R 将缺少的列值替换为'；匹配'；柱

这是我的后续问题 这已经得到了充分的解决。我的问题是关于多个匹配列

示例数据集：

s <- data.frame(ID=c(191, 282, 202, 210), 
            Group.1=c(NA, "A", NA, "B"), 
            Back.1=c("DD", "AA", "DD", "BB"), 
            Group.2=c("D","A", NA, "B"),
            Back.2=c("DD", "BB", "CC", "AA"),
            stringsAsFactors=FALSE)

   ID Group.1 Back.1 Group.2 Back.2
1 191    <NA>     DD       D     DD
2 282       A     AA       A     BB
3 202    <NA>     DD    <NA>     CC
4 210       B     BB       B     AA

所以，如果我有许多匹配的列，我希望能够在它们之间映射、循环或应用任何内容。尝试循环函数会产生：

for (i in 1:2){
  s[paste0("Group.", i)] <- ifelse(test = !is.na(s[paste0("Group.", i)]), 
                                   yes = s[paste0("Group.", i)], 
                                   no = s[paste0("Back.", i)])
}

Warning messages:
1: In `[<-.data.frame`(`*tmp*`, paste0("Group.", i), value = list(c("DD",  :
  provided 4 variables to replace 1 variables
2: In `[<-.data.frame`(`*tmp*`, paste0("Group.", i), value = list(c("D",  :
  provided 4 variables to replace 1 variables
> s
   ID Group.1 Back.1 Group.2 Back.2
1 191      DD     DD       D     DD
2 282      AA     AA       A     BB
3 202      DD     DD    <NA>     CC
4 210      BB     BB       B     AA

for（1:2中的i）{
s[paste0（“Group.”，i）]我们可以使用

gr1 <- grep("Group", names(s), value = TRUE)
bc1 <- grep("Back", names(s), value = TRUE)
setDT(s)
for(j in seq_along(gr1)){
    s[is.na(get(gr1[j])), (gr1[j]) := get(bc1[j])]
}

s
#    ID Group.1 Back.1 Group.2 Back.2
#1: 191      DD     DD       D     DD
#2: 282       A     AA       A     BB
#3: 202      DD     DD      CC     CC
#4: 210       B     BB       B     AA

gr1为什么你有两列的名称是Back.1如果你用长格式做这件事会变得非常简单
long预期的'Donk.1'是什么？抱歉@waterling拾起了Back.1。1添加了两次。应该只添加一次。以后会编辑吗？在上一个示例中，不清楚应该从哪个列替换哪个列？你有Back.1、Back.2和B确认1.1存在第1组、第2组和第1组时，用
gr1@waterling抱歉，我不明白为什么需要它。我是根据逻辑
setDT（s）[is.na（Group.1），Group.1:=Back.1]尝试的
在OP的codeOP中提供，OP希望能够用
Back.1
替换
Donk.1
，或者用
Back.Number
替换一些
arbitrarytext.Number
。这句话的意思是：“推广到其他命名列，以便与Back.x进行数字匹配的列也可以有Back.x插补的缺失值。”@谢谢，我错过了那部分。我更新了帖子，假设专栏是交替的
for (i in 1:2){
  s[paste0("Group.", i)] <- ifelse(test = !is.na(s[paste0("Group.", i)]), 
                                   yes = s[paste0("Group.", i)], 
                                   no = s[paste0("Back.", i)])
}

Warning messages:
1: In `[<-.data.frame`(`*tmp*`, paste0("Group.", i), value = list(c("DD",  :
  provided 4 variables to replace 1 variables
2: In `[<-.data.frame`(`*tmp*`, paste0("Group.", i), value = list(c("D",  :
  provided 4 variables to replace 1 variables
> s
   ID Group.1 Back.1 Group.2 Back.2
1 191      DD     DD       D     DD
2 282      AA     AA       A     BB
3 202      DD     DD    <NA>     CC
4 210      BB     BB       B     AA

s <- data.frame(ID=c(191, 282, 202, 210), 
            Group.1=c(NA, "A", NA, "B"), 
            Back.1=c("DD", "AA", "DD", "BB"), 
            Group.2=c("D","A", NA, "B"),
            Back.2=c("DD", "BB", "CC", "AA"),
            Donk.1 =c("PP", "ZZ", NA, "QQ"),
            stringsAsFactors=FALSE)

gr1 <- grep("Group", names(s), value = TRUE)
bc1 <- grep("Back", names(s), value = TRUE)
setDT(s)
for(j in seq_along(gr1)){
    s[is.na(get(gr1[j])), (gr1[j]) := get(bc1[j])]
}

s
#    ID Group.1 Back.1 Group.2 Back.2
#1: 191      DD     DD       D     DD
#2: 282       A     AA       A     BB
#3: 202      DD     DD      CC     CC
#4: 210       B     BB       B     AA

gr1 <- names(s)[seq(2, ncol(s), by = 2)]
bc1 <- names(s)[seq(3, ncol(s), by = 2)]

setDT(s)
for(j in seq_along(gr1)){
    s[is.na(get(gr1[j])), (gr1[j]) := get(bc1[j])][]
}
s
#    ID Group.1 Back.1 Group.2 Back.2 Donk.1 Back.1.1
#1: 191      DD     DD       D     DD     PP       DD
#2: 282       A     AA       A     BB     ZZ       AA
#3: 202      DD     DD      CC     CC     DD       DD
#4: 210       B     BB       B     AA     QQ       BB

s <- data.frame(ID=c(191, 282, 202, 210), 
        Group.1=c(NA, "A", NA, "B"), 
        Back.1=c("DD", "AA", "DD", "BB"), 
        Group.2=c("D","A", NA, "B"),
        Back.2=c("DD", "BB", "CC", "AA"),
        Donk.1 =c("PP", "ZZ", NA, "QQ"),
        Back.1=c("DD", "AA", "DD", "BB"), 
        stringsAsFactors=FALSE)