Work with unique values ​​in scale (for loops, applications, or plyr)

Question

I'm not sure if this is possible, but if it is, it will make life much more efficient.

A common problem that will be of interest to the wider SO community: for loops (and basic functions such as apply) are applicable for general / consistent operations, such as adding X to each column or row of a data frame. I have a common / consistent operation that I want to perform, but with unique values ​​for each element of the data frame.

Is there a way to do this more efficiently than a subset of my data frame for each grouping, applying a function with specific numbers relative to that grouping, and then recombining? I don't care if it's a for loop or apply, but bonus points if it uses plyr functionality.

---

Here is a more specific problem I'm working on: I have the data below. Ultimately, what I want is a data frame for a time series with a date, and each column represents the relation of the region to some reference.

Problem: The measure of interest for each region is different, and this is also a benchmark. Here is the data:

library(dplyr)
library(reshape2)

data <- data.frame(
    region = sample(c("northeast","midwest","west"), 100, replace = TRUE),
    date = rep(seq(as.Date("2010-02-01"), length=10, by = "1 day"),10),
    population = sample(50000:100000, 10, replace = T),
    skiers = sample(1:100),
    bearsfans = sample(1:100),
    dudes = sample(1:100)
)

and the composite frame I'm working on:

data2 <- data %.%
    group_by(date, region) %.%
    summarise(skiers = sum(skiers), 
            bearsfans= sum(bearsfans), 
            dudes = sum(dudes), 
            population = sum(population)) %.%
    mutate(ppl_per_skier = population/skiers,
            ppl_per_bearsfan = population/bearsfans,
            ppl_per_dude = population/dudes) %.%
    select(date, region, ppl_per_skier, ppl_per_bearsfan , ppl_per_dude)

Here's the tricky part:

• In the northeast, I only care about "ppl_per_skier", and the standard is 3500
• In the Midwest, I only care about ppl_per_bearsfan, and the benchmark is 1200
• "ppl_per_dude", 5000 -

, , , ... . :

midwest <- data2 %.% 
            filter(region == "midwest") %.%
            select(date, region, ppl_per_bearsfan) %.%
            mutate(bmark = 1200, against_bmk = bmark/ppl_per_bearsfan-1) %.%
            select(date, against_bmk)

, , . , - , (, , ):

        date midwest_againstbmk northeast_againstbmk west_againstbmk
1 2010-02-10          0.9617402            0.6008032       0.3403260
2 2010-02-11          0.5808621            0.5119942       0.7787559
3 2010-02-12          0.4828346            0.6560053       0.3747920
4 2010-02-13          0.6499841            0.7567194       0.8387461
5 2010-02-14          0.6367520            0.4564254       0.7269161

, X , ?

Answer 1

mapply:

> mapply(function(d,y,b) {(b/d[,y])-1},
         split(data2,data2$region), 
         c('ppl_per_bearsfan','ppl_per_skier','ppl_per_dude'), 
         c(1200,3500,5000))
          midwest   northeast      west
 [1,] -0.26625428 -0.02752186 3.5881957
 [2,]  0.48715638  1.89169295 2.6928546
 [3,] -0.94222992  1.26065537 4.0388343
 [4,] -0.38116663  0.79572184 1.4118364
 [5,] -0.05937874  2.05459482 1.8822015
 [6,] -0.41463925  1.60668461 1.5914408
 [7,] -0.31211391  1.21093777 2.7517886
 [8,] -0.88923466  0.44917981 1.2251965
 [9,] -0.02781965 -0.24637182 2.7143103
[10,] -0.46643682  1.28944776 0.6246315