Yes and no. Different functional programming languages solve them differently. In Haskell (a very pure one) all this stuff has to happen in something called the I/O Monad – see here. You can think of it as getting another input (and output) into your function (the world-state) or easier as a place where “impureness” like … Read more
This is because of the order of execution. If you pass a function call to setTimeout, the function will be executed immediately, i.e. the function is put on javascript’s execution stack immediately. If you pass a function name, i.e. a reference to a function, the function is only put in the javascript thread’s execution stack … Read more
def subs(l): if l == []: return [[]] x = subs(l[1:]) return x + [[l[0]] + y for y in x] Results: >>> print (subs([1, 2, 3])) [[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]
You just don’t want to do that, don’t even start. OO sure has its merits, but “classic examples” like your C++ one are almost always contrived structures designed to hammer the paradigm into undergraduate students’ brains so they won’t start complaining about how stupid the languages are they’re supposed to use†. The idea seems basically … Read more
Use reduce(): reduce(dict.__getitem__, l, d) or better still, using operator.getitem(): from operator import getitem reduce(getitem, l, d) Demo: >>> d = {“a”: {“b”: {“c”: 4}}} >>> l = [“a”, “b”, “c”] >>> from operator import getitem >>> reduce(getitem, l, d) 4 Python 3 moved the reduce() function out of the built-ins and into functools.reduce().
My first ugly approach was: allTerms :: Omega T allTerms = do which <- each [ 1,2,3 ] if which == 1 then return A else if which == 2 then do x <- allTerms return $ B x else do x <- allTerms y <- allTerms return $ C x y But then, after … Read more
try like this: reduce(enumerate(a), [String:UIView]()) { (var dict, enumeration) in dict[“\(enumeration.index)”] = enumeration.element return dict } Xcode 8 • Swift 2.3 extension Array where Element: AnyObject { var indexedDictionary: [String:Element] { var result: [String:Element] = [:] for (index, element) in enumerate() { result[String(index)] = element } return result } } Xcode 8 • Swift 3.0 … Read more
zip2 xs ys = foldr step done xs ys where done ys = [] step x zipsfn [] = [] step x zipsfn (y:ys) = (x, y) : (zipsfn ys) How this works: (foldr step done xs) returns a function that consumes ys; so we go down the xs list building up a nested composition … Read more
Do it the old-fashioned way, without Guava. (Speaking as a Guava developer.) List<Person> filtered = Lists.newArrayList(); for(Person p : allPersons) { if(acceptedNames.contains(p.getName())) { filtered.add(p); } } You can do this with Guava, but Java isn’t Python, and trying to make it into Python is just going to perpetuate awkward and unreadable code. Guava’s functional utilities … Read more
It is doable, and in fact I just found two snippets for this. Solution #1 The first snippet by cotton substitutes RegexURLPattern and RegexURLResolver with custom implementations that inject given decorator during resolve call. from django.core.urlresolvers import RegexURLPattern, RegexURLResolver from django.conf.urls.defaults import patterns, url, include from django.contrib import admin from myproject.myapp.decorators import superuser_required class DecoratedURLPattern(RegexURLPattern): … Read more