))
x_sub3 <- list(x6=c(9,6,7))


I tried this:



x_sub <- list()
for(j in 1:(k-1)){
x_sub[[j]] <- x[[i]]
}


and of course, it is not what I expected.



Any idea, please? How I generate it for an arbitrary number of vectors? for example, how I can apply the split idea over n vectors?



Many thanks for all helps.







r






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 11 at 5:47

























asked Nov 11 at 5:33









Maryam

18211




18211












  • Sorry, it is not clear to me
    – akrun
    Nov 11 at 5:45










  • @MrFlick, thank you so much. This is a typo. I will edit it.
    – Maryam
    Nov 11 at 5:48










  • @akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
    – Maryam
    Nov 12 at 4:48






  • 1




    @akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
    – Maryam
    Nov 13 at 9:28


















  • Sorry, it is not clear to me
    – akrun
    Nov 11 at 5:45










  • @MrFlick, thank you so much. This is a typo. I will edit it.
    – Maryam
    Nov 11 at 5:48










  • @akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
    – Maryam
    Nov 12 at 4:48






  • 1




    @akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
    – Maryam
    Nov 13 at 9:28
















Sorry, it is not clear to me
– akrun
Nov 11 at 5:45




Sorry, it is not clear to me
– akrun
Nov 11 at 5:45












@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 at 5:48




@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 at 5:48












@akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
– Maryam
Nov 12 at 4:48




@akrun Thank you so much for your comment and answer. I just meant how I can generate it to the n number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
– Maryam
Nov 12 at 4:48




1




1




@akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
– Maryam
Nov 13 at 9:28




@akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
– Maryam
Nov 13 at 9:28












2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










We can use rep to split the list into a list of lists



lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))


and extract the nested list with the name



lst[["x_sub1"]]


It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env



list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3

#$x2
#[1] 1 4 3

#$x3
#[1] 3 4 6





share|improve this answer





















  • Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
    – Maryam
    Nov 11 at 5:42










  • @Maryam In the question, the input is a list 'x'
    – akrun
    Nov 11 at 5:43


















up vote
1
down vote













x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))


You can try the function split() to partition your list x by certain format. I don't understand what you mean for n, k, J, so you need to define them by yourself. I just simply set J as 1:3.



J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)

x_sub1
x_sub2
x_sub3





share|improve this answer





















  • Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
    – Maryam
    Nov 11 at 8:42











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










We can use rep to split the list into a list of lists



lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))


and extract the nested list with the name



lst[["x_sub1"]]


It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env



list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3

#$x2
#[1] 1 4 3

#$x3
#[1] 3 4 6





share|improve this answer





















  • Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
    – Maryam
    Nov 11 at 5:42










  • @Maryam In the question, the input is a list 'x'
    – akrun
    Nov 11 at 5:43















up vote
0
down vote



accepted










We can use rep to split the list into a list of lists



lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))


and extract the nested list with the name



lst[["x_sub1"]]


It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env



list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3

#$x2
#[1] 1 4 3

#$x3
#[1] 3 4 6





share|improve this answer





















  • Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
    – Maryam
    Nov 11 at 5:42










  • @Maryam In the question, the input is a list 'x'
    – akrun
    Nov 11 at 5:43













up vote
0
down vote



accepted







up vote
0
down vote



accepted






We can use rep to split the list into a list of lists



lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))


and extract the nested list with the name



lst[["x_sub1"]]


It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env



list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3

#$x2
#[1] 1 4 3

#$x3
#[1] 3 4 6





share|improve this answer












We can use rep to split the list into a list of lists



lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))


and extract the nested list with the name



lst[["x_sub1"]]


It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env



list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3

#$x2
#[1] 1 4 3

#$x3
#[1] 3 4 6






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 11 at 5:35









akrun

392k13181253




392k13181253












  • Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
    – Maryam
    Nov 11 at 5:42










  • @Maryam In the question, the input is a list 'x'
    – akrun
    Nov 11 at 5:43


















  • Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
    – Maryam
    Nov 11 at 5:42










  • @Maryam In the question, the input is a list 'x'
    – akrun
    Nov 11 at 5:43
















Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
– Maryam
Nov 11 at 5:42




Thank you so much for your answer. So what about for n vectors. Can I use lst <- split(x, rep(paste0("x_sub", 1:n), n:1)). Also, I have edited my question, if you would like to have a look. Thank you again.
– Maryam
Nov 11 at 5:42












@Maryam In the question, the input is a list 'x'
– akrun
Nov 11 at 5:43




@Maryam In the question, the input is a list 'x'
– akrun
Nov 11 at 5:43












up vote
1
down vote













x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))


You can try the function split() to partition your list x by certain format. I don't understand what you mean for n, k, J, so you need to define them by yourself. I just simply set J as 1:3.



J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)

x_sub1
x_sub2
x_sub3





share|improve this answer





















  • Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
    – Maryam
    Nov 11 at 8:42















up vote
1
down vote













x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))


You can try the function split() to partition your list x by certain format. I don't understand what you mean for n, k, J, so you need to define them by yourself. I just simply set J as 1:3.



J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)

x_sub1
x_sub2
x_sub3





share|improve this answer





















  • Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
    – Maryam
    Nov 11 at 8:42













up vote
1
down vote










up vote
1
down vote









x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))


You can try the function split() to partition your list x by certain format. I don't understand what you mean for n, k, J, so you need to define them by yourself. I just simply set J as 1:3.



J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)

x_sub1
x_sub2
x_sub3





share|improve this answer












x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))


You can try the function split() to partition your list x by certain format. I don't understand what you mean for n, k, J, so you need to define them by yourself. I just simply set J as 1:3.



J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)

x_sub1
x_sub2
x_sub3






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 11 at 7:12









Darren Tsai

776116




776116












  • Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
    – Maryam
    Nov 11 at 8:42


















  • Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
    – Maryam
    Nov 11 at 8:42
















Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 at 8:42




Thank you so much for your answer. n,k, and J are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 at 8:42


















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