eval() (and parse() )

‘If eval is the answer, you’re probably asking the wrong question’- or is it ?

• Very early on in my careeR I came across the need to create file names on the fly.
  
• What I needed to do was run some code on a file, take the results and save it to a file whose name depended on data dependent on the previous code.
  
• Further I was doing this in a huge loop and creating literally hundreds of files whose name couldn’t be know prior to the code running.
  
• The solution I had was eval (and parse).

• Investigating eval, which has steep learning curve syntax (imho), I came across the quote.
• If eval is the answer, then you are asking the wrong question.
  
• I’ve also seen very respected, by me anyway, R coders make the same claim about parse().
  
• It was from a respected person, so it gave me pause. I’ve asked just about everyone I talk to about R what their opinion is and
• I have gotten answers ranging from it’s dangerous but use it if you want to I don’t use it.

• Nonetheless I will state two things- (a year later, errr 4 things)

1. I still think it is useful.
2. At least one other person agrees with me, see Rchaeology: Idioms of R Programming Paul E. Johnson (available on the web).
3. There are times when it can help explain what is going on with the code.
4. I’ve certainly seen it used in the guts of serious code.

What is eval

• eval() evaluates an R expression. Let me give some examples

x = 3
y = 4
eval(3+4)

## [1] 7

eval(x+y)

## [1] 7

eval(paste0(x,"+",y))

## [1] "3+4"

eval(paste0('x',"+","y"))

## [1] "x+y"

• The last two may be surprising.
• What is going on is that R doesn’t understand that the last expression, “eval(paste0(‘x’,”+“,”y“))” is a formula.
• and that is where parse comes in.
  
• More correctly, parse(text = )

eval(parse(text = paste0('x',"+","y")))

## [1] 7

• I think that is what we expected to see.
• The moral is that eval evaluates an expression in R, but an expression may not be a formula.

• I could end this here and everyone would leave and think if eval is the answer, then the question may or may not be wrong, but it is probably dumb.
• The syntax of eval(parse(text=)) can be daunting, particularly if the paste is complicated,
• There have been times when I have had to evaluate code on the fly and run different code depending on the output.
• At such times, I have been unable to find a better answer than eval.

An example

The idea is -

1. Create an text expression from many variable pieces using paste or paste0
2. Use parse(text =) to turn that expression that R can evaluate and
3. Use eval to evaluate the expression.

• I’ve also used it in functions. Below are two pieces

g = function(a,b){ans = a+ 2*b
ans}
paste0('z=',y,'*cos(x)' )

## [1] "z=4*cos(x)"

paste0('g(exp(z)*y+',2,'+z,z)')

## [1] "g(exp(z)*y+2+z,z)"

Now we can put it together

eval(parse(text =  paste0('z=',y,'*cos(x)')  ))
eval(parse(text =  paste0('g(exp(z)*y+',2,'+z,z)')  ))

## [1] -9.803655

I’ve used this for when I need to save files with different names based on parameter choices

# for( i in some range){
## do some ML stuff and evaluate the results
## Classify the results as say good bad or indifferent in a variable 'outcome' tod = Sys.Date()
# eval(parse( text = paste0("write.csv(","test on ",tod,"",outcome,".txt",")""))
#}

• Imagine one is imagine one if running a random forest.
• If you don’t know what it is, imagine it as a black box.
• The important part is that there are two ‘tuning parameters’, ntree and mtry.
• I want to run the test a number of times and save each object created.

library(MASS) # my favorite data set Boston
library(randomForest) # running a random forest =

## randomForest 4.6-14

## Type rfNews() to see new features/changes/bug fixes.

data(Boston)
summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

• Usually one predicts medv, the median value of a house based on the other variables.
• We will let mtry, the number of variables used run from 2 - 4 and the number of trees run from 100 - 500.
• Neither of these assumptions are particularly good, I just want to create a set of models.
• First I’ll run the paste statement -

i= 2; j=100
flnm = paste0("bost_mtr_",i,"_ntree_",j,".rds")
 flnm    

## [1] "bost_mtr_2_ntree_100.rds"

for(i in 2:4){ # i runs over number of variables
  for(j in (1:5)){ # j is ntree
      tmp = randomForest(medv~., data = Boston, mtry = i, ntree = j)
      flnm = paste0("bost_mtr_",i,"_ntree_",j,".rds")
    eval(parse( text =  paste0('saveRDS(tmp,file =\' ',flnm,' \' )' ) )) # 
    
    
  }
  
}

• I find eval and parse to be functions that I don’t often use, but when necessary they can be very helpful.