CODEWITHSUNDEEP
rmax
Arushi
Arushi

Reputation: 271

Maximum value in column of a matrix

I am using R for my project.I am totally new to R. I have following data 

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)

I want for each place, which product is giving maximum profit at each location and in output it will add one more column with a binary entries, where 1 corresponds to position in the vector where profit is max, like in following way:

solution<-c(0,1,1,0,0,1,0,0)

Hope my question is clear. Thanks in advance.

Upvotes: 0

Views: 144

Answers (2)

Sven Hohenstein
Sven Hohenstein

Reputation: 81683

You can use 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

Upvotes: 2

Julien Navarre
Julien Navarre

Reputation: 7830

Is it this vector you expect for this example ? How can "solution" contains only 3 "1" if you have 2 differents location for 2 differents places ?

Here is my solution :

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

Hope this help.

Upvotes: 0

Related Questions