我想通过匹配值从基础创建组。
我有以下数据表:
now<-c(1,2,3,4,24,25,26,5,6,21,22,23)
before<-c(0,1,2,3,23,24,25,4,5,0,21,22)
after<-c(2,3,4,5,25,26,0,6,0,22,23,24)
df<-as.data.frame(cbind(now,before,after))
再现以下数据:
now before after
1 1 0 2
2 2 1 3
3 3 2 4
4 4 3 5
5 24 23 25
6 25 24 26
7 26 25 0
8 5 4 6
9 6 5 0
10 21 0 22
11 22 21 23
12 23 22 24
我想得到:
now before after group
1 1 0 2 A
2 2 1 3 A
3 3 2 4 A
4 4 3 5 A
5 5 4 6 A
6 6 5 0 A
7 21 0 22 B
8 22 21 23 B
9 23 22 24 B
10 24 23 25 B
11 25 24 26 B
12 26 25 0 B
我想在不使用“for”循环的情况下得到答案,因为实际数据太大了。
任何你能提供的将不胜感激。
答案 0 :(得分:0)
这是一种方法。很难避免for循环,因为这是一个非常棘手的算法。对他们的反对通常是基于优雅而不是速度,但有时他们是完全合适的。
df$group <- seq_len(nrow(df)) #assign each row to its own group
stop <- FALSE #indicates convergence
while(!stop){
pre <- df$group #group column at start of loop
for(i in seq_len(nrow(df))){
matched <- which(df$before==df$now[i] | df$after==df$now[i]) #check matches in before and after columns
group <- min(df$group[i], df$group[matched]) #identify smallest group no of matching rows
df$group[i] <- group #set to smallest group
df$group[matched] <- group #set to smallest group
}
if(identical(df$group, pre)) stop <- TRUE #stop if no change
}
df$group <- LETTERS[match(df$group, sort(unique(df$group)))] #convert groups to letters
#(just use match(...) to keep them as integers - e.g. if you have more than 26 groups)
df <- df[order(df$group, df$now),] #reorder as required
df
now before after group
1 1 0 2 A
2 2 1 3 A
3 3 2 4 A
4 4 3 5 A
8 5 4 6 A
9 6 5 0 A
10 21 0 22 B
11 22 21 23 B
12 23 22 24 B
5 24 23 25 B
6 25 24 26 B
7 26 25 0 B