I have a list and I want to remove a single element from it. How can I do this?
I've tried looking up what I think the obvious names for this function would be in the reference manual and I haven't found anything appropriate.
If you don't want to modify the list in-place (e.g. for passing the list with an element removed to a function), you can use indexing: negative indices mean "don't include this element".
x <- list("a", "b", "c", "d", "e"); # example list
x[-2]; # without 2nd element
x[-c(2, 3)]; # without 2nd and 3rd
Also, logical index vectors are useful:
x[x != "b"]; # without elements that are "b"
This works with dataframes, too:
df <- data.frame(number = 1:5, name = letters[1:5])
df[df$name != "b", ]; # rows without "b"
df[df$number %% 2 == 1, ] # rows with odd numbers only
I don't know R at all, but a bit of creative googling led me here: http://tolstoy.newcastle.edu.au/R/help/05/04/1919.html
The key quote from there:
I do not find explicit documentation for R on how to remove elements from lists, but trial and error tells me
myList[[5]] <- NULL
will remove the 5th element and then "close up" the hole caused by deletion of that element. That suffles the index values, So I have to be careful in dropping elements. I must work from the back of the list to the front.
A response to that post later in the thread states:
For deleting an element of a list, see R FAQ 7.1
And the relevant section of the R FAQ says:
... Do not set x[i] or x[[i]] to NULL, because this will remove the corresponding component from the list.
Which seems to tell you (in a somewhat backwards way) how to remove an element.
I would like to add that if it's a named list you can simply use within.
l <- list(a = 1, b = 2)
> within(l, rm(a))
$b
[1] 2
So you can overwrite the original list
l <- within(l, rm(a))
to remove element named a from list l.
Here is how the remove the last element of a list in R:
x <- list("a", "b", "c", "d", "e")
x[length(x)] <- NULL
If x might be a vector then you would need to create a new object:
x <- c("a", "b", "c", "d", "e")
x <- x[-length(x)]
Work for lists and vectors
Removing Null elements from a list in single line :
x=x[-(which(sapply(x,is.null),arr.ind=TRUE))]
Cheers
If you have a named list and want to remove a specific element you can try:
lst <- list(a = 1:4, b = 4:8, c = 8:10)
if("b" %in% names(lst)) lst <- lst[ - which(names(lst) == "b")]
This will make a list lst with elements a, b, c. The second line removes element b after it checks that it exists (to avoid the problem #hjv mentioned).
or better:
lst$b <- NULL
This way it is not a problem to try to delete a non-existent element (e.g. lst$g <- NULL)
Use - (Negative sign) along with position of element, example if 3rd element is to be removed use it as your_list[-3]
Input
my_list <- list(a = 3, b = 3, c = 4, d = "Hello", e = NA)
my_list
# $`a`
# [1] 3
# $b
# [1] 3
# $c
# [1] 4
# $d
# [1] "Hello"
# $e
# [1] NA
Remove single element from list
my_list[-3]
# $`a`
# [1] 3
# $b
# [1] 3
# $d
# [1] "Hello"
# $e
[1] NA
Remove multiple elements from list
my_list[c(-1,-3,-2)]
# $`d`
# [1] "Hello"
# $e
# [1] NA
my_list[c(-3:-5)]
# $`a`
# [1] 3
# $b
# [1] 3
my_list[-seq(1:2)]
# $`c`
# [1] 4
# $d
# [1] "Hello"
# $e
# [1] NA
There's the rlist package (http://cran.r-project.org/web/packages/rlist/index.html) to deal with various kinds of list operations.
