higher-order-functions

Scope functions apply/with/run/also/let: Where do the names come from?

感情迁移 提交于 2019-12-02 18:27:52
There are quite a few blog posts (like this ) on usages of the standard library functions apply / with / run / also / let available that make it a bit easier to distingish when to actually use which of those pretty functions. For a few weeks now, the official docs even provide guidelines on that topic finally: https://kotlinlang.org/docs/reference/coding-conventions.html#using-scope-functions-applywithrunalsolet Nevertheless, I think it is pretty hard to memorize the function's individual use cases by the function names . I mean, for me they seem to be interchangeable, why isn't let called run

Please explain me this higher-order function javascript code

瘦欲@ 提交于 2019-12-02 16:06:18
问题 I'm studying higher order functions following the Eloquent JavaScript book. I haven't been able to understand this code, why is "Boolean" passed as noisy first argument? This is supposed to be function that changes other function, I just don't get how it works! function noisy(f) { return function(arg) { console.log("calling with", arg); var val = f(arg); console.log("called with", arg, "- got", val); return val; }; } noisy(Boolean)(0); // → calling with 0 // → called with 0 - got false 回答1:

FoldLeft using FoldRight in scala

核能气质少年 提交于 2019-12-02 15:48:37
While going through Functional Programming in Scala , I came across this question: Can you right foldLeft in terms of foldRight? How about the other way around? In solution provided by the authors they have provided an implementation as follows: def foldRightViaFoldLeft_1[A,B](l: List[A], z: B)(f: (A,B) => B): B = foldLeft(l, (b:B) => b)((g,a) => b => g(f(a,b)))(z) def foldLeftViaFoldRight[A,B](l: List[A], z: B)(f: (B,A) => B): B = foldRight(l, (b:B) => b)((a,g) => b => g(f(b,a)))(z) Can somebody help me trace through this solution and make me understand how this actually gets the foldl

What are some interesting uses of higher-order functions?

筅森魡賤 提交于 2019-12-02 14:10:31
I'm currently doing a Functional Programming course and I'm quite amused by the concept of higher-order functions and functions as first class citizens. However, I can't yet think of many practically useful, conceptually amazing, or just plain interesting higher-order functions. (Besides the typical and rather dull map , filter , etc functions). Do you know examples of such interesting functions? Maybe functions that return functions, functions that return lists of functions (?), etc. I'd appreciate examples in Haskell, which is the language I'm currently learning :) Well, you notice that

Why is Haskell (GHC) so darn fast?

放肆的年华 提交于 2019-12-02 13:47:00
Haskell (with the GHC compiler) is a lot faster than you'd expect . Used correctly, it can get close-ish to low-level languages. (A favorite thing for Haskellers to do is to try and get within 5% of C (or even beat it, but that means you are using an inefficient C program, since GHC compiles Haskell to C).) My question is, why? Haskell is declarative and based on lambda calculus. Machine architectures are clearly imperative, being based on turing machines, roughly. Indeed, Haskell doesn't even have a specific evaluation order. Also, instead of dealing with machine data types, you make

Sorting in haskell with parameter using higher order function

丶灬走出姿态 提交于 2019-12-02 13:16:46
问题 Hi I'm a Haskell beginner and I'm really lost. This is for my assignment, and it asks me to do something like below using higer order function Main> mySort (<) [1,5,3,6,4,1,3,3,2] [1,1,2,3,3,3,4,5,6] Main> mySort (>) [1,5,3,6,4,1,3,3,2] [6,5,4,3,3,3,2,1,1] Main> mySort longerWord [“Hello”, “The”, “a”, “Daniel”, “Declarative”] [“Declarative”, “Daniel”, “Hello”, “The”, “a”] First of all, I thought I should make a function that distinguish whether it's < , > or longerWord checkConditionStr:

Please explain me this higher-order function javascript code

自古美人都是妖i 提交于 2019-12-02 12:26:02
I'm studying higher order functions following the Eloquent JavaScript book. I haven't been able to understand this code, why is "Boolean" passed as noisy first argument? This is supposed to be function that changes other function, I just don't get how it works! function noisy(f) { return function(arg) { console.log("calling with", arg); var val = f(arg); console.log("called with", arg, "- got", val); return val; }; } noisy(Boolean)(0); // → calling with 0 // → called with 0 - got false noisy accepts any one-argument function as its argument. It returns a new function that calls that function,

higher order function python walkthrough

陌路散爱 提交于 2019-12-02 10:08:31
def tracer(fn): def traced(x): print('Calling',fn,'(',x,')') result=fn(x) print('Got',result,'from',fn,'(',x,')') return result return traced def fact(n): if n ==0: return 1 return n * fact(n-1) new_fact = tracer(fact) new_fact(2) I used this code on pythontutor.com to better understand higher order functions, but i'm having difficulty understanding why new_fact(2) in step 8 is mapped to traced? In other words, how does the traced function know the argument is 2? In Python, functions are objects too. When you call the tracer() function, it returns the nested traced() function; it is in fact

Map value in particular key in array of dictionary

我们两清 提交于 2019-12-02 09:04:24
Map value the particular key wherever presents in array of dictionary and replace it in same array. We need to update pan_card key 0 to 1 , in occurence of array of dictionary. let keyToUpdate = "pan_card" var arrayOfDictionary = [[String:Any]]() var firstDict = [String:Any]() firstDict["passport"] = 0 firstDict["ration_card"] = 0 firstDict["pan_card"] = 0 var arrayDict = [String : Any]() arrayDict["currentObject"] = firstDict arrayDict["title"] = "Documents list" var secondDict = [String:Any]() secondDict["dl"] = 0 secondDict["voter"] = 0 secondDict["pan_card"] = 0 //let dic = secondDict

Why doesn't Haskell accept arguments after a function composition?

元气小坏坏 提交于 2019-12-02 06:18:51
问题 Since Haskell functions have only one argument, where the rest of arguments stay like lambdas, then we can do this: foo a b = a + b -- this is like foo a = \b -> a + b foo 1 2 -- ok Well, I noticed that if I declare the function returning a lambda, just like in comment, the foo 1 2 will works fine the same way. But when I compose these functions, like this: foo a b = a + b bar x = x * x bar . foo 1 2 -- oh, it's wrong, I need do '(bar . foo 1) 2' ...This returns an error. Ok, the question is: