R is definitely our first choice go-to analysis system. In our opinion you really shouldn’t use something else until you have an articulated reason (be it a need for larger data scale, different programming language, better data source integration, or something else). The advantages of R are numerous:

• Single integrated work environment.
• Powerful unified scripting/programming environment.
• Many many good tutorials and books available.
• Wide range of machine learning and statistical libraries.
• Very solid standard statistical libraries.
• Excellent graphing/plotting/visualization facilities (especially ggplot2).
• Schema oriented data frames allowing batch operations, plus simple row and column manipulation.
• Unified treatment of missing values (regardless of type).

For all that we always end up feeling just a little worried and a little guilty when introducing a new user to R. R is very powerful and often has more than one way to perform a common operation or represent a common data type. So you are never very far away from a strange and painful corner case. This why when you get R training you need to make sure you get an R expert (and not an R apologist). One of my favorite very smart experts is Norm Matloff (even his most recent talk title is smart: “What no one else will tell you about R”). Also, buy his book; we are very happy we purchased it.

But back to corner cases. For each method in R you really need to double check if it actually works over the common R base data types (numeric, integer, character, factor, and logical). Not all of them do and and sometimes you get a surprise.

Recent corner case problems we ran into include:

• randomForest regression fails on character arguments, but works on factors.
• mgcv gam() model doesn’t convert strings to formulas.
• R maps can’t use the empty string as a key (that is the string of length 0, not a NULL array or NA value).

These are all little things, but can be a pain to debug when you are in the middle of something else.

For our concrete example let’s concentrate on the pain generated by the empty string.

In R strings represent free-form text and factors represent strings from a pre-defined finite set of possibility (actually called “levels”). The difference can be subtle and you may not always know which one you have (R may have converted for you) and which one will work (R may fail to convert for you).

Take for example the simple case of building a linear model mapping a string or factor valued x to numeric y-values:

d <- data.frame(y=c(1,2,3),x=c('a','b','c')) lmModel <- lm(y~0+x,data=d) d$lmPred <- predict(lmModel,newdata=d) print(d) y x lmPred 1 1 a 1 2 2 b 2 3 3 c 3

We have used x to predict y. This works (under the covers) because R converts the string-values of x into factor levels and then uses. For the basic details of how this works try: help('data.matrix') or help('contrasts'). For a bit more on conversion to indicators (which is pretty much automatic in R, but can be a painful manual step in some other machine learning frameworks) see chapter 2 section 2.2.3 of Practical Data Science with R.

Of course it is silly of us to use the entire lm() framework to model an expected value conditioned on a single categorical variable. That is what table() and aggregate() are for. lm() gets the right answer but it has to do some unnecessary steps (such as forming and inverting a design matrix, which happens to be diagonal- but probably wastes space in a non-sparse representation). To directly build a model that maps level to expected value we do the following:

mkMapModel <- function(yVarName,xVarName,data) { means <- aggregate(as.formula(paste(yVarName,xVarName,sep='~')), data=data,FUN=mean) model <- as.list(means[[yVarName]]) names(model) <- means[[xVarName]] model } mapModel <- mkMapModel('y','x',d) d$mapPred <- as.numeric(mapModel[d$x]) print(d) y x lmPred mapPred 1 1 a 1 1 2 2 b 2 2 3 3 c 3 3

This works great. It makes sense, and is much more efficient for variables that have a very large number of levels (see Modeling Trick: Impact Coding of Categorical Variables with Many Levels for more details). However it turns out this is only working because the x-variable is encoded as a factor. As the code below shows, when the x-variable takes on string values (called character in R) the lm() works (likely as it triggers a conversion to factor at some point), but our hand-rolled mkMapModel() fails:

d <- data.frame(y=c(1,2,3),x=c('a','b',''),stringsAsFactors=FALSE) lmModel <- lm(y~0+x,data=d) d$lmPred <- predict(lmModel,newdata=d) print(d) y x lmPred 1 1 a 1 2 2 b 2 3 3 3

mapModel <- mkMapModel('y','x',d) d$mapPred <- as.numeric(mapModel[d$x]) Error: (list) object cannot be coerced to type 'double'

This is because the empty string "" (not null, just the string of length 0) isn’t a legal map-key in R. Likely this is due R’s linkages between evaluation environments and maps (and while the empty string may be a traditional string, it isn’t a traditional variable name; see the “environment->list coercion” example in help('as.list') for the connection).

One work around is to make sure the x-variable is a factor (not a character array). We demonstrate this in the working code below. Another fix would be to use paste() to prefix all strings (so none are empty).

d <- data.frame(y=c(1,2,3),x=c('a','b',''),stringsAsFactors=TRUE) mapModel <- mkMapModel('y','x',d) d$mapPred <- as.numeric(mapModel[d$x]) print(d) y x mapPred 1 1 a 1 2 2 b 2 3 3 3

The requirement to ensure strings are already converted to factors is not unique to our function mkMapModel() (for example the randomForest package has similar issues in some circumstances, but at least is does work with factors unlike some Python scikit-learn packages). Even with such issues I think you are net-ahead using R (notice we didn’t have to write any for-loops due to the vectorized nature of the operator []). With some defensive coding you certainly are ahead in using R’s built in models (like lm()).

Related posts:

1. Survive R
2. Why I don’t like Dynamic Typing
3. Selection in R