我有一个列表，我想从其中删除一个元素。我该怎么做呢?
我试着在参考手册中查找我认为这个函数的明显名称，但我没有找到任何合适的名称。
当前回答
这里有一个简单的解决方案，可以使用底数r。它从原始的数字列表中删除数字5。您可以使用相同的方法从列表中删除任何您想要的元素。
#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
}
}
new_list变量不再包含5。
new_list
# [1] 1 2 3 4 6 7 8 9 10
其他回答
rlist包(http://cran.r-project.org/web/packages/rlist/index.html)用于处理各种列表操作。
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"))
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
这个怎么样?还是使用下标
> 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
使用lapply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))]
下面是如何在R中删除列表的最后一个元素:
x <- list("a", "b", "c", "d", "e")
x[length(x)] <- NULL
如果x可能是一个向量，那么你需要创建一个新对象:
x <- c("a", "b", "c", "d", "e")
x <- x[-length(x)]
处理列表和向量
这里有一个简单的解决方案，可以使用底数r。它从原始的数字列表中删除数字5。您可以使用相同的方法从列表中删除任何您想要的元素。
#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
}
}
new_list变量不再包含5。
new_list
# [1] 1 2 3 4 6 7 8 9 10
