采取以下簡單的鏈接ID數(shù)據(jù)框：test <- data.frame(id1=c(10,10,1,1,24,8),id2=c(1,36,24,45,300,11))> test id1 id21 10 12 10 363 1 244 1 455 24 3006 8 11我現(xiàn)在要將所有鏈接的ID組合在一起?！版溄印笔侵赴凑真溄渔溸M行操作，以便將一組中的所有ID一起標記。一種分支結(jié)構(gòu)。即：Group 110 --> 1, 1 --> (24,45) 24 --> 300 300 --> NULL 45 --> NULL10 --> 36, 36 --> NULL,Final group members: 10,1,24,36,45,300Group 28 --> 11 11 --> NULLFinal group members: 8,11現(xiàn)在，我大致了解了我想要的邏輯，但是不知道如何優(yōu)雅地實現(xiàn)它。我正在考慮遞歸地使用match或%in%遍歷每個分支，但是這次確實很困惑。我要追求的最終結(jié)果是：result <- data.frame(group=c(1,1,1,1,1,1,2,2),id=c(10,1,24,36,45,300,8,11))> result group id1 1 102 1 13 1 244 1 365 1 456 1 3007 2 88 2 11
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不使用包：
# 2 sets of test data
mytest <- data.frame(id1=c(10,10,3,1,1,24,8,11,32,11,45),id2=c(1,36,50,24,45,300,11,8,32,12,49))
test <- data.frame(id1=c(10,10,1,1,24,8),id2=c(1,36,24,45,300,11))
grouppairs <- function(df){
# from wide to long format; assumes df is 2 columns of related id's
test <- data.frame(group = 1:nrow(df),val = unlist(df))
# keep moving to next pair until all same values have same group
i <- 0
while(any(duplicated(unique(test)$val))){
i <- i+1
# get group of matching values
matches <- test[test$val == test$val[i],'group']
# change all groups with matching values to same group
test[test$group %in% matches,'group'] <- test$group[i]
}
# renumber starting from 1 and show only unique values in group order
test$group <- match(test$group, sort(unique(test$group)))
unique(test)[order(unique(test)$group), ]
}
# test
grouppairs(test)
grouppairs(mytest)
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