Looping with Multiple Parameters without the for loop

Question

Below is a micro-version of my data set. I would like to loop through the three variables without using the 'for' function.

Reproducible Version of the Subject Data Set

cat1 <- rep(c("A","S"),each=40)# First variable
cat2 <- rep(c("C","M","Y","K"), each=10, times=2) # Second variable
num1 <- rep(c(0,10,20,30,40,45,65,80,90,100),8) # Third variable
rspns <-structure(list(dataSet = c(0.2484375, 0.3959375, 0.4875, 0.57875, 0.6696875, 0.694375,
    0.816875, 0.879375, 0.9121875, 0.93125, 0.25125, 0.3796875, 0.4609375, 0.5396875, 0.6159375,
    0.6515625, 0.7696875, 0.8384375, 0.864375, 0.8865625, 0.271875, 0.39875, 0.4821875,
    0.5628125, 0.6284375, 0.650625, 0.7553125, 0.8003125, 0.8103125, 0.8125, 0.251875, 0.3775,
    0.4703125, 0.5725, 0.6996875, 0.7378125, 0.945, 1.055625, 1.1021875, 1.1140625, 0.25125,
    0.4203125, 0.5215625, 0.615625, 0.71, 0.74, 0.865625, 0.9246875, 0.9603125, 0.9734375,
    0.256875, 0.3953125, 0.4775, 0.5528125, 0.62875, 0.65875, 0.78375, 0.8384375, 0.8653125,
    0.8740625, 0.2790625, 0.4215625, 0.515, 0.6009375, 0.6665625, 0.693125, 0.7959375, 0.83875,
    0.8490625, 0.8575, 0.2571875, 0.3759375, 0.4665625, 0.56375, 0.68875, 0.725, 0.9259375,
    1.0328125, 1.085625, 1.1096875)), .Names = "rspns",
  class = "data.frame", row.names = c(NA, -80L))
gain <- data.frame(cat1,cat2,num1, rspns=rspns)

tint <-  function(x,y,z) gain[cat1 == x & cat2 == y & num1 == z, 4]
dgain <- function(x,y,z){(100* (1-10^-(tint(x,y,z) - tint(x,y,0))) / (1-10^-(tint(x,y,100)- tint(x,y,0)))) - z}

How can I get the dgain function, which takes three parameters, to loop through cat1, cat2 and num1 variables without using for loops?

Your help and technical expertise is greatly appreciated.

@juba I think your mapply answer was actually ok (i.e. I'm not sure their function is already vectorized). — joran
– joran, Commented Mar 12, 2013 at 16:09
@joran Thanks, but I deleted my answer because it seems that both give the same values. — juba
– juba, Commented Mar 12, 2013 at 16:10
juba/joran, i am interested in the mapply and sapply answers, can you please re-post it. — Ragy Isaac
– Ragy Isaac, Commented Mar 12, 2013 at 16:37

joran · Accepted Answer · 2013-03-12 17:02:03Z

1

Here's the basic idea of what I was getting at with my comments:

#With a basic for loop for reference
out1 <- rep(NA,80)
for (i in 1:80){
    out1[i] <- dgain(cat1[i],cat2[i],num1[i])
}

out2 <- mapply(dgain,x = cat1,y = cat2,z = num1)

out3 <- dgain(cat1,cat2,num1)

all.equal(out1,unname(out2))
[1] TRUE
> all.equal(out1,out3)
[1] "Mean relative difference: 0.2579067"

It's probably possible to refactor your code such that the final idiom (dgain(cat1,cat2,num1)) would return the correct result, but it would take more tinkering than I have time for. The issue is that you're subsetting a data frame from your global environment inside one of your functions using ==, which won't return what you want if you pass is vector arguments.

Nothing special is going on with mapply. It's almost literally doing the same thing as the for loop, only perhaps slightly more efficiently and perhaps in a more aesthetically pleasing manner.

answered Mar 12, 2013 at 17:02

joran

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3 Comments

Ragy Isaac Over a year ago

joran, you are correct, the idiom (dgain(cat1,cat2,num1)) does not provide the correct answer. That is why I am interested in mapply or sammply. How should i rewrite the functions? Inputting the three parameters in the current functions manually provides the correct answer. Can you please outline what i should be doing?

joran Over a year ago

@RagyIsaac I showed you the mapply solution. I'm not really interested in re-writing your code for you. You'll probably have to use match instead of == inside tint, though.

Ragy Isaac Over a year ago

Thanks joran, this help me quite a bit

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