Question

这是一个后续问题 replace missing values with a value from another column已得到充分解决。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

如果我想用匹配的'Back'列替换NA，我会使用：

s$Group.1 <- ifelse(test = !is.na(s$Group.1), yes = s$Group.1, no = s$Back.1)
s$Group.2 <- ifelse(test = !is.na(s$Group.2), yes = s$Group.2, no = s$Back.2)
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

由Akrun发布，另一种方法是：

library(data.table)
setDT(s)[is.na(Group.1), Group.1:= Back.1]
setDT(s)[is.na(Group.2), Group.2:= Back.2]

因此，如果我有许多匹配的列，我希望能够映射，循环或应用或跨越它们。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

哪个似乎适用于Group.1和Back.1但不适用于Group.2，从我的角度来看很难理解警告信息。

如果有人能用适当的循环来解决这个问题，那将非常感激。更有帮助的是能够推广到其他命名列，以便Back.x的数字匹配列也可以具有Back.x推测的缺失值。即。

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)

Answer 1

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

对于更新的数据集

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

数据

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)

使用“匹配”列

1 个答案:

数据