Mixing
Creating alternate series in r
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I have a list of -0.5, -0.6, 0.7, 1, 1.5, 3, -5 and I would like to sort it as 3, -5, 1.5, -0.6, 1, -0.5, 0.7. In other words, I would like to sparate the list into positive and negative lists, then sort them from largest to smallest, but alternating.
How can I do this?
r sorting
edited Jan 31 at 6:10
Darren Tsai
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
add a comment |
I have a list of -0.5, -0.6, 0.7, 1, 1.5, 3, -5 and I would like to sort it as 3, -5, 1.5, -0.6, 1, -0.5, 0.7. In other words, I would like to sparate the list into positive and negative lists, then sort them from largest to smallest, but alternating.
How can I do this?
r sorting
edited Jan 31 at 6:10
Darren Tsai
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57
add a comment |
I have a list of -0.5, -0.6, 0.7, 1, 1.5, 3, -5 and I would like to sort it as 3, -5, 1.5, -0.6, 1, -0.5, 0.7. In other words, I would like to sparate the list into positive and negative lists, then sort them from largest to smallest, but alternating.
How can I do this?
r sorting
edited Jan 31 at 6:10
Darren Tsai
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
I have a list of -0.5, -0.6, 0.7, 1, 1.5, 3, -5 and I would like to sort it as 3, -5, 1.5, -0.6, 1, -0.5, 0.7. In other words, I would like to sparate the list into positive and negative lists, then sort them from largest to smallest, but alternating.
How can I do this?
r sorting
r sorting
edited Jan 31 at 6:10
Darren Tsai
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
edited Jan 31 at 6:10
Darren Tsai
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
edited Jan 31 at 6:10
Darren Tsai
2,6472531
edited Jan 31 at 6:10
Darren Tsai
2,6472531
edited Jan 31 at 6:10
Darren Tsai
2,6472531
2,6472531
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
asked Jan 30 at 15:17
Math AvengersMath Avengers
516
516
Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57
add a comment |
Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57
Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57
add a comment |
4 Answers
4
active
oldest
votes
Data
x <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
Use rbind() to create an alternate indices vector.
ind <- seq_along(x)
sort(x, decreasing = T)[c(rbind(ind, rev(ind)))][ind]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
This method above should be memory consuming because it will create a twice longer vector. To conquer it, you can set a midpoint as the end index. In this case, it will be ceiling(7 / 2) = 4.
n <- length(x)
ind <- 1:ceiling(n / 2) # [1] 1 2 3 4
sort(x, decreasing = T)[c(rbind(ind, (n:1)[ind]))][1:n]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
add a comment |
We can split the list based on sign and sort them. We then create a new list by taking alternating elements from positive and negative lists making sure to reverse the positive part of the list.
new_list <- sapply(split(x, sign(x)), sort)
c(rbind(rev(new_list[['1']]), new_list[['-1']]))[1:length(x)]
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
x <- c(-0.5, -.6, 0.7, 1, 1.5, 3, -5 )
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
add a comment |
One option is to split the vector by the sign of the vector into a list of vectors, loop through the vector, order the absolute value of the elements in decreasing order, get the lengths of the list elements same with padding NA at the end, rbind them together as a matrix, convert it to vector with c and remove the NA elements with na.omit
lst1 <- sapply(split(v1, sign(v1)), function(x)
x[order(abs(x), decreasing = TRUE)])[2:1]
c(na.omit(c(do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1)))))))
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
v1 <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
answered Jan 30 at 15:22
akrunakrun
419k13207283
add a comment |
How about
pos <- sort(x[x>0], decreasing = T)
neg <- sort(x[x<0], decreasing = T)
c(rbind(pos,neg))[1:length(x)]
#[1] 3.0 -0.5 1.5 -0.6 1.0 -5.0 0.7
answered Jan 30 at 16:30
989989
9,08251835
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Data
x <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
Use rbind() to create an alternate indices vector.
ind <- seq_along(x)
sort(x, decreasing = T)[c(rbind(ind, rev(ind)))][ind]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
This method above should be memory consuming because it will create a twice longer vector. To conquer it, you can set a midpoint as the end index. In this case, it will be ceiling(7 / 2) = 4.