Example (http://cran.r-project.org/web/packages/rlist/vignettes/Filtering.html):
library(rlist)
devs <-
list(
p1=list(name="Ken",age=24,
interest=c("reading","music","movies"),
lang=list(r=2,csharp=4,python=3)),
p2=list(name="James",age=25,
interest=c("sports","music"),
lang=list(r=3,java=2,cpp=5)),
p3=list(name="Penny",age=24,
interest=c("movies","reading"),
lang=list(r=1,cpp=4,python=2)))
list.remove(devs, c("p1","p2"))
Results in:
# $p3
# $p3$name
# [1] "Penny"
#
# $p3$age
# [1] 24
#
# $p3$interest
# [1] "movies" "reading"
#
# $p3$lang
# $p3$lang$r
# [1] 1
#
# $p3$lang$cpp
# [1] 4
#
# $p3$lang$python
# [1] 2
Don't know if you still need an answer to this but I found from my limited (3 weeks worth of self-teaching R) experience with R that, using the NULL assignment is actually wrong or sub-optimal especially if you're dynamically updating a list in something like a for-loop.
To be more precise, using
myList[[5]] <- NULL
will throw the error
myList[[5]] <- NULL : replacement has length zero
or
more elements supplied than there are to replace
What I found to work more consistently is
myList <- myList[[-5]]
Just wanted to quickly add (because I didn't see it in any of the answers) that, for a named list, you can also do l["name"] <- NULL. For example:
l <- list(a = 1, b = 2, cc = 3)
l['b'] <- NULL
In the case of named lists I find those helper functions useful
member <- function(list,names){
## return the elements of the list with the input names
member..names <- names(list)
index <- which(member..names %in% names)
list[index]
}
exclude <- function(list,names){
## return the elements of the list not belonging to names
member..names <- names(list)
index <- which(!(member..names %in% names))
list[index]
}
aa <- structure(list(a = 1:10, b = 4:5, fruits = c("apple", "orange"
)), .Names = c("a", "b", "fruits"))
> aa
## $a
## [1] 1 2 3 4 5 6 7 8 9 10
## $b
## [1] 4 5
## $fruits
## [1] "apple" "orange"
> member(aa,"fruits")
## $fruits
## [1] "apple" "orange"
> exclude(aa,"fruits")
## $a
## [1] 1 2 3 4 5 6 7 8 9 10
## $b
## [1] 4 5
Using lapply and grep:
lst <- list(a = 1:4, b = 4:8, c = 8:10)
# say you want to remove a and c
toremove<-c("a","c")
lstnew<-lst[-unlist(lapply(toremove, function(x) grep(x, names(lst)) ) ) ]
#or
pattern<-"a|c"
lstnew<-lst[-grep(pattern, names(lst))]
You can also negatively index from a list using the extract function of the magrittr package to remove a list item.
a <- seq(1,5)
b <- seq(2,6)
c <- seq(3,7)
l <- list(a,b,c)
library(magrittr)
extract(l,-1) #simple one-function method
[[1]]
[1] 2 3 4 5 6
[[2]]
[1] 3 4 5 6 7
There are a few options in the purrr package that haven't been mentioned:
pluck and assign_in work well with nested values and you can access it using a combination of names and/or indices:
library(purrr)
l <- list("a" = 1:2, "b" = 3:4, "d" = list("e" = 5:6, "f" = 7:8))
# select values (by name and/or index)
all.equal(pluck(l, "d", "e"), pluck(l, 3, "e"), pluck(l, 3, 1))
[1] TRUE
# or if element location stored in a vector use !!!
pluck(l, !!! as.list(c("d", "e")))
[1] 5 6
# remove values (modifies in place)
pluck(l, "d", "e") <- NULL
# assign_in to remove values with name and/or index (does not modify in place)
assign_in(l, list("d", 1), NULL)
$a
[1] 1 2
$b
[1] 3 4
$d
$d$f
[1] 7 8
Or you can remove values using modify_list by assigning zap() or NULL:
all.equal(list_modify(l, a = zap()), list_modify(l, a = NULL))
[1] TRUE
You can remove or keep elements using a predicate function with discard and keep:
# remove numeric elements
discard(l, is.numeric)
$d
$d$e
[1] 5 6
$d$f
[1] 7 8
# keep numeric elements
keep(l, is.numeric)
$a
[1] 1 2
$b
[1] 3 4
Here is a simple solution that can be done using base R. It removes the number 5 from the original list of numbers. You can use the same method to remove whatever element you want from a list.
