I was wondering if somebody could help with this problem. I have a list coming out from a function similar to the following:
lis<-vector("list",3)
lis[[1]]<-c(1,2,3)
lis[[2]]<-c(1,2,3)
lis[[3]]<-c(1,2,3)
so it looks like
[[1]]
[1] 1 2 3
[[2]]
[1] 1 2 3
[[3]]
[1] 1 2 3
What I want to do is remove, for example, the first element from each component of the list so it ends up like:
[[1]]
[1] 2 3
[[2]]
[1] 2 3
[[3]]
[1] 2 3
Any ideas would be most welcome.
You can use lapply() and do the index function for each element of the list. The index -1 means without the first element:
lis <- list(a=1:3, b=11:13, c=21:23)
lapply(lis, '[', -1)
# $a
# [1] 2 3
#
# $b
# [1] 12 13
#
# $c
# [1] 22 23
Related
Consider a vector:
vec <- c(1, 3, 4, 3, 3, 1, 1)
I'd like to get, for each element of the vector, a subset of the values in between the nth element and its previous occurrence.
The expected output is:
f(vec)
# [[1]]
# [1] 1
#
# [[2]]
# [1] 3
#
# [[3]]
# [1] 4
#
# [[4]]
# [1] 3 4 3
#
# [[5]]
# [1] 3 3
#
# [[6]]
# [1] 1 3 4 3 3 1
#
# [[7]]
# [1] 1 1
We may loop over the sequence of the vector, get the index of the last match of the same element ('i1') from the previous elements of the vector and get the sequence (:) to subset the vector
lapply(seq_along(vec), function(i) {
i1 <- tail(which(vec[1:(i-1)] == vec[i]), 1)[1]
i1[is.na(i1)] <- i
vec[i1:i]
})
-output
[[1]]
[1] 1
[[2]]
[1] 3
[[3]]
[1] 4
[[4]]
[1] 3 4 3
[[5]]
[1] 3 3
[[6]]
[1] 1 3 4 3 3 1
[[7]]
[1] 1 1
I have a list of values called squares and would like to replace all values which are 0 to a 40.
I tried:
replace(squares, squares==0, 40)
but the list remains unchanged
If it is a list, then loop through the list with lapply and use replace
squares <- lapply(squares, function(x) replace(x, x==0, 40))
squares
#[[1]]
#[1] 40 1 2 3 4 5
#[[2]]
#[1] 1 2 3 4 5 6
#[[3]]
#[1] 40 1 2 3
data
squares <- list(0:5, 1:6, 0:3)
I think for this purpose, you can just treat it as if it were a vector as follows:
squares=list(2,4,6,0,8,0,10,20)
squares[squares==0]=40
Output:
[[1]]
[1] 2
[[2]]
[1] 4
[[3]]
[1] 6
[[4]]
[1] 40
[[5]]
[1] 8
[[6]]
[1] 40
[[7]]
[1] 10
[[8]]
[1] 20
How to remove outliers using a criterion that a value cannot be more than 2-fold higher then its preceding one.
Here is my try:
x<-c(1,2,6,4,10,20,50,10,2,1)
remove_outliers <- function(x, na.rm = TRUE, ...) {
for(i in 1:length(x))
x < (x[i-1] + 2*x)
x
}
remove_outliers(y)
expected outcome: 1,2,4,10,20,2,1
Thanks!
I think the first 10 should be removed in your data because 10>2*4. Here's a way to do what you want without loops. I'm using the dplyr version of lag.
library(dplyr)
x<-c(1,2,6,4,10,20,50,10,2,1)
x[c(TRUE,na.omit(x<=dplyr::lag(x)*2))]
[1] 1 2 4 20 10 2 1
EDIT
To use this with a data.frame:
df <- data.frame(id=1:10, x=c(1,2,6,4,10,20,50,10,2,1))
df[c(TRUE,na.omit(df$x<=dplyr::lag(df$x,1)*2)),]
id x
1 1 1
2 2 2
4 4 4
6 6 20
8 8 10
9 9 2
10 10 1
A simple sapply:
bool<-sapply(seq_along(1:length(x)),function(i) {ifelse(x[i]<2*x[i-1],FALSE,TRUE)})
bool
[[1]]
logical(0)
[[2]]
[1] TRUE
[[3]]
[1] TRUE
[[4]]
[1] FALSE
[[5]]
[1] TRUE
[[6]]
[1] TRUE
[[7]]
[1] TRUE
[[8]]
[1] FALSE
[[9]]
[1] FALSE
[[10]]
[1] FALSE
resulting in:
x[unlist(bool)]
[1] 1 2 4 10 20 1
Say, I have the following recursive list:
rec_list <- list(list(rep(1,5), 10), list(rep(100, 4), 20:25))
rec_list
[[1]]
[[1]][[1]]
[1] 1 1 1 1 1
[[1]][[2]]
[1] 10
[[2]]
[[2]][[1]]
[1] 100 100 100 100
[[2]][[2]]
[1] 20 21 22 23 24 25
Now, I would like to replace all the values of the list, say, with the vector seq_along(unlist(rec_list)), and keep the structure of the list. I tried using the empty index subsetting like
rec_list[] <- seq_along(unlist(rec_list))
But this doesn't work.
How can I achieve the replacement while keeping the original structure of the list?
You can use relist:
relist(seq_along(unlist(rec_list)), skeleton = rec_list)
# [[1]]
# [[1]][[1]]
# [1] 1 2 3 4 5
#
# [[1]][[2]]
# [1] 6
#
#
# [[2]]
# [[2]][[1]]
# [1] 7 8 9 10
#
# [[2]][[2]]
# [1] 11 12 13 14 15 16
If you wanted to uniquely index each element of a nested list, you could start with the rapply() function which is the recursive form of the apply() family. Here I use a special function that can uniquely index across a list of any structure
rapply(rec_list,
local({i<-0; function(x) {i<<-i+length(x); i+seq_along(x)-length(x)}}),
how="replace")
other functions are simplier, for example if you just wanted to seq_along each subvector
rapply(rec_list, seq_along, how="replace")
how do I get the x value of each element in the list.
example:
list1 <- list(1:3,4:6)
list1
#[[1]]
#[1] 1 2 3
#
#[[2]]
#[1] 4 5 6
Imaginary function I'm looking for:
function(list1, 1)
# [1] 1 4
function(list2, 2)
# [1] 2 5
How can I do this?
Use sapply or lapply, in combination with the [ extraction function (see ?Extract for more info) like so:
> sapply(list1,"[",1)
[1] 1 4
...or with a list output:
> lapply(list1,"[",1)
[[1]]
[1] 1
[[2]]
[1] 4