n <- length(x)
ind <- 1:ceiling(n / 2) # [1] 1 2 3 4
sort(x, decreasing = T)[c(rbind(ind, (n:1)[ind]))][1:n]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
add a comment |
Data
x <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
Use rbind() to create an alternate indices vector.
ind <- seq_along(x)
sort(x, decreasing = T)[c(rbind(ind, rev(ind)))][ind]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
This method above should be memory consuming because it will create a twice longer vector. To conquer it, you can set a midpoint as the end index. In this case, it will be ceiling(7 / 2) = 4.
n <- length(x)
ind <- 1:ceiling(n / 2) # [1] 1 2 3 4
sort(x, decreasing = T)[c(rbind(ind, (n:1)[ind]))][1:n]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
add a comment |
Data
x <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
Use rbind() to create an alternate indices vector.
ind <- seq_along(x)
sort(x, decreasing = T)[c(rbind(ind, rev(ind)))][ind]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
This method above should be memory consuming because it will create a twice longer vector. To conquer it, you can set a midpoint as the end index. In this case, it will be ceiling(7 / 2) = 4.
n <- length(x)
ind <- 1:ceiling(n / 2) # [1] 1 2 3 4
sort(x, decreasing = T)[c(rbind(ind, (n:1)[ind]))][1:n]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
Data
x <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
Use rbind() to create an alternate indices vector.
ind <- seq_along(x)
sort(x, decreasing = T)[c(rbind(ind, rev(ind)))][ind]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
This method above should be memory consuming because it will create a twice longer vector. To conquer it, you can set a midpoint as the end index. In this case, it will be ceiling(7 / 2) = 4.
n <- length(x)
ind <- 1:ceiling(n / 2) # [1] 1 2 3 4
sort(x, decreasing = T)[c(rbind(ind, (n:1)[ind]))][1:n]
# [1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
edited Jan 30 at 17:39
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
answered Jan 30 at 15:30
Darren TsaiDarren Tsai
2,6472531
2,6472531
add a comment |
add a comment |
We can split the list based on sign and sort them. We then create a new list by taking alternating elements from positive and negative lists making sure to reverse the positive part of the list.
new_list <- sapply(split(x, sign(x)), sort)
c(rbind(rev(new_list[['1']]), new_list[['-1']]))[1:length(x)]
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
x <- c(-0.5, -.6, 0.7, 1, 1.5, 3, -5 )
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
add a comment |
We can split the list based on sign and sort them. We then create a new list by taking alternating elements from positive and negative lists making sure to reverse the positive part of the list.
new_list <- sapply(split(x, sign(x)), sort)
c(rbind(rev(new_list[['1']]), new_list[['-1']]))[1:length(x)]
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
x <- c(-0.5, -.6, 0.7, 1, 1.5, 3, -5 )
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
add a comment |
We can split the list based on sign and sort them. We then create a new list by taking alternating elements from positive and negative lists making sure to reverse the positive part of the list.
new_list <- sapply(split(x, sign(x)), sort)
c(rbind(rev(new_list[['1']]), new_list[['-1']]))[1:length(x)]
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
x <- c(-0.5, -.6, 0.7, 1, 1.5, 3, -5 )
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
We can split the list based on sign and sort them. We then create a new list by taking alternating elements from positive and negative lists making sure to reverse the positive part of the list.
