Useful little functions in R?

Question

What are the functions that you wrote, don't quite deserve a package, but you wish to share?

I will throw in some of mine:

destring <- function(x) {
    ## convert factor to strings
    if (is.character(x)) {
        as.numeric(x)
    } else if (is.factor(x)) {
        as.numeric(levels(x))[x]
    } else if (is.numeric(x)) {
        x
    } else {
        stop("could not convert to numeric")
    }
}

pad0 <- function(x,mx=NULL,fill=0) {
  ## pad numeric vars to strings of specified size
  lx <- nchar(as.character(x))
  mx.calc <- max(lx,na.rm=TRUE)
  if (!is.null(mx)) {
    if (mx<mx.calc) {
      stop("number of maxchar is too small")
    }
  } else {
    mx <- mx.calc
  }
  px <- mx-lx
  paste(sapply(px,function(x) paste(rep(fill,x),collapse="")),x,sep="")
}


.eval <- function(evaltext,envir=sys.frame()) {
  ## evaluate a string as R code
  eval(parse(text=evaltext), envir=envir)
}

## trim white space/tabs
## this is marek's version
trim<-function(s) gsub("^[[:space:]]+|[[:space:]]+$","",s)

Eduardo, this is a topic more fit for a blog rather than SO. — Paul Sasik, Aug 14 '10 at 22:57
Paul - I agree. But I thought a community wiki here would help me find some gems. Base R is "missing" a few of this helper functions. — Eduardo Leoni, Aug 14 '10 at 23:00
A good place for such code is at [http://wiki.r-project.org] — Aniko, Aug 15 '10 at 20:15
For your `trim` function I would do `trim<-function(s) gsub("^[[:space:]]+|[[:space:]]+$","",s)` which removes other blank characters (see `?regex`) and combinations like space+tab+normal_text. — Marek, Aug 17 '10 at 08:51

score 28 · Accepted Answer · edited Aug 06 '19 at 09:06

Here's a little function to plot overlapping histograms with pseudo-transparency:

_{(source: chrisamiller.com)}

plotOverlappingHist <- function(a, b, colors=c("white","gray20","gray50"),
                            breaks=NULL, xlim=NULL, ylim=NULL){

  ahist=NULL
  bhist=NULL

  if(!(is.null(breaks))){
    ahist=hist(a,breaks=breaks,plot=F)
    bhist=hist(b,breaks=breaks,plot=F)
  } else {
    ahist=hist(a,plot=F)
    bhist=hist(b,plot=F)

    dist = ahist$breaks[2]-ahist$breaks[1]
    breaks = seq(min(ahist$breaks,bhist$breaks),max(ahist$breaks,bhist$breaks),dist)

    ahist=hist(a,breaks=breaks,plot=F)
    bhist=hist(b,breaks=breaks,plot=F)
  }

  if(is.null(xlim)){
    xlim = c(min(ahist$breaks,bhist$breaks),max(ahist$breaks,bhist$breaks))
  }

  if(is.null(ylim)){
    ylim = c(0,max(ahist$counts,bhist$counts))
  }

  overlap = ahist
  for(i in 1:length(overlap$counts)){
    if(ahist$counts[i] > 0 & bhist$counts[i] > 0){
      overlap$counts[i] = min(ahist$counts[i],bhist$counts[i])
    } else {
      overlap$counts[i] = 0
    }
  }

  plot(ahist, xlim=xlim, ylim=ylim, col=colors[1])
  plot(bhist, xlim=xlim, ylim=ylim, col=colors[2], add=T)
  plot(overlap, xlim=xlim, ylim=ylim, col=colors[3], add=T)
}

An example of how to run it:

a = rnorm(10000,5)
b = rnorm(10000,3)
plotOverlappingHist(a,b)

Update: FWIW, there's a potentially simpler way to do this with transparency that I've since learned:

a=rnorm(1000, 3, 1)
b=rnorm(1000, 6, 1)
hist(a, xlim=c(0,10), col="red")
hist(b, add=T, col=rgb(0, 1, 0, 0.5)

that's very neat chris. i will accept this answer, as it also got the highest number of votes. — Eduardo Leoni, Aug 30 '10 at 04:18

score 14 · Answer 2 · edited Aug 06 '19 at 09:07

The output of the fft (Fast Fourier Transform) function in R can be a little bit tedious to process. I wrote this plotFFT function in order to do a frequency vs power plot of the FFT. The getFFTFreqs function (used internally by plotFFT) returns the frequency associated to each FFT value.

This was mostly based on the very interesting discussion at http://tolstoy.newcastle.edu.au/R/help/05/08/11236.html

# Gets the frequencies returned by the FFT function
getFFTFreqs <- function(Nyq.Freq, data)
    {
    if ((length(data) %% 2) == 1) # Odd number of samples
        {
        FFTFreqs <- c(seq(0, Nyq.Freq, length.out=(length(data)+1)/2), 
               seq(-Nyq.Freq, 0, length.out=(length(data)-1)/2))
        }
    else # Even number
        {
        FFTFreqs <- c(seq(0, Nyq.Freq, length.out=length(data)/2), 
               seq(-Nyq.Freq, 0, length.out=length(data)/2))
        }

    return (FFTFreqs)
    }

# FFT plot
# Params:
# x,y -> the data for which we want to plot the FFT 
# samplingFreq -> the sampling frequency
# shadeNyq -> if true the region in [0;Nyquist frequency] will be shaded
# showPeriod -> if true the period will be shown on the top
# Returns a list with:
# freq -> the frequencies
# FFT -> the FFT values
# modFFT -> the modulus of the FFT
plotFFT <- function(x, y, samplingFreq, shadeNyq=TRUE, showPeriod = TRUE)
    {
    Nyq.Freq <- samplingFreq/2
    FFTFreqs <- getFFTFreqs(Nyq.Freq, y)

