2

どちらか一方の df で発生する na 値を持つ 2 つの data.frame をマージするのに苦労しています。

sampleA <- structure(list(Nom_xp = "A1MRJ", Rep = 1L, GB05 = 102L, GB05.1 = 102L, 
    GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 169L, 
    GB27.1 = 169L, GB24 = 240L, GB24.1 = 242L, GB28 = NA_integer_, 
    GB28.1 = NA_integer_, GB15 = 142L, GB15.1 = 144L, GB02 = 197L, 
    GB02.1 = 197L, GB10 = 126L, GB10.1 = 134L, GB14 = 181L, GB14.1 = 193L), .Names = c("Nom_xp", 
"Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1", 
"GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15", 
"GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"
), row.names = 32L, class = "data.frame")


sampleB <- structure(list(Nom_xp = "A1MRJ", Rep = 2L, GB05 = NA, GB05.1 = NA, 
    GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 169L, 
    GB27.1 = 169L, GB24 = 240L, GB24.1 = 242L, GB28 = 390L, GB28.1 = 390L, 
    GB15 = 142L, GB15.1 = 144L, GB02 = 197L, GB02.1 = 197L, GB10 = 126L, 
    GB10.1 = 134L, GB14 = 181L, GB14.1 = 193L), .Names = c("Nom_xp", 
"Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1", 
"GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15", 
"GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"
), row.names = 33L, class = "data.frame")

data.frame として必要な出力。「Nom_xp」に一致するのは 1 行ごとに 1 行だけなので、値がいずれかの DF に存在する場合、NA get は A または B のいずれかの値に置き換えられます。

Nom_xp  GB05  GB05  GB18  GB18  GB06  GB06  GB27  GB27  GB24  GB24  GB28    GB28    GB15  GB15  GB02  GB02  GB10  GB10  GB14  GB14
A1MRJ   102 102 177 177 240 240 169 169 240 242 390 390 142 144 197 197 126 134 181 193

私はそれを考えていただろう：

output <- merge(A,B,by="Nom_xp",all.x=T,all.y=T)

また

output <- join(A,B,by="Nom_xp",match="all")

私が必要なものを私に与えるでしょうが、今のところ運がありません...何が欠けていますか? 実際の data.frame には複数の行があります。

4

2 に答える 2

1

一列だけですか？では、これで十分ではないでしょうか。次のように結果を取得できますsampleB。

sampleB[, is.na(sampleB)] <- sampleA[, is.na(sampleB)]

いいえ、適用、結合、マージはここでは必要ないと思います。テストされていませんが、これは機能します。

sampleB[is.na(sampleB)] <- sampleA[is.na(sampleB)]
于 2013-02-13T22:05:06.920 に答える
0

データセット全体がどのように見えるか完全にはわかりませんが、2つだけでなく、同じ「Nom_xp」を持つサンプルがいくつかあると思いますか? そして、おそらくすべてのデータが大きなデータフレームなどにあると思いますか?

もしそうなら、このコードは良いスタートになるかもしれません (誰かが助けて、これをより効率的に書き直すことができるでしょうか?)。とにかく：

sampleA <- structure(list(Nom_xp = "A1MRJ", Rep = 1L, GB05 = 102L, GB05.1 = 102L, 
                          GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 169L, 
                          GB27.1 = 169L, GB24 = 240L, GB24.1 = 242L, GB28 = NA_integer_, 
                          GB28.1 = NA_integer_, GB15 = 142L, GB15.1 = 144L, GB02 = 197L, 
                          GB02.1 = 197L, GB10 = 126L, GB10.1 = 134L, GB14 = 181L, GB14.1 = 193L), .Names = c("Nom_xp", "Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1","GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15","GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"), row.names = 32L, class = "data.frame")

sampleB <- structure(list(Nom_xp = "A1MRJ", Rep = 2L, GB05 = NA, GB05.1 = NA, 
                          GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 169L, 
                          GB27.1 = 169L, GB24 = 240L, GB24.1 = 242L, GB28 = 390L, GB28.1 = 390L, 
                          GB15 = 142L, GB15.1 = 144L, GB02 = 197L, GB02.1 = 197L, GB10 = 126L, 
                          GB10.1 = 134L, GB14 = 181L, GB14.1 = 193L), .Names = c("Nom_xp","Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1", "GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15", "GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"  ), row.names = 33L, class = "data.frame")

sampleC <- structure(list(Nom_xp = "ASDF", Rep = 2L, GB05 = NA, GB05.1 = NA, 
                          GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 12349L, 
                          GB27.1 = 3, GB24 = 234112, GB24.1 = 242L, GB28 = 234, GB28.1 = 390L, 
                          GB15 = NA, GB15.1 = 144L, GB02 = 197L, GB02.1 = 197L, GB10 = 126L, 
                          GB10.1 = 134L, GB14 = NA, GB14.1 = 193L), .Names = c("Nom_xp", "Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1", "GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15", "GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"), row.names = 34L, class = "data.frame")

sampleD <- structure(list(Nom_xp = "ASDF", Rep = 2L, GB05 = 214, GB05.1 = 34, 
                          GB18 = 177L, GB18.1 = 177L, GB06 = 240L, GB06.1 = 240L, GB27 = 169L, 
                          GB27.1 = 3, GB24 = NA, GB24.1 = 242L, GB28 = 234, GB28.1 = 390L, 
                          GB15 = 56, GB15.1 = 144L, GB02 = 197L, GB02.1 = 197L, GB10 = 15466L, 
                          GB10.1 = 134L, GB14 = 34, GB14.1 = 193L), .Names = c("Nom_xp", "Rep", "GB05", "GB05.1", "GB18", "GB18.1", "GB06", "GB06.1", "GB27", "GB27.1", "GB24", "GB24.1", "GB28", "GB28.1", "GB15", "GB15.1", "GB02", "GB02.1", "GB10", "GB10.1", "GB14", "GB14.1"), row.names = 35L, class = "data.frame")

cdat<-rbind(sampleA,sampleB,sampleC,sampleD) #simulating your data set (?)
dcols<-dim(cdat)[2]

mat<-matrix(nrow=length(unique(cdat$Nom_xp)),ncol=dcols)
colnames(mat)<-colnames(cdat)
for (j in 1:length(unique(cdat$Nom_xp))) 
{
  g<-grep(unique(cdat$Nom_xp)[j],cdat$Nom_xp)   #Get the Nom_xp rows that match
  mat[j,1]<-cdat[g[1],1]                        #Fill in the "Nom_xp"
  mat[j,2]<-paste(g,collapse=" ")               #Fill in the "rep"
  mat[j,3:dcols]<-apply(cdat[g,3:dcols],2,      #Calculate a mean for each column
   function(x){as.numeric(mean(x,na.rm=T))})          
}
于 2013-02-13T23:18:11.383 に答える