evaluation

Background I am learning the sicp according to an online course and got confused by its lecture notes. In the lecture notes, the applicative order seems to equal cbv and normal order to cbn. Confusion But the wiki points out that, beside evaluation orders(left to right, right to left, or simultaneous), there is a difference between the applicative order and cbv: Unlike call-by-value, applicative order evaluation reduces terms within a function body as much as possible before the function is applied. I don't understand what does it mean by reduced. Aren't applicative order and cbv both going to

Say I have an expression as follows (where ⨁ and ⨂ are binary operators which have the same precedence level but not the same associativity): x ⨁ y ⨂ z Would y belong to ⨁ or ⨂ , and based on what criteria? According to the Edsgar Dijkstra's Shunting-yard algorithm if neighboring two operators in an expressions have the same precedence level then the expression is disambiguated based on the associativity of the second operator. If the second operator is left associative then the operand belongs to the first operator. If the second operator is right associative then the operand belongs to the

If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]:=x^y; f[3+4] (* -> f[7] *) f[Unevaluated[3+4]] (* ->81 *) A level of Unevaluated is stripped off with every

I made really time consuming algorithm which produces a short string as the result. When I try to print it (via putStrLn) it appears on the screen character by character. I did understand why that happened, and I tried to force evaluation of the string before actual printing. myPrint !str = putStrLn str But this help very little. When I ran the program in debug I noticed that the !str forced evaluation only for the first character. Does anyone know why is that, and how to deal with this? (!) translates into seq , which evaluates strictly to Weak Head Normal Form -- that is, it only evaluates

In Haskell , I might implement if like this: if' True x y = x if' False x y = y spin 0 = () spin n = spin (n - 1) This behaves how I expect : haskell> if' True (spin 1000000) () -- takes a moment haskell> if' False (spin 1000000) () -- immediate In Racket , I could implement a flawed if like this: (define (if2 cond x y) (if cond x y)) (define (spin n) (if (= n 0) (void) (spin (- n 1)))) This behaves how I expect : racket> (if2 #t (spin 100000000) (void)) -- takes a moment racket> (if2 #f (spin 100000000) (void)) -- takes a moment In Idris , I might implement if like this: if' : Bool -> a -> a

Could someone please help me with the following: I need to change my xgboost training model with caret package to an undefault metric RMSLE. By default caret and xgboost train and measure in RMSE. Here are the lines of code: create custom summary function in caret format custom_summary = function(data, lev = NULL, model = NULL){ out = rmsle(data[, "obs"], data[, "pred"]) names(out) = c("rmsle") out } create control object control = trainControl(method = "cv", number = 2, summaryFunction = custom_summary) create grid of tuning parameters grid = expand.grid(nrounds = 100, max_depth = 6, eta = 0