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When programming in Stata I often find myself using the loop index in the programming. For example, I'll loop over a list of the variables nominalprice and realprice: 

local list = "nominalprice realprice"
foreach i of local list {
  summarize `i'
  twoway (scatter `i' time)
  graph export "C:\TimePlot-`i'.png"
}

This will plot the time series of nominal and real prices and export one graph called TimePlot-nominalprice.png and another called TimePlot-realprice.png.

In R the method I've come up with to do the same thing would be: 

clist <- c("nominalprice", "realprice")
for (i in clist) {
  e <- paste("png(\"c:/TimePlot-",i,".png\")", sep="")
  eval(parse(text=e))
  plot(time, eval(parse(text=i)))
  dev.off() 
}

This R code looks unintuitive and messy to me and I haven't found a good way to do this sort of thing in R yet. Maybe I'm just not thinking about the problem the right way? Can you suggest a better way to loop using strings?

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4 Answers 4

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24

As other people have intimated, this would be easier if you had a dataframe with columns named nominalprice and realprice. If you do not, you could always use get. You shouldn't need parse at all here.

clist <- c("nominalprice", "realprice")
for (i in clist) {
   png(paste("c:/TimePlot-",i,".png"), sep="")
   plot(time, get(i))
   dev.off() 
}
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2 Comments

Thanks Jonathon. I like the get(i) option and I'll try working with that. The data I was using is dummy data so could be structured in a dataframe. What would be different if using a dataframe?
Because then it's easy to access by string. Say you have a data frame df with a column named nominalprice. Then you can just write df[,"nominalprice"] to obtain that column.
2

If your main issue is the need to type eval(parse(text=i)) instead of ``i'`, you could create a simpler-to-use functions for evaluating expressions from strings:

e = function(expr) eval(parse(text=expr))

Then the R example could be simplified to:

clist <- c("nominalprice", "realprice")
for (i in clist) {
  png(paste("c:/TimePlot-", i, ".png", sep=""))
  plot(time, e(i))
  dev.off() 
}
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1 Comment

I think you should use get() as shown by Jonathan Chang instead.
1

Using ggplot2 and reshape: 

library(ggplot2)
library(reshape)
df <- data.frame(nominalprice=rexp(10), time=1:10)
df <- transform(df, realprice=nominalprice*runif(10,.9,1.1))
dfm <- melt(df, id.var=c("time"))
qplot(time, value, facets=~variable, data=dfm)
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Comments

1

I don't see what's especially wrong with your original solution, except that I don't know why you're using the eval() function. That doesn't seem necessary to me.

You can also use an apply function, such as lapply. Here's a working example. I created dummy data as a zoo() time series (this isn't necessary, but since you're working with time series data anyway):

# x <- some time series data
time <- as.Date("2003-02-01") + c(1, 3, 7, 9, 14) - 1
x <- zoo(data.frame(nominalprice=rnorm(5),realprice=rnorm(5)), time)
lapply(c("nominalprice", "realprice"), function(c.name, x) { 
  png(paste("c:/TimePlot-", c.name, ".png", sep=""))
  plot(x[,c.name], main=c.name)
  dev.off()
}, x=x)
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2 Comments

In the function above what does c.name refer to? I'm trying to understand how I could apply this technique to other situations.
c.name is a variable name that I'm assigning to what get's passed by lapply (lapply, in this instance, is just passing each element of that vector, one at a time). Try playing with this to see how it works: lapply(c(1,2,3), function(x) print(x)). Also look at ?lapply.

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