new_list <- sapply(split(x, sign(x)), sort)
c(rbind(rev(new_list[['1']]), new_list[['-1']]))[1:length(x)]
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
x <- c(-0.5, -.6, 0.7, 1, 1.5, 3, -5 )
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
edited Jan 30 at 15:34
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
answered Jan 30 at 15:26
Ronak ShahRonak Shah
45.1k104266
45.1k104266
add a comment |
add a comment |
One option is to split the vector by the sign of the vector into a list of vectors, loop through the vector, order the absolute value of the elements in decreasing order, get the lengths of the list elements same with padding NA at the end, rbind them together as a matrix, convert it to vector with c and remove the NA elements with na.omit
lst1 <- sapply(split(v1, sign(v1)), function(x)
x[order(abs(x), decreasing = TRUE)])[2:1]
c(na.omit(c(do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1)))))))
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
v1 <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
answered Jan 30 at 15:22
akrunakrun
419k13207283
add a comment |
One option is to split the vector by the sign of the vector into a list of vectors, loop through the vector, order the absolute value of the elements in decreasing order, get the lengths of the list elements same with padding NA at the end, rbind them together as a matrix, convert it to vector with c and remove the NA elements with na.omit
lst1 <- sapply(split(v1, sign(v1)), function(x)
x[order(abs(x), decreasing = TRUE)])[2:1]
c(na.omit(c(do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1)))))))
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
v1 <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
answered Jan 30 at 15:22
akrunakrun
419k13207283
add a comment |
One option is to split the vector by the sign of the vector into a list of vectors, loop through the vector, order the absolute value of the elements in decreasing order, get the lengths of the list elements same with padding NA at the end, rbind them together as a matrix, convert it to vector with c and remove the NA elements with na.omit
lst1 <- sapply(split(v1, sign(v1)), function(x)
x[order(abs(x), decreasing = TRUE)])[2:1]
c(na.omit(c(do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1)))))))
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
v1 <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
answered Jan 30 at 15:22
akrunakrun
419k13207283
One option is to split the vector by the sign of the vector into a list of vectors, loop through the vector, order the absolute value of the elements in decreasing order, get the lengths of the list elements same with padding NA at the end, rbind them together as a matrix, convert it to vector with c and remove the NA elements with na.omit
lst1 <- sapply(split(v1, sign(v1)), function(x)
x[order(abs(x), decreasing = TRUE)])[2:1]
c(na.omit(c(do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1)))))))
#[1] 3.0 -5.0 1.5 -0.6 1.0 -0.5 0.7
data
v1 <- c(-0.5, -0.6, 0.7, 1, 1.5, 3, -5)
answered Jan 30 at 15:22
akrunakrun
419k13207283
answered Jan 30 at 15:22
akrunakrun
419k13207283
answered Jan 30 at 15:22
akrunakrun
419k13207283
answered Jan 30 at 15:22
akrunakrun
419k13207283
419k13207283
add a comment |
add a comment |
How about
pos <- sort(x[x>0], decreasing = T)
neg <- sort(x[x<0], decreasing = T)
c(rbind(pos,neg))[1:length(x)]
#[1] 3.0 -0.5 1.5 -0.6 1.0 -5.0 0.7
answered Jan 30 at 16:30
989989
9,08251835
add a comment |
How about
pos <- sort(x[x>0], decreasing = T)
neg <- sort(x[x<0], decreasing = T)
c(rbind(pos,neg))[1:length(x)]
#[1] 3.0 -0.5 1.5 -0.6 1.0 -5.0 0.7
answered Jan 30 at 16:30
989989
9,08251835
add a comment |
How about
pos <- sort(x[x>0], decreasing = T)
neg <- sort(x[x<0], decreasing = T)
c(rbind(pos,neg))[1:length(x)]
#[1] 3.0 -0.5 1.5 -0.6 1.0 -5.0 0.7
answered Jan 30 at 16:30
989989
9,08251835
How about
pos <- sort(x[x>0], decreasing = T)
neg <- sort(x[x<0], decreasing = T)
c(rbind(pos,neg))[1:length(x)]
#[1] 3.0 -0.5 1.5 -0.6 1.0 -5.0 0.7
answered Jan 30 at 16:30
989989
9,08251835
answered Jan 30 at 16:30
989989
9,08251835
answered Jan 30 at 16:30
989989
9,08251835
answered Jan 30 at 16:30
989989
9,08251835
9,08251835
add a comment |
add a comment |
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Is it a list or a vector?
– Sotos
Jan 30 at 15:20
Do you know that the number of positive and negative numbers are equal?
– Henry
Jan 30 at 21:34
@Henry they are not actually ! But I can't really change them, so I will just leave them as what they are.
– Math Avengers
Jan 31 at 6:57