如何从列表中删除元素?

I have a list and I want to remove a single element from it. How can I do this?

我有一个列表，我想从其中删除一个元素。我该怎么做呢?

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.

我已经试过了，我认为这个函数的明显名称应该在参考手册中，我还没有找到合适的。

12 个解决方案

#1

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

我根本不知道R，但有一点有创意的谷歌搜索把我带到了这里:http://托尔斯泰。newcastle.edu.au/r/help/05/04/1919.html。

The key quote from there:

这里的关键词是:

I do not find explicit documentation for R on how to remove elements from lists, but trial and error tells me

关于如何从列表中删除元素，我没有找到关于R的显式文档，但是尝试和错误告诉我。

myList[[5]] <- NULL

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.

将删除第5个元素，然后“关闭”该元素所造成的漏洞。这就满足了索引值，所以我必须小心删除元素。我必须从名单的后面到前线工作。

A response to that post later in the thread states:

稍后在线程状态中对该post的响应:

For deleting an element of a list, see R FAQ 7.1

要删除列表中的元素，请参见R FAQ 7.1。

And the relevant section of the R FAQ says:

R FAQ的相关部分说:

... Do not set x[i] or x[[i]] to NULL, because this will remove the corresponding component from the list.

…不要将x[i]或x[[i]]设置为NULL，因为这将从列表中删除相应的组件。

Which seems to tell you (in a somewhat backwards way) how to remove an element.

这似乎告诉你(在某种程度上向后)如何移除一个元素。

Hope that helps, or at least leads you in the right direction.

希望这能有所帮助，或者至少引导你走向正确的方向。

#2

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".

如果您不想对列表进行修改(例如，将列表中的元素删除为一个函数)，则可以使用索引:负索引表示“不包含该元素”。

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:

此外，逻辑索引向量是有用的:

x[x != "b"]; # without elements that are "b"

This works with dataframes, too:

这也适用于dataframes:

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

#3

Here is how the remove the last element of a list in R:

下面是在R中删除列表最后一个元素的方法:

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:

如果x可能是一个向量那么你需要创建一个新的对象:

x <- c("a", "b", "c", "d", "e")
x <- x[-length(x)]

• Work for lists and vectors
• 为列表和向量工作。

#4

Removing Null elements from a list in single line :

从单行中删除列表中的Null元素:

x=x[-(which(sapply(x,is.null),arr.ind=TRUE))]

x = x(((酸式焦磷酸钠(x,is.null)arr.ind = TRUE)))

Cheers

干杯

#5

If you have a named list and want to remove a specific element you can try:

如果您有一个指定的列表，并且想要删除一个特定的元素，您可以尝试:

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).

这将使用元素a、b、c列出清单1。第二行删除元素b后，检查它是否存在(以避免@hjv提到的问题)。

or better:

或更好:

lst$b <- NULL

This way it is not a problem to try to delete a non-existent element (e.g. lst$g <- NULL)

这样，删除一个不存在的元素就不是问题了(例如，lst$g <- NULL)

#6

There's the rlist package (http://cran.r-project.org/web/packages/rlist/index.html) to deal with various kinds of list operations.

这里有一个rlist包(http://cran.r project.org/web/packages/rlist/index.html)来处理各种类型的列表操作。

Example (http://cran.r-project.org/web/packages/rlist/vignettes/Filtering.html):

示例(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:

结果:

# $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

#7

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.

不知道您是否还需要一个答案，但是我从我有限的(3周的自学R)经验中发现，使用NULL赋值实际上是错误的或者是次优化的，特别是当你在一个for循环中动态更新一个列表时。

To be more precise, using

更确切地说，使用。

myList[[5]] <- NULL

will throw the error

将抛出错误

myList[[5]] <- NULL : replacement has length zero

myList[[5]] <- NULL:替换长度为0。

or

或

more elements supplied than there are to replace

提供的元素多于替换的元素。

What I found to work more consistently is

我发现工作更一致的是!

myList <- myList[[-5]]

#8

In the case of named lists I find those helper functions useful

在命名列表的情况下，我发现这些helper函数是有用的。

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

#9

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:

只是想快速添加(因为我没有在任何答案中看到它)，对于一个已命名的列表，您也可以使用l["name"] <- NULL。例如:

l <- list(a = 1, b = 2, cc = 3)
l['b'] <- NULL

#10

Using lapply and grep:

使用拉普和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))]

#11

How about this? Again, using indices

这个怎么样?再次,使用指数

> m <- c(1:5)
> m
[1] 1 2 3 4 5

> m[1:length(m)-1]
[1] 1 2 3 4

or

或

> m[-(length(m))]
[1] 1 2 3 4

#12

You can use which.

您可以使用。

x<-c(1:5)
x
#[1] 1 2 3 4 5
x<-x[-which(x==4)]
x
#[1] 1 2 3 5