I need to create named lists dynamically in R as follows.
Suppose there is an array of names.
name_arr<-c("a","b")
And that there is an array of values.
value_arr<-c(1,2,3,4,5,6)
What I want to do is something like this:
list(name_arr[1]=value_arr[1:3])
But R throws an error when I try to do this. Any suggestions as to how to get around this problem?
you can use [[...]] to assign values to keys given by strings:
my.list <- list()
my.list[[name_arr[1]]] <- value_arr[1:3]
You could use setNames. Examples:
setNames(list(value_arr[1:3]), name_arr[1])
#$a
#[1] 1 2 3
setNames(list(value_arr[1:3], value_arr[4:6]), name_arr)
#$a
#[1] 1 2 3
#
#$b
#[1] 4 5 6
Or without setNames:
mylist <- list(value_arr[1:3])
names(mylist) <- name_arr[1]
mylist
#$a
#[1] 1 2 3
mylist <- list(value_arr[1:3], value_arr[4:6])
names(mylist) <- name_arr
mylist
#$a
#[1] 1 2 3
#
#$b
#[1] 4 5 6
Your code will throw a error. Because in list(A = B), A must be a name instead of an object.
You could convert a object to a name by function eval. Here is the example.
eval(parse(text = sprintf('list(%s = value_arr[1:3])',name_arr[1])))
Related
I have a vector like below
tmp <- c(a=1, b=2, c=3)
a b c
1 2 3
I want to flatten this vector to get only 1, 2, 3.
I tried unlist(tmp) but it still gives me the same result.
How to achieve that efficiently?
You just want to remove the names attribute from tmp. There are a number of ways to do that.
You can unname it.
unname(tmp)
# [1] 1 2 3
Or use a very common method for removing names, by setting them to NULL.
names(tmp) <- NULL
Or strip the attributes with as.vector.
as.vector(tmp)
# [1] 1 2 3
Or re-concatenate it without the names.
c(tmp, use.names=FALSE)
# [1] 1 2 3
Or use setNames.
setNames(tmp, NULL)
# [1] 1 2 3
There is a use case that the above does not cover:
tmp <- c(1,2,3)
names(tmp) <- c("a","b","c")
In this case you need to use both:
unlist(unname(tmp))
I have a vector like this.
filenames <- c("kisyu2_mst.csv", "kisyu3_mst.csv", "kisyu2_mst.csv",
"kisyu3_mst.csv", "kisyu3_mst.csv")
I need to get indices from filenames vector for each unique value.output look like this
for "kisyu2_mst.csv" indices vector c(1,3)
for "kisyu3_mst.csv" indices vector c(2,4,5)
Finally, I need to insert it to a list like this:
final <- list("kisyu2_mst.csv" = c(1,3), "kisyu3_mst.csv"=c(2,4,5))
How to get the indices of unique value from the vector?
We can use split
split(seq_along(filenames), filenames)
#$kisyu2_mst.csv
#[1] 1 3
#$kisyu3_mst.csv
#[1] 2 4 5
We could try which:
sapply(unique(filenames), function(i) which(filenames %in% i))
# $kisyu2_mst.csv
# [1] 1 3
#
# $kisyu3_mst.csv
# [1] 2 4 5
We can use tapply
tapply(seq_along(filenames), filenames, FUN = I)
#$kisyu2_mst.csv
#[1] 1 3
#$kisyu3_mst.csv
#[1] 2 4 5
I've like to remove elements in a list, if the number of elements are smaller than 3.
For this I try:
#Create a list
my_list <- list(a = c(3,5,6), b = c(3,1,0), c = 4, d = NA)
my_list
$a
[1] 3 5 6
$b
[1] 3 1 0
$c
[1] 4
$d
[1] NA
# Thant I create a function for remove the elements by my condition:
delete.F <- function(x.list){
x.list[unlist(lapply(x.list, function(x) ncol(x)) < 3)]}
delete.F(my_list)
And I have as output:
Error in unlist(lapply(x.list, function(x) ncol(x)) < 3) :
(list) object cannot be coerced to type 'double'
Any ideas, please?