#the original list
original_list = c(1:10)
#the list element to remove
remove = 5
#the new list (which will not contain whatever the `remove` variable equals)
new_list = c()
#go through all the elements in the list and add them to the new list if they don't equal the `remove` variable
counter = 1
for (n in original_list){
if (n != ){
new_list[[counter]] = n
counter = counter + 1
}
}
The new_list variable no longer contains 5.
new_list
# [1] 1 2 3 4 6 7 8 9 10
How about this? Again, using indices
> m <- c(1:5)
> m
[1] 1 2 3 4 5
> m[1:length(m)-1]
[1] 1 2 3 4
or
> m[-(length(m))]
[1] 1 2 3 4
You can use which.
x<-c(1:5)
x
#[1] 1 2 3 4 5
x<-x[-which(x==4)]
x
#[1] 1 2 3 5
if you'd like to avoid numeric indices, you can use
a <- setdiff(names(a),c("name1", ..., "namen"))
to delete names namea...namen from a. this works for lists
> l <- list(a=1,b=2)
> l[setdiff(names(l),"a")]
$b
[1] 2
as well as for vectors
> v <- c(a=1,b=2)
> v[setdiff(names(v),"a")]
b
2
Related
It is straight forward to obtain unique values of a column using unique. However, I am looking to do the same but for multiple columns in a dataframe and store them in a list, all using base R. Importantly, it is not combinations I need but simply unique values for each individual column. I currently have the below:
# dummy data
df = data.frame(a = LETTERS[1:4]
,b = 1:4)
# for loop
cols = names(df)
unique_values_by_col = list()
for (i in cols)
{
x = unique(i)
unique_values_by_col[[i]] = x
}
The problem comes when displaying unique_values_by_col as it shows as empty. I believe the problem is i is being passed to the loop as a text not a variable.
Any help would be greatly appreciated. Thank you.
Why not avoid the for loop altogether using lapply:
lapply(df, unique)
Resulting in:
> $a
> [1] A B C D
> Levels: A B C D
> $b
> [1] 1 2 3 4
Or you have also apply that is specifically done to be run on column or line:
apply(df,2,unique)
result:
> apply(df,2,unique)
a b
[1,] "A" "1"
[2,] "B" "2"
[3,] "C" "3"
[4,] "D" "4"
thought if you want a list lapply return you a list so may be better
Your for loop is almost right, just needs one fix to work:
# for loop
cols = names(df)
unique_values_by_col = list()
for (i in cols) {
x = unique(df[[i]])
unique_values_by_col[[i]] = x
}
unique_values_by_col
# $a
# [1] A B C D
# Levels: A B C D
#
# $b
# [1] 1 2 3 4
i is just a character, the name of a column within df so unique(i) doesn't make sense.
Anyhow, the most standard way for this task is lapply() as shown by demirev.
Could this be what you're trying to do?
Map(unique,df)
Result:
$a
[1] A B C D
Levels: A B C D
$b
[1] 1 2 3 4
In python we can do this..
numbers = [1, 2, 3]
characters = ['foo', 'bar', 'baz']
for item in zip(numbers, characters):
print(item[0], item[1])
(1, 'foo')
(2, 'bar')
(3, 'baz')
We can also unpack the tuple rather than using the index.
for num, char in zip(numbers, characters):
print(num, char)
(1, 'foo')
(2, 'bar')
(3, 'baz')
How can we do the same using base R?
To do something like this in an R-native way, you'd use the idea of a data frame. A data frame has multiple variables which can be of different types, and each row is an observation of each variable.
d <- data.frame(numbers = c(1, 2, 3),
characters = c('foo', 'bar', 'baz'))
d
## numbers characters
## 1 1 foo
## 2 2 bar
## 3 3 baz
You then access each row using matrix notation, where leaving an index blank includes everything.
d[1,]
## numbers characters
## 1 1 foo
You can then loop over the rows of the data frame to do whatever you want to do, presumably you actually want to do something more interesting than printing.
for(i in seq_len(nrow(d))) {
print(d[i,])
}
## numbers characters
## 1 1 foo
## numbers characters
## 2 2 bar
## numbers characters
## 3 3 baz
For another option, how about mapply, which is the closest analog to zip I can think of in R. Here I'm using the c function to make a new vector, but you could use any function you'd like:
numbers<- c(1, 2, 3)
characters<- c('foo', 'bar', 'baz')
mapply(c,numbers, characters, SIMPLIFY = FALSE)
[[1]]
[1] "1" "foo"
[[2]]
[1] "2" "bar"
[[3]]
[1] "3" "baz"
Which way is of most use depends on what you want to do with your output, but as the other answers mention, a dataframe is the most natural approach in R (and pandas dataframe probably in python).
