list R + 将向量列表合并为一个向量
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R + combine a list of vectors into a single vector
提问by Rachit Agrawal
I have a single list of numeric vector and I want to combine them into one vector. But I am unable to do that. This list can have one element common across the list element. Final vector should not add them twice. Here is an example:
我有一个数字向量列表，我想将它们组合成一个向量。但我无法做到这一点。该列表可以有一个跨列表元素通用的元素。最终向量不应将它们添加两次。下面是一个例子：
>lst
`1`
[1] 1 2
`2`
[2] 2 4 5
`3`
[3] 5 9 1
I want final result as this
我想要这样的最终结果
>result
[1] 1 2 4 5 9 1
I tried doing following things, without worrying about the repition:
我尝试做以下事情，而不用担心 reition：
>vec<-vector()
>sapply(lst, append,vec)
and
和
>vec<-vector()
>sapply(lst, c, vec)
None of them worked. Can someone help me on this?
他们都没有工作。有人可以帮我吗？
Thanks.
谢谢。
采纳答案by Rachit Agrawal
A solution that is faster than the one proposed above:
一种比上面提出的更快的解决方案：
vec<-unlist(lst)
vec[which(c(1,diff(vec)) != 0)]
回答by Paul Rougieux
Another answer using Reduce().
另一个答案使用Reduce().
Create the list of vectors:
创建向量列表：
lst <- list(c(1,2),c(2,4,5),c(5,9,1))
Combine them into one vector
将它们合并为一个向量
vec <- Reduce(c,lst)
vec
# [1] 1 2 2 4 5 5 9 1
Keep the repeated ones only once:
重复的只保留一次：
unique(Reduce(c,lst))
#[1] 1 2 4 5 9
If you want to keep that repeated one at the end, You might want to use vec[which(c(1,diff(vec)) != 0)]as in @Rachid's answer
如果你想在最后保留那个重复的，你可能想vec[which(c(1,diff(vec)) != 0)]在@Rachid 的回答中使用
回答by 0mn1
stack will do this nicely too, and looks more concise:
stack 也可以很好地做到这一点，并且看起来更简洁：
stack(lst)$values
回答by Matthew Lundberg
You want rle:
你想要rle：
rle(unlist(lst))$values
> lst <- list(`1`=1:2, `2`=c(2,4,5), `3`=c(5,9,1))
> rle(unlist(lst))$values
## 11 21 22 31 32 33
## 1 2 4 5 9 1
回答by MartijnVanAttekum
Doing it the tidyverse way:
以 tidyverse 的方式进行：
library(tidyverse)
lst %>% reduce(c) %>% unique
This uses the (uncapitalized) reduceversion from purrr in combination with pipes. Also note that if the list contains namedvectors, the final naming will be different depending on whether unlistor reducemethods are used.
这将reducepurrr的（未大写）版本与管道结合使用。另请注意，如果列表包含命名向量，则最终命名将根据是否使用unlist或reduce方法而有所不同。
回答by Prradep
Benchmarking the two answers by Rachitand Martijn
rbenchmark::benchmark(
"unlist" = {
vec<-unlist(a)
vec[which(diff(vec) != 0)]
},
"reduce" = {
a %>% reduce(c) %>% unique
}
)
Output:
输出：
test replications elapsed relative user.self sys.self user.child sys.child
2 reduce 100 0.036 3 0.036 0.000 0 0
1 unlist 100 0.012 1 0.000 0.004 0 0
Thisone clearly beat the other one.
这一个明显击败了另一个。
