如何基于对象索引合并两个列表-保留属性? [英] How to merge two lists based on object indices - keeping attributes?

问题描述

我想合并两个列表，保留每个对象的索引:

I want to merge two lists keeping the index of each object:

mylist<-list(1,NULL,2)
otherlist<-list(NULL,3,NULL,4,5,6)

# Desired

list(1,3,2,4,5,6)

# my try:

suppressWarnings(mapply(c, mylist, otherlist) )

答案应该是普遍的

为了避免类似问题的泛滥.我决定在这里要求保留属性(最好是保留基数)的可能性.

In order to avoid proliferation of similar questions. I decided to request here also the possibility of keeping attributes (preferably with base).

mylist<-list(1,NULL,2)
attr(mylist[[1]],"at")<-"a"
attr(mylist[[3]],"at")<-"c"

otherlist<-list(NULL,3,NULL,4,5,6)

attr(otherlist[[2]],"at")<-"b"
attr(otherlist[[4]],"at")<-"d"
attr(otherlist[[5]],"at")<-"e"
attr(otherlist[[6]],"at")<-"f"

推荐答案

这里是一个选项，其中我们使用lengths创建逻辑索引(当存在NULL时将返回0)，并使用mylist不公开

Here is an option where we create a logical index with lengths (which will return 0 when there is NULL) and use to assign the elements with mylist unlisted

otherlist[lengths(otherlist) == 0] <- unlist(mylist)
otherlist
#[[1]]
#[1] 1

#[[2]]
#[1] 2

#[[3]]
#[1] 3

#[[4]]
#[1] 4

#[[5]]
#[1] 5

#[[6]]
#[1] 6

---

如果需要使用Map，请确保相应元素的lengths相同

---

If we need to use Map, make sure the lengths are the same for the corresponding elements

otherlist[seq_along(mylist)] <- Map(c, otherlist[seq_along(mylist)], mylist)

更新

对于更新后的示例

Update

For the updated example

i1 <- sapply(otherlist, is.null)
i2 <- !sapply(mylist, is.null)
otherlist[i1] <- mylist[i2]

otherlist
#[[1]]
#[1] 1
#attr(,"at")
#[1] "a"

#[[2]]
#[1] 3
#attr(,"at")
#[1] "b"

#[[3]]
#[1] 2
#attr(,"at")
#[1] "c"

#[[4]]
#[1] 4
#attr(,"at")
#[1] "d"

#[[5]]
#[1] 5
#attr(,"at")
#[1] "e"

#[[6]]
#[1] 6
#attr(,"at")
#[1] "f"

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