To index a vector in R, where the vector is variable x would be x[1]. This would return the first element of the vector. R element numbering starts at 1 in contrast to Python which starts at 0.
For this problem it would be:
x = seq(1,10)
j = seq(11,20)
for (i in 1:length(x)){
print (c(x[i],j[i]))
}
Many functions in R are vectorized and don't require loops:
numbers = c(1, 2, 3)
characters = c('foo', 'bar', 'baz')
myList <- list(numbers, characters)
myDF <- data.frame(numbers,characters, stringsAsFactors = F)
print(myList)
print(myDF)
This is the conceptual equivalent:
for (item in Map(list,numbers,characters)){ # though most of the time you would actually do all your work inside Map
print(item[c(1,2)])
}
# [[1]]
# [1] 1
#
# [[2]]
# [1] "a"
#
# [[1]]
# [1] 2
#
# [[2]]
# [1] "b"
#
# [[1]]
# [1] 3
#
# [[2]]
# [1] "c"
#
# [[1]]
# [1] 4
#
# [[2]]
# [1] "d"
#
# [[1]]
# [1] 5
#
# [[2]]
# [1] "e"
Though most of the time you would actually do all your work inside Map and do something like this:
Map(function(nu,ch){print(data.frame(nu,ch))},numbers,characters)
This is the closest I could get to a clone:
zip <- function(...){ Map(list,...)}
print2 <- function(...){do.call(cat,c(list(...),"\n"))}
for (item in zip(numbers,characters)){
print2(item[[1]],item[[2]])
}
# 1 a
# 2 b
# 3 c
# 4 d
# 5 e
to be able to call items by their names (still works with indices):
zip <- function(...){
names <- sapply(substitute(list(...))[-1],deparse)
Map(function(...){setNames(list(...),names)}, ...)
}
for (item in zip(numbers,characters)){
print2(item[["numbers"]],item[["characters"]])
}
The tidyverse solution would be to use purrr::map2 function. Ex:
numbers <- c(1, 2, 3)
characters <- c('foo', 'bar', 'baz')
map2(numbers, characters, ~paste0(.x, ',', .y))
#[[1]]
#[1] "1,foo"
#[[2]]
#[1] "2,bar"
#[[3]]
#[1] "3,baz"
See API here
Other scalable alternatives: Store the vectors in the list and iterate over.
vect1 <- c(1, 2, 3)
vect1 <- c('foo', 'bar', 'baz')
vect2 <- c('a', 'b', 'c')
idx_list <- list(vect1, vect2)
idx_vect <- c(1:length(idx_list[[1]]))
for(i in idx_vect){
x <- idx_list[[1]][i]
j <- idx_list[[2]][i]
print(c(i, x, j))
}
I have a list of mixed types, vector and data frames of 2 columns.
> my.list
$a
[1] 1
$df1
key value
1 b 2
2 c 3
$df2
key value
1 d 4
2 e 5
I would like to end up with a list of vectors only, each data frame row would become a list element with column value as value and column key as element name.
