根据向量重叠从列表中删除元素并将元素添加到列表 [英] removing and appending elemets from and to a list based on a vector overlap
问题描述
我有一个包含约500个模型对象的列表.该对象的名称为v1:
I have a list with ~ 500 model objects. The names of this objects are v1:
existing.list <- vector("list", 3)
v1 <- names(existing.list) <- c("A", "B", "C")
我现在得到了不同的数据集,我也需要对其进行建模,并将其保存在同一列表中.此新数据集中的对象与existing.list
中的某些对象重叠.因为这非常耗时,所以我确实想保留旧的结果.此新数据集的名称为v2:
I now get different dataset, which i need to model, too, and save in the same list. The objects in this new dataset are overlapping with the some of the objects in existing.list
. Because it is very time-consuming, i do want to keep the old results. The names of this new dataset are v2:
v2 <- c("B", "C", "D")
我首先要删除v1中的对象,而不是v2中的对象,然后将其附加到现存.列出v2中所有新的唯一名称. 我可以用相当复杂的方式完成第一个任务:
I first want to remove the objects in v1, which are not in v2, then append to existing.list all the new, unique names from v2. I can do the first task in a rather complicated way:
rm <- v1[!v1 %in% v2]
rm.i <- which(v1 %in% rm)
v1 <- v1[-rm.i]
但是后来我无法追加新对象,这取决于v2中的唯一元素:
But then i fail at appending the new objects, as determined by the unique elements in v2:
new.elements <- v2[!v2 %in% v1]
所需的输出是修改后的existing.list
,具有完整的元素"B"和"C"以及新的空元素"D".基本上,它是一个列表,其中的元素由v2
中的名称确定,但是由于多种原因,仅创建一个新列表并将existing.list
的一部分复制到其中会很复杂.
由于我需要针对多个列表执行此操作,因此比我现在更简单的方法将很方便.
The desired output is a modified existing.list
, with intact elements "B" and "C" and a new empty element "D". Basically, its a list with elements determined by the names in v2
, but for a number of reasons it would be complicated to just create a new list and copy parts of existing.list
to it.
Since i need to do this for a number of lists, a less complicated way than i am doing now would be handy.
非常感谢您!这是项目的最后一分钟,因此对您的帮助深表感谢!
Thank you very much in advance! This is a last minute addition to a project, so any help is highly appreciated!
此问题基于上一个问题,我用草率的措词造成了混乱.感谢那些仍在努力帮助我的用户.
this question is based on a previous question, which i sloppily worded and thus created confusion. My thanks to those users, who still tried to help me.
推荐答案
如果我对您的理解正确,那么您可以首先获取v2
中而不是v1
If I understood you correctly you can first get the names of elements that are in v2
but not in v1
tmp <- setdiff(v2, v1) # "D"
然后是子集existing.list
并按如下所示附加它
And then subset existing.list
and append it as follows
existing.list <- c(existing.list[v1 %in% v2],
setNames(vector("list", length(tmp)), tmp))
结果
existing.list
#$B
#NULL
#$C
#NULL
#$D
#NULL
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