An option is to create a logical expression with lengths and use that for subsetting the list
my_list[lengths(my_list) >=3]
#$a
#[1] 3 5 6
#$b
#[1] 3 1 0
Note that in the example, it is a list of vectors and not a list of data.frame. the ncol/nrow is when there is a dim attribute - matrix checks TRUE for that, as do data.frame
If we want to somehow use lapply (based on some constraints), create the logic with length
unlist(lapply(my_list, function(x) if(length(x) >=3 ) x))
If we need to create the index with lapply, use length (but it would be slower than lengths)
my_list[unlist(lapply(my_list, length)) >= 3]
Here are few more options. Using Filter in base R
Filter(function(x) length(x) >=3, my_list)
#$a
#[1] 3 5 6
#$b
#[1] 3 1 0
Or using purrr's keep and discard
purrr::keep(my_list, ~length(.) >= 3)
purrr::discard(my_list, ~length(.) < 3)
Let's assume we have a list with three sublists: list1 <- [[1,2],[4,5],[7,8]]
and a vector: vector1 <- c(3,6,9)
Is there a way in R, without using loops, to append vector's values to the list, so we could get the result list2 = [[1,2,3],[4,5,6],[7,8,9]]
?
Thanks for all comments
Use Map
Map(c, list1, vector1)
#[[1]]
#[1] 1 2 3
#[[2]]
#[1] 4 5 6
#[[3]]
#[1] 7 8 9
Or lapply
lapply(seq_along(list1), function(x) c(list1[[x]], vector1[[x]]))
The equivalent purrr variants can be
purrr::map2(list1, vector1, c)
purrr::map(seq_along(list1), ~c(list1[[.]], vector1[[.]]))
data
list1 <- list(c(1,2),c(4,5),c(7,8))
vector1 <- c(3,6,9)
I would like to subset an environment by its variable names.
e <- new.env(parent=emptyenv())
e$a <- 1
e$b <- 2
e$d <- 3
e[ls(e) %in% c("a","b", "c")]
### if e was a list, this would return the subset list(a=1, b=2)
I could not figure out how to subset elements of an environment by their names. Using lapply or eapply does not work either. What is the proper or easy way to subset an environment by its variable names?
Thank you.
Okay, after thinking this through a bit more, may I suggest:
mget(c("a","b"), envir=e)
#$a
#[1] 1
#
#$b
#[1] 2
My original solution is to use get() / mget() (maybe OP saw my deleted comment earlier). Then I noticed that OP had tried eapply(), so I thought about possible solutions with that. Here it is (with help of #thelatemail).
# try some different data type
e <- new.env(parent=emptyenv())
e$a <- 1:3
e$b <- matrix(1:4, 2)
e$c <- data.frame(x=letters[1:2],y=LETTERS[1:2])
You can use either of the following to collect objects in environment e into a list:
elst <- eapply(e, "[") ## my idea
elst <- eapply(e, identity) ## thanks to #thelatemail
elst <- as.list.environment(e) ## thanks to #thelatemail
#$a
#[1] 1 2 3
#$b
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
#$c
# x y
#1 a A
#2 b B
The as.list.environment() can be seen as the inverse operation of list2env(). It is mentioned in the "See Also" part of ?list2env.
The result elst is just an ordinary list. There are various way to subset this list. For example:
elst[names(elst) %in% c("a","b")] ## no need to use "ls(e)" now
#$a
#[1] 1 2 3
#$b
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
mget(ls(e)[ls(e) %in% c('a','b','d')], e)
The [ operator usually returns the same type of object as the original, so I guess you're expecting an environment, rather than a list. The same environment but with a different set of elements, or a new environment with the specified elements? Either way I think you'll end up iterating, e.g.,
f = new.env(parent=emptyenv())
for (elt in c("a", "b"))
f[[elt]] = e[[elt]]
Working with environments is not very idiomatic R code, which might explain why there is not a more elegant solution.
You can use rlang::env_get_list() to get a list of the bindings:
rlang::env_get_list(env=e, c("a","b"))
#$a
#[1] 1
#
#$b
#[1] 2
If you're trying to get an environment, rather than a list, I'm not sure how you would do that, other than just creating a new environment using the output of rlang::env_get_list().
If you want to include elements in your list that might not exist in the environment (like "c"), you have to specify a default value - otherwise you'll get an error:
env_get_list(env = e, c("a","b","c"))
#Error in env_get_list(env = e, c("a", "b", "c")) : argument "default" is missing, with no default
env_get_list(env = e, c("a","b","c"),default=NULL)
#$a
#[1] 1
#
#$b
#[1] 2
#
#$c
#NULL
I assume you don't want c at all, so I'd do something like:
temp <- c("a","b","c")[c("a","b","c") %in% env_names(e)]
temp
[1] "a" "b"
env_get_list(env=e,temp)
#$a
#[1] 1
#
#$b
#[1] 2