    FFT <- fft(y)
    modFFT <- Mod(FFT)
    FFTdata <- cbind(FFTFreqs, modFFT)
    plot(FFTdata[1:nrow(FFTdata)/2,], t="l", pch=20, lwd=2, cex=0.8, main="",
        xlab="Frequency (Hz)", ylab="Power")
    if (showPeriod == TRUE)
        {
        # Period axis on top        
        a <- axis(3, lty=0, labels=FALSE)
        axis(3, cex.axis=0.6, labels=format(1/a, digits=2), at=a)
        }
    if (shadeNyq == TRUE)
        {
        # Gray out lower frequencies
        rect(0, 0, 2/max(x), max(FFTdata[,2])*2, col="gray", density=30)
        }

    ret <- list("freq"=FFTFreqs, "FFT"=FFT, "modFFT"=modFFT)
    return (ret)
    }

As an example you can try this

# A sum of 3 sine waves + noise
x <- seq(0, 8*pi, 0.01)
sine <- sin(2*pi*5*x) + 0.5 * sin(2*pi*12*x) + 0.1*sin(2*pi*20*x) + 1.5*runif(length(x))
par(mfrow=c(2,1))
plot(x, sine, "l")
res <- plotFFT(x, sine, 100)

or

linearChirp <- function(fr=0.01, k=0.01, len=100, samplingFreq=100)
    {
    x <- seq(0, len, 1/samplingFreq)
    chirp <- sin(2*pi*(fr+k/2*x)*x) 

    ret <- list("x"=x, "y"=chirp)
    return(ret)
    }

chirp <- linearChirp(1, .02, 100, 500)
par(mfrow=c(2,1))
plot(chirp, t="l")
res <- plotFFT(chirp$x, chirp$y, 500, xlim=c(0, 4))

Which give

_{(source: nicolaromano.net)}

score 10 · Answer 3 · answered Aug 16 '10 at 09:06

Very simple but i use it a lot:

setdiff2 <- function(x,y) {
    #returns a list of the elements of x that are not in y 
     #and the elements of y that are not in x (not the same thing...)

    Xdiff = setdiff(x,y)
    Ydiff = setdiff(y,x)
    list(X_not_in_Y=Xdiff, Y_not_in_X=Ydiff)
}

score 6 · Answer 4 · answered Aug 16 '10 at 06:03

# Create a circle with n number of "sides" (kudos to Barry Rowlingson, r-sig-geo).
circle <-  function(x = 0, y = 0, r = 100, n = 30){
    t <- seq(from = 0, to = 2 * pi, length = n + 1)[-1]
    t <- cbind(x = x + r * sin(t), y = y + r * cos(t))
    t <- rbind(t, t[1,])
    return(t)
}
# To run it, use
plot(circle(x = 0, y = 0, r = 50, n = 100), type = "l")

score 5 · Answer 5 · answered Aug 16 '10 at 10:57

It was annoying to me how data.frame with many columns is printed, I mean this split over columns. So I wrote my own version:

print.data.frame <- function(x, ...) {
    oWidth <- getOption("width")
    oMaxPrint <- getOption("max.print")
    on.exit(options(width=oWidth, max.print=oMaxPrint))
    options(width=10000, max.print=300)
    base::print.data.frame(x, ...)
}

score 4 · Answer 6 · answered Aug 14 '10 at 23:42

I frequently want to use sum contrasts in regressions, and I usually want the terms to be meaningfully named. So I wrote this recontrast function.

recontrast<-function(data,type = "sum"){
    data.type <-class(data)
    if(data.type == "factor"&!is.ordered(data)&nlevels(data)>1&nlevels(data)<1000){
        if(type == "sum"){
            contrasts(data)<-contr.sum(levels(data))
            colnames(contrasts(data))<-levels(data)[-nlevels(data)]
        }else if(type == "treatment"){
            contrasts(data)<-contr.treatment(levels(data))
        }
    }else if(data.type == "data.frame"){
        for(i in 1:ncol(data)){
            if(is.factor(data[,i]) &     !is.ordered(data[,i])&nlevels(data[,i])>1&nlevels(data[,i])<1000){
                if(type == "sum"){
                    contrasts(data[,i])<-contr.sum(levels(data[,i]))
                    colnames(contrasts(data[,i]))<-levels(data[,i])[-    nlevels(data[,i])]
                }else if(type == "treatment"){
                    contrasts(data[,i])<-    contr.treatment(levels(data[,i]))
                }
            }
        }
    }
return(data)
}

It take both entire dataframes and factors as arguments. If it's a data frame, it'll convert all contrasts of unordered factors with <1000 levels to either treatment or sum contrasts. With sum contrasts, it meaningfully names the columns, so you'll have meaningful labels in the regression output.

score 1 · Answer 7 · answered Nov 12 '10 at 15:10

1

In the most useful R trick posting I saw a post by Keving from Nov 3 '09 bout dropping unused levels. The first function was provided there. and I took the best step in the second function to drop levels from a subset.

drop.levels <- function (dat) {if (is.factor(dat)) dat <- dat[, drop = TRUE] else dat[] <- lapply(dat, function(x) x[, drop = TRUE]); return(dat) ;};

subset.d    <- function (...) drop.levels(subset(...)); # function to drop levels of subset

answered Nov 12 '10 at 15:10

Adam SO

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For notice: in R-2.12.0 is new function `droplevels`. It used `factor(x)` instead of `x[,drop=TRUE]` to drop levels. – Marek Nov 27 '10 at 21:55

Useful little functions in R?

7 Answers7

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