How to splitting a list of vectors to small lists in decreasing order in r
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
add a comment |
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
Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 at 5:48
@akrun Thank you so much for your comment and answer. I just meant how I can generate it to then
number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
– Maryam
Nov 12 '18 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 '18 at 9:28
add a comment |
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
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
r
edited Nov 11 '18 at 5:47
Maryam
asked Nov 11 '18 at 5:33
MaryamMaryam
18211
18211
Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 at 5:48
@akrun Thank you so much for your comment and answer. I just meant how I can generate it to then
number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
– Maryam
Nov 12 '18 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 '18 at 9:28
add a comment |
Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 at 5:48
@akrun Thank you so much for your comment and answer. I just meant how I can generate it to then
number of vectors. Then, I found that you delete your answer. I am really so sorry if I misunderstand you.
– Maryam
Nov 12 '18 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 '18 at 9:28
Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 at 5:48
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 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 '18 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 '18 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 '18 at 9:28
@akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
– Maryam
Nov 13 '18 at 9:28
add a comment |
2 Answers
2
active
oldest
votes
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
#$x3
#[1] 3 4 6
Thank you so much for your answer. So what about forn
vectors. Can I uselst <- split(x, rep(paste0("x_sub", 1:n), n:1))
. Also, I have edited my question, if you would like to have a look. Thank you again.
– Maryam
Nov 11 '18 at 5:42
@Maryam In the question, the input is alist
'x'
– akrun
Nov 11 '18 at 5:43
add a comment |
x <- list(x1=c(1,2,3), x2=c(1,4,3), x3=c(3,4,6),
x4=c(4,8,4), x5=c(4,33,4),
x6=c(9,6,7))
You can try the function split()
to partition your list x
by certain format. I don't understand what you mean for n
, k
, J
, so you need to define them by yourself. I just simply set J
as 1:3
.
J <- 1:3
breaks <- rep(J, rev(J)) # [1] 1 1 1 2 2 3
y <- split(x, f = breaks)
names(y) <- paste0("x_sub", J)
list2env(y, envir = .GlobalEnv)
x_sub1
x_sub2
x_sub3
Thank you so much for your answer.n
,k
, andJ
are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 '18 at 8:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We can use rep
to split
the list
into a list
of lists
lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))
and extract the nested list
with the name
lst[["x_sub1"]]
It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env
list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3
#$x2
#[1] 1 4 3
#$x3
#[1] 3 4 6
Thank you so much for your answer. So what about forn
vectors. Can I uselst <- 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 '18 at 5:42
@Maryam In the question, the input is alist
'x'
– akrun
Nov 11 '18 at 5:43
add a comment |
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
#$x3
#[1] 3 4 6
Thank you so much for your answer. So what about forn
vectors. Can I uselst <- split(x, rep(paste0("x_sub", 1:n), n:1))
. Also, I have edited my question, if you would like to have a look. Thank you again.
– Maryam
Nov 11 '18 at 5:42
@Maryam In the question, the input is alist
'x'
– akrun
Nov 11 '18 at 5:43
add a comment |
We can use rep
to split
the list
into a list
of lists
lst <- split(x, rep(paste0("x_sub", 1:3), 3:1))
and extract the nested list
with the name
lst[["x_sub1"]]
It is better not to create multiple objects in the global environment. But, if it is needed, then use list2env
list2env(lst, envir = .GlobalEnv)
x_sub1
#$x1
#[1] 1 2 3
#$x2
#[1] 1 4 3
#$x3
#[1] 3 4 6
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
answered Nov 11 '18 at 5:35
akrunakrun
402k13194266
402k13194266
Thank you so much for your answer. So what about forn
vectors. Can I uselst <- 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 '18 at 5:42
@Maryam In the question, the input is alist
'x'
– akrun
Nov 11 '18 at 5:43
add a comment |
Thank you so much for your answer. So what about forn
vectors. Can I uselst <- 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 '18 at 5:42
@Maryam In the question, the input is alist
'x'
– akrun
Nov 11 '18 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 '18 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 '18 at 5:42
@Maryam In the question, the input is a
list
'x'– akrun
Nov 11 '18 at 5:43
@Maryam In the question, the input is a
list
'x'– akrun
Nov 11 '18 at 5:43
add a comment |
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
Thank you so much for your answer.n
,k
, andJ
are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 '18 at 8:42
add a comment |
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
Thank you so much for your answer.n
,k
, andJ
are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 '18 at 8:42
add a comment |
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
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
answered Nov 11 '18 at 7:12
Darren TsaiDarren Tsai
1,8601323
1,8601323
Thank you so much for your answer.n
,k
, andJ
are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 '18 at 8:42
add a comment |
Thank you so much for your answer.n
,k
, andJ
are just numbers that help me to define the number of my list and their number of vectors.
– Maryam
Nov 11 '18 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 '18 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 '18 at 8:42
add a comment |
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Sorry, it is not clear to me
– akrun
Nov 11 '18 at 5:45
@MrFlick, thank you so much. This is a typo. I will edit it.
– Maryam
Nov 11 '18 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 '18 at 4:48
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@akrun I am always proud of your work. It is correct. I really surprised why someone downvoted it.
– Maryam
Nov 13 '18 at 9:28