合并两个data.frame，将所有匹配行保留在R中_R

合并两个data.frame，将所有匹配行保留在R中

合并两个data.frame，将所有匹配行保留在R中,r,R,我正在努力合并两个data.frame，其中一个df中出现na值 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_

我正在努力合并两个data.frame，其中一个df中出现na值

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")

我会认为：

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

output您只有一行吗？那么，这还不够吗？您可以在sampleB
中获得以下结果：
sampleB[, is.na(sampleB)] <- sampleA[, is.na(sampleB)]

sampleB[，is.na（sampleB）]不能完全确定您的整个数据集是什么样子，但我假设您可以有几个具有相同“Nom_xp”的样本，而不仅仅是2个？你可能把所有的数据都放在一个大数据框里
如果是这样的话，也许这段代码是一个好的开始（也许有人可以帮忙，重新编写这段代码会更有效率？）。无论如何：
不，我有很多行。我没有想过使用其他东西，但加入或合并。。。我可能走错方向了…如果Samam辩诉和sampleB在相同的顺序中有相同的行，这就行了？是的，你是对的。我建议这样做是因为OP之前对执行apply感兴趣，这让我认为他是在逐行比较替换值。我认为逐行比较是可能的，因为两个数据帧具有相同顺序的相同行数。明天早上我会尽快试一试。又来了一堆！如果忽略NA的话，Sam辩诉和sampleB中的行是否完全相同？
output <- join(A,B,by="Nom_xp",match="all")

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

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

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))})          
}