我正在使用R作为我的项目。我是R的新手。我有以下数据
place<-c("S1","S1","S1","S1","S2","S2","S2","S2")
product<-c("P1","P2","P3","P1","P2","P3","P1","P2")
location<-c("loc1","loc1","loc2","loc2","loc1","loc1","loc2","loc2")
profit<-c(55,80,70,90,30,40,15,20)
data<-data.frame(place,product,location,profit)
我想要每个地方，哪个产品在每个位置给出最大利润，在输出中它将添加一个带有二进制条目的列，其中1对应于向量中利润最大的位置，如下面的方式：
solution<-c(0,1,1,0,0,1,0,0)
希望我的问题很明确。 提前谢谢。
答案 0 :(得分：2)
您可以使用ave：
transform(data, solution = ave(profit, place, location,
FUN = function(x) as.integer(x == max(x))))
place product location profit solution
1 S1 P1 loc1 55 0
2 S1 P2 loc1 80 1
3 S1 P3 loc2 70 0
4 S1 P1 loc2 90 1
5 S2 P2 loc1 30 0
6 S2 P3 loc1 40 1
7 S2 P1 loc2 15 0
8 S2 P2 loc2 20 1
答案 1 :(得分：0)
这个例子你期望这个向量吗？ 如果2个不同位置有2个不同的位置，“解决方案”如何只包含3“1”？
这是我的解决方案：
place<-c("S1","S1","S1","S1","S2","S2","S2","S2")
product<-c("P1","P2","P3","P1","P2","P3","P1","P2")
location<-c("loc1","loc1","loc2","loc2","loc1","loc1","loc2","loc2")
profit<-c(55,80,70,90,30,40,15,20)
data<-data.frame(place,product,location,profit)
# Returns a data frame with the profit max for each place at each location
df <- aggregate(data$profit, by = list(place = data$place, location = data$location), max)
# Formating names
names(df)[3] <- c("profit")
# All the lines returned are thoses you want to index with "1" in "solution
df$solution <- rep(1, nrow(df))
# Right outter join, we keep all lines of data which don't respect the join criteria (we dont precise by.x and by.y, it's a natural join on the names, it will produce NA in "solution" for missing correspondances)
data <- merge(df, data, all.y = T)
# The join produced NA solutions for lines which didn't exist in "data", we replace them by 0
data$solution[which(is.na(data$solution))] <- 0
> data
place location profit solution product
1 S1 loc1 55 0 P1
2 S1 loc1 80 1 P2
3 S1 loc2 70 0 P3
4 S1 loc2 90 1 P1
5 S2 loc1 30 0 P2
6 S2 loc1 40 1 P3
7 S2 loc2 15 0 P1
8 S2 loc2 20 1 P2
> data$solution
[1] 0 1 0 1 0 1 0 1
希望得到这个帮助。