So the result in this example would be :
$a
[1] 1
$b
[1] 2
$c
[1] 3
$d
[1] 4
$e
[1] 5
Actually here is how I achieve this :
my.list <- list(a = 1,
df1 = data.frame(key = c("b", "c"), value = 2:3),
df2 = data.frame(key = c("d", "e"), value = 4:5))
unlist(lapply(seq_along(my.list), function(i) {
if (is.data.frame(my.list[[i]])) {
with(my.list[[i]], as.list(setNames(value, key), all.names = TRUE))
} else {
setNames(my.list[i], names(my.list[i]))
}
}), recursive = FALSE)
But I don't realy like this solution. Do you have smarter ideas to achieve this please ? Thanks
You can do it in two steps base R:
x = do.call(rbind, Filter(is.data.frame, my.list))
c(Filter(Negate(is.data.frame), my.list), as.list(setNames(x$value, x$key)))
$a
[1] 1
$b
[1] 2
$c
[1] 3
$d
[1] 4
$e
[1] 5
One option (based on the example) would be to melt (from reshape2) to 'long' format, convert to data.table (setDT), replace the NA elements in 'key' with the corresponding values from 'L1', and split the 'value' based on 'key'.
library(data.table)
library(reshape2)
with(setDT(melt(my.list))[is.na(key), key := L1], split(value, key))
If I have the following list:
a <- list(1:3, 4:5, 6:9)
a
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5
[[3]]
[1] 6 7 8 9
I want to determine which element of the list a specific value is in. For example, I might want to find which element the number 5 falls under. In this case it would be [[2]].
My goal is to have something like
match(5,a)
return the value 2.
However, this code only checks whether the selected number exists as a complete element of a given element
match(5,a)
[1] NA
Further, unlist only tells me where in the entire length of all values my number of interest falls:
match(5,unlist(a))
[1] 5
Thoughts?
You can use grep function
grep(5, a)
# [1] 2
grep(9, a)
# [1] 3
Updated Answer
After reading #nicola 's comment came to know that grep command works only for the numbers that belong to start and end of the list and not for the numbers that are in between.
You can try the below mentioned code for the complete solution,
a <- list(1:3, 4:5, 6:9)
df <- data.frame(unlist(a))
df$group <- 0
k <- 1
i<-0
for(i in 1:length(a))
{
x[i] <- length(unlist(a[i]))
for(j in 1:x[i])
{
df$group[k] <- i
k <- k+1
}
}
colnames(df)[1] <- "num"
df[df$num == 5, ]$group
# [1] 2
> df[df$num == 9, ]$group
#[1] 3
df[df$num == 8, ]$group
# [1] 3
I have this toy character vector:
a = c("a","b","c","d","e","d,e","f")
in which some elements are concatenated with a comma (e.g. "d,e")
and a list that contains the unique elements of that vector, where in case of comma concatenated elements I do not keep their individual components.
So this is the list:
l = list("a","b","c","d,e","f")
I am looking for an efficient way to obtain the indices of the elements of a in the l list. For elements of a that are represented by the comma concatenated elements in l it should return the indices of the these comma concatenated elements in l.
So the output of this function would be:
c(1,2,3,4,4,4,5)
As you can see it returns index 4 for a elements: "d", "e", and "d,e"
I would make your search vector into a set of regular expressions, by substituting the comma with a pipe. Add names to the search vector too, according to its position in the list.
L <- setNames(lapply(l, gsub, pattern = ",", replacement = "|"), seq_along(l))
Then you can do:
lapply(L, function(x) grep(x, a, value = TRUE))
# $`1`
# [1] "a"
#
# $`2`
# [1] "b"
#
# $`3`
# [1] "c"
#
# $`4`
# [1] "d" "e" "d,e"
#
# $`5`
# [1] "f"
The names are important, because you can now use stack to get what you are looking for.
stack(lapply(L, function(x) grep(x, a, value = TRUE)))
# values ind
# 1 a 1
# 2 b 2
# 3 c 3
# 4 d 4
# 5 e 4
# 6 d,e 4
# 7 f 5
You could use a strategy with factors. First, find the index for each element in your list with
l <- list("a","b","c","d,e","f")
idxtr <- Map(function(x) unique(c(x, strsplit(x, ",")[[1]])), unlist(l))
This build a list for each item in l along with all possible matches for each element. Then we take the vector a and create a factor with those levels, and then reassign based on the list we just build
a <- c("a","b","c","d","e","d,e","f")
a <- factor(a, levels=unlist(idxtr));
levels(a) <- idxtr
as.numeric(a)
# [1] 1 2 3 4 4 4 5
finally, to get the index, we use as.numeric on the factor