在if中循环并重新编码

2023年5月22日 165次阅读

我知道我的问题很简单但不适合我.这是一个小数据集.

mark1 <- c("AB", "BB", "AB", "BB", "BB", "AB", "--", "BB")
mark2 <- c("AB", "AB", "AA", "BB", "BB", "AA", "--", "BB")
mark3 <- c("BB", "AB", "AA", "BB", "BB", "AA", "--", "BB")
mark4 <- c("AA", "AB", "AA", "BB", "BB", "AA", "--", "BB")
mark5 <- c("AB", "AB", "AA", "BB", "BB", "AA", "--", "BB")
mark6 <- c("--", "BB", "AA", "BB", "BB", "AA", "--", "BB")
mark7 <- c("AB", "--", "AA", "BB", "BB", "AA", "--", "BB")
mark8 <- c("BB", "AA", "AA", "BB", "BB", "AA", "--", "BB")
mymark <- data.frame (mark1, mark2, mark3, mark4, mark5, mark6, mark7, mark8)
tmymark <- data.frame (t(mymark))
names (tmymark) <- c("P1", "P2","I1", "I2", "I3", "I4", "KL", "MN")

因此数据集变为：

      P1 P2 I1 I2 I3 I4 KL MN
mark1 AB BB AB BB BB AB -- BB
mark2 AB AB AA BB BB AA -- BB
mark3 BB AB AA BB BB AA -- BB
mark4 AA AB AA BB BB AA -- BB
mark5 AB AB AA BB BB AA -- BB
mark6 -- BB AA BB BB AA -- BB
mark7 AB -- AA BB BB AA -- BB
mark8 BB AA AA BB BB AA -- BB

我想基于P1和P2比较对mark1：8进行分类,并提供一个代码,它将创建一个新的变量：

loctype <- NULL

if (tmymark$P1 == "AB" &  tmymark$P2 == "AB"){
       loctype = "<hkxhk>"
       } else {
if (tmymark$P1== "AB" & tmymark$P2 == "BB") {
      loctype = "<lmxll>"
      } else {
      if (tmymark$P1 == "AA" & tmymark$P2 == "AB") {
       loctype = "<nnxnp>"
       } else {
        if (tmymark$P1 == "AA" & tmymark$P2 == "BB") {
        loctype = "MN"
        } else {
        if (tmymark$P1 == "BB" & tmymark$P2 == "AA"){
         loctype = "MN"
         } else {
         if (tmymark$P1 == "--" & tmymark$P2 == "AA"){
         loctype = "NR"
         }  else {
if (tmymark$P1 == "AA" & tmymark$P2 == "--"){
          loctype = "NR"
} else {
    cat ("error wrong input in P1 or P2")
    }} }}}}}

在这里我要做的是比较P1和P2值并生成一个新变量.
例如,如果tmymark $P1 ==“AB”& tmymark $P2 ==“AB”loctype应为“”.如果不是第二个条件将是应用程序等等.

这是我的错误消息.

Warning messages:
1: In if (tmymark$P1 == "AB" & tmymark$P2 == "AB") { :
  the condition has length > 1 and only the first element will be used
2: In if (tmymark$P1 == "AB" & tmymark$P2 == "BB") { :
  the condition has length > 1 and only the first element will be used

一旦生成了loctype向量,我想用这个变量中的信息重新编码tmymark：

tmymark1 <- data.frame (loctype, tmymark)      
require(car) 
for(i in 2:length(tmymark)){

        if (loctype = "<hkxhk>") {
       tmymark[[i]] <- recode (x, "AB" = "hk", "BA" = "hk", "AA" = "hh", "BB" = "kk")
       } else {
       if (loctype = "<lmxll>") {
       tmymark[[i]] <- recode ((x, "AB" = "lm", "BA" = "lm", "AA" = "--", "BB" = "kk")
       } else {

        if (loctype = "<nnxnp>") {
       tmymark[[i]] <- recode ((x, "AB" = "np", "BA" = "np", "AA" = "nn", "BB" = "--")
             } else {
       if (loctype = "MN") {
        tmymark[[i]] <- "--"
       } esle {
      if (loctype = "NR") {
        tmymark[[i]] <- "NA"
       } else {
       cat ("error wrong input code")
      } } }}}

我是在正确的轨道上吗？

编辑：预期输出

      loctype  P1 P2 I1 I2 I3 I4 KL MN 
mark1  <lmxmm> lm mm lm mm mm lm -- mm 
mark2  <hkxhk> hk hk hh kk kk hh -- kk 
mark3 <nnxnp> nn np nn -- -- nn -- -- 
 and so on

最佳答案 匹配绝对是要走的路.我将两个数据帧作为键,如下所示：

key <- data.frame(
             P1=c("AB", "AB", "AA", "AA", "BB", "--", "AA"),
             P2=c("AB", "BB", "AB", "BB", "AA", "AA", "--"),
             loctype=c("<hkxhk>", "<lmxll>", "<nnxnp>", "MN", "MN", "NR", "NR"))

key2 <- cbind(
  `<hkxhk>` = c("hk","hk","hh","kk"),
  `<lmxll>` = c("lm", "lm", "--", "kk"),
  `<nnxnp>` = c("np", "np", "nn", "--"),
  MN = rep("--", 4),
  NR = rep("NA", 4) )
rownames(key2) = c("AB","BA", "AA", "BB")

然后使用key1上的匹配来获取loctype(正如Justin所推荐的那样),并且还使用key2的rownames和列来获得所需的替换,使用矩阵索引从密钥中获取所需的值.

loctype <- key$loctype[match(with(tmymark, paste(P1, P2, sep="\b")), 
                             with(key, paste(P1, P2, sep="\b")))]
ii <- match(as.vector(as.matrix(tmymark)), rownames(key2))
jj <- rep(match(loctype, colnames(key2)), nrow(tmymark))
out <- as.data.frame(matrix(key2[cbind(ii,jj)], nrow=nrow(tmymark)))
colnames(out) <- colnames(tmymark)
rownames(out) <- rownames(tmymark)
out$loctype <- loctype

然后结果看起来像这样,缺少的值是因为我的键中没有这些组合的值.

> print(out, na="")
      P1 P2 I1 I2 I3 I4 KL MN loctype
mark1 lm kk lm kk kk lm    kk <lmxll>
mark2 hk hk hh kk kk hh    kk <hkxhk>
mark3                                
mark4 nn np nn -- -- nn    -- <nnxnp>
mark5 hk hk hh kk kk hh    kk <hkxhk>
mark6                                
mark7                                
mark8 -- -- -- -- -- --    --      MN
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