No, you can’t use yield inside of the inner function. But in your case you don’t need it. You can always use for-of loop instead of forEach method. It will look much prettier, and you can use continue, break, yield inside it: var trivialGenerator = function *(array) { for (var item of array) { // … Read more
The next method exists on the generator that the function returns, not on the generator function itself. export default class GeneratorClass { constructor() { const iterator = this.generator(10); iterator.next(); } *generator(count:number): IterableIterator<number> { while(true) yield count++; } }
Just use the .union() method. >>> l = [set([1,2,3]), set([4,5,6]), set([1,4,9])] >>> frozenset().union(*l) frozenset([1, 2, 3, 4, 5, 6, 9]) This works for any iterable of iterables.
The problem was not with the recursion. Your function did call itself recursively, it just didn’t yield values outside. When you call helper(), you get an iterator as a return value, but you wanted the iterated values of that iterator to be yielded. If you want to yield recursively, you need to yield *. Try … Read more
Do you understand list comprehensions? If so, a generator expression is like a list comprehension, but instead of finding all the items you’re interested and packing them into list, it waits, and yields each item out of the expression, one by one. >>> my_list = [1, 3, 5, 9, 2, 6] >>> filtered_list = [item … Read more
If you are using C++ and are concerned about good distribution you can use TR1 C++11 <random>. #include <random> std::random_device rseed; std::mt19937 rgen(rseed()); // mersenne_twister std::uniform_int_distribution<int> idist(0,100); // [0,100] std::cout << idist(rgen) << std::endl;
If your intention is to some other scope, iterate over it, do some other stuff, then be able to iterate over it again later on in that same scope. Then the only thing you shouldn’t try doing is passing the iterator, instead pass the generator: var generator = function*() { yield 1; yield 2; yield … Read more
You can use itertools.cycle (source included on linked page). import itertools a = [1, 2, 3] for element in itertools.cycle(a): print element # -> 1 2 3 1 2 3 1 2 3 1 2 3 …
http://staticsitegenerators.net is a crowd-sourced definitive listing of all the static site generators, their github stars, their website, their language, created and updated dates, etc. +1 for DocPad, I’ve found Jekyll to be quite crippling with it’s lack of extensibility (not enough markups supported, difficult to filter documents in content listings based on certain criteria, hard … Read more
>>> from itertools import groupby >>> L = [0, 0, 0, 3, 3, 2, 5, 2, 6, 6] >>> grouped_L = [(k, sum(1 for i in g)) for k,g in groupby(L)] >>> # Or (k, len(list(g))), but that creates an intermediate list >>> grouped_L [(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, … Read more