CODEWITHSUNDEEP
raggregateweighted-average
Rodrigo
Rodrigo

Reputation: 5139

R weighted arithmetic mean

Suppose I have this data.frame in R:

ages <- data.frame(Indiv = numeric(),
    Age = numeric(),
    W = numeric())
ages[1,] <- c(1,10,2)
ages[2,] <- c(1,15,5)
ages[3,] <- c(2,5,1)
ages[4,] <- c(2,100,2)

ages

  Indiv Age W
1     1  10 2
2     1  15 5
3     2   5 1
4     2 100 2

If I do:

meanAge <- aggregate(ages$Age,list(ages$Indiv),mean)

I get the mean Age (x) for each Indiv (Group.1):

  Group.1    x
1       1 12.5
2       2 52.5

But I want to calculate the weighted arithmetic mean of Age (weight being W). If I do:

WmeanAge <- aggregate(ages$Age,list(ages$Indiv),weighted.mean,ages$W)

I get:

Error in weighted.mean.default(X[[1L]], ...) : 
  'x' and 'w' must have the same length

I think I should have:

  Group.1           x
1       1 13.57142857
2       2 68.33333333

What am I doing wrong? Thanks in advance!

Upvotes: 2

Views: 7966

Answers (4)

Jeffrey Evans
Jeffrey Evans

Reputation: 2417

Your number of weight values do not match your number of groups and so aggregate cannot collapse the groups properly. Here is a very inelegant solution using a for loop. 

ages = data.frame(Indiv=c(1,1,2,2),Age=c(10,15,5,100),W=c(2,5,1,2))

age.Indiv <- vector()
  for(i in unique(ages$Indiv)){
  age.Indiv <- append(age.Indiv, weighted.mean( ages[ages$Indiv == i ,]$Age, 
                      ages[ages$Indiv == i ,]$W))
    } 
  names(age.Indiv) <- unique(ages$Indiv)
    age.Indiv

Upvotes: 1

rbatt
rbatt

Reputation: 4807

The problem is that aggregate does not split up the w arguments – so weighted.mean is receiving subsets of ages$Age, but it is not receiving the equivalent subsets of ages$W.

Try the plyr package!! It's great. I use it in 95% of the scripts that I write.

library("plyr")

# the plyr package has functions that come in the format of  _ _ ply
# the first blank is the input format, and the second is the output format
# d = data.frame, l = list, a = array, etc.
# thus, with ddply(), you supply a data.frame (ages), and it returns a data.frame (WmeanAge)

# .data is your data set
# .variables is the name of the column (or columns!) to be used to split .data
# .fun is the function you want to apply to each subset of .data

new.weighted.mean <- function(x, ...){
   weighted.mean(x=x[,"Age"], w=x[,"W"], ...)
}

WmeanAge <- ddply(.data=ages, .variables="Indiv", .fun=new.weighted.mean, na.rm=TRUE)
print(WmeanAge)

Upvotes: 2

MrFlick
MrFlick

Reputation: 206546

If you want to use base functions, here's one possibility

as.vector(by(ages[c("Age","W")],
    list(ages$Indiv),
     function(x) {
         do.call(weighted.mean, unname(x))
     }
))

Since aggregate won't subset multiple columns, i user the more general by and simplified the result to a vector.

Upvotes: 2

jaradniemi
jaradniemi

Reputation: 618

Doh, you beat me to it. But anyway, here is my answer using both plyr and dplyr:

ages = data.frame(Indiv = c(1,1,2,2),
              Age = c(10,15,5,100),
              W = c(2,5,1,2))

library(plyr)
ddply(ages, .(Indiv), summarize, 
      mean = mean(Age),
      wmean = weighted.mean(Age, w=W))


library(dplyr)
ages %.% 
  group_by(Indiv) %.% 
  summarise(mean = mean(Age), wmean = weighted.mean(Age, W))

Upvotes: 11

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