r - 在 R : How to replace NA in a Vector found between two integers

Question

我有以下向量:

A:(NA NA NA NA 1 NA NA 4 NA NA 1 NA NA NA NA NA 4 NA 1 NA 4)

我想用 2 替换 1 和 4 之间的所有 Nas(但不是 4 和 1 之间的 Nas)

您是否会为此任务推荐/使用任何方法？

它也可以作为数据帧进行管理:

 A 
----
 NA 
 NA 
 NA 
 NA 
 1 
 NA 
 NA 
 4 
 NA 
 NA 
 1 
 NA 
 NA 
 NA 
 NA 
 NA
 4 
 NA 
 1
 NA 
 4
----

编辑 :
1.我将字符串“Na”更改为NA。

解决方案/更新
感谢大家的见解。我从他们那里学到了针对我的案例提出以下解决方案。我希望它对其他人有用:

A <- c(df$A)

index.1<-which(df$A %in% c(1)) # define location for 1s in A
index.14<-which(df$A %in% c(1,4)) # define location for 1s and 4s in A

loc.1<-which(index.14 %in% index.1) # location of 1s in  index.14
loc.4<-loc.1+1 # location of 4s relative to 1s in index.14

start.i<-((index.14[loc.1])+1) # starting index for replacing with 2
end.i<-((index.14[loc.4])-1) # ending index for replacing with 2 in index

fill.v<-sort(c(start.i, end.i))# sequence of indexes to fill-in with # 2

# create matrix of beginning and ending sequence
fill.m<-matrix(fill.v,nrow = (length(fill.v)/2),ncol = 2, byrow=TRUE) 

# create a list with indexes to replace
list.1<-apply(fill.m, MARGIN=1,FUN=function(x) seq(x[1],x[2])) 

# unlist list to use as the indexes for replacement
list.2<-unlist(list.1) 

df$A[list.2] <- 2 # replace indexed location with 2

Answer 1

假设 A如末尾的注释中可重复显示的那样，显示的 cumsum 的差异为 1 和 4 之间的元素提供了 TRUE，并且下一个条件消除了端点。最后，我们将剩下的位置为 TRUE 的位置替换为 2。

replace(A, (cumsum(A == 1) - cumsum(A == 4)) & (A == "Na"), 2)

给予:

 [1] "Na" "Na" "Na" "Na" "1"  "2"  "2"  "4"  "Na" "Na" "1"  "2"  "2"  "2"  "2" 
[16] "2"  "4"  "Na" "1"  "2"  "4"

NA值

R 区分大小写，Na 与 NA 不同。问题中的样本数据显示的是 Na 值而不是 NA 值，但如果实际上意味着是具有 NA 值的数值向量，如 AA在下面的注释中，然后将表达式修改为如下所示:

replace(AA, cumsum(!is.na(AA) & AA == 1) - cumsum(!is.na(AA) & AA == 4) & is.na(AA), 2)

给予:

[1] NA NA NA NA  1  2  2  4 NA NA  1  2  2  2  2  2  4 NA  1  2  4

笔记

A <- c("Na", "Na", "Na", "Na", "1", "Na", "Na", "4", "Na", "Na", 
"1", "Na", "Na", "Na", "Na", "Na", "4", "Na", "1", "Na", "4")

AA <- as.numeric(replace(A, A == "Na", NA))