Aggregate function in R with multiple function arguments

Question

I have sample data.set containing climate data for different seasons:

df <- data.frame(season=rep(1:5,2),year=rep(1:2,each=5),
      temp=c(2,4,3,5,2,4,1,5,4,3),ppt=c(4,3,1,5,6,2,1,2,2,2),
      samples=c(22,25,24,31,31,29,28,31,30,32))

I can determine the mean of my climate variables for each season for each year simply:

aggregate(df[,c('temp','ppt')], by = list(df$season,df$year), function(x) mean(x,na.rm=T))

However, I want to determine the weighted mean of each season|year combo using variable samples as my weights.

Essentially I want to replace my mean function in aggregate() with weighted.mean. That would require adding a second argument to my function that needs to change with my x.

    function(x,w) weighted.mean(x,w,na.rm=T))

Though, I'm not sure how to let the weight argument ('w') of weighted.mean() vary with each subset of the aggregated data.

Can I do this all within an aggregate function?

Any advice would be great!

Pierre L · Accepted Answer · 2016-01-26 00:04:57Z

3

Try summarise_each from dplyr. It allows for the prior grouping with group_by and application to multiple columns:

library(dplyr)
df %>% group_by(season, year) %>%
        summarise_each(funs(weighted.mean(., samples,na.rm=T)), temp,ppt)
# Source: local data frame [10 x 5]
# Groups: season, year [10]
# 
#    season  year  temp   ppt samples
#    (int) (int) (dbl) (dbl)   (dbl)
# 1       1     1     2     4      22
# 2       2     1     4     3      25
# 3       3     1     3     1      24
# 4       4     1     5     5      31
# 5       5     1     2     6      31
# 6       1     2     4     2      29
# 7       2     2     1     1      28
# 8       3     2     5     2      31
# 9       4     2     4     2      30
# 10      5     2     3     2      32

edited Jan 26, 2016 at 0:04

answered Jan 25, 2016 at 20:28

Pierre L

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2 Comments

theforestecologist Over a year ago

Can this be done using aggregate or any other function in the base package of R?

Pierre L Over a year ago

I have no idea why you want a complicated base solution when this works, but here you go. The explanation would take very long to go through

df[,c("temp", "ppt")] <- matrix(ncol=2, unlist(do.call(rbind, lapply(split(df, list(df$season, df$year)),function(df) {   lapply(df[,c("temp", "ppt")], function(cols) weighted.mean(cols, df$samples, na.rm=T))}))))

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