Summarizing the comments: you can’t use yield inside a regular function, so you can’t use yield with forEach. Here an example of “generatorized” foreach: function * foreach (arr, fn) { var i for (i = 0; i < arr.length; i++) { yield * fn(arr[i]) } } function * gen (number) { yield number + 1 … Read more
Something like http://jscc.phorward-software.com/, maybe? JS/CC is the first available parser development system for JavaScript and ECMAScript-derivates. It has been developed, both, with the intention of building a productive compiler development system and with the intention of creating an easy-to-use academic environment for people interested in how parse table generation is done general in bottom-up parsing. … Read more
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.
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
There is lots of information about this available. For the “official word” on the issue, read the (closed) Python bugtracker issue. The core reasoning, by one of the people who made the decision, is detailed on this blog: Since a generator is essentially a souped-up function, we would need to save its bytecode, which is … Read more
It’s not thread-safe; simultaneous calls may interleave, and mess with the local variables. The common approach is to use the master-slave pattern (now called farmer-worker pattern in PC). Make a third thread which generates data, and add a Queue between the master and the slaves, where slaves will read from the queue, and the master … Read more
Had the same need so wrote a little class for it. Here are some examples: Generator<Integer> simpleGenerator = new Generator<Integer>() { public void run() throws InterruptedException { yield(1); // Some logic here… yield(2); } }; for (Integer element : simpleGenerator) System.out.println(element); // Prints “1”, then “2”. Infinite generators are also possible: Generator<Integer> infiniteGenerator = new … Read more
Try: var g = f(); // this question is over this value. g.next(); // returns: { value: 1, done: false } g.next(123); // returns: { value: 2, done: false } g.next(456); // returns: { value: [123, 456], done: true }
Try this generator: def generator_two_img(X1, X2, y, batch_size): genX1 = gen.flow(X1, y, batch_size=batch_size, seed=1) genX2 = gen.flow(X2, y, batch_size=batch_size, seed=1) while True: X1i = genX1.next() X2i = genX2.next() yield [X1i[0], X2i[0]], X1i[1] Generator for 3 inputs: def generator_three_img(X1, X2, X3, y, batch_size): genX1 = gen.flow(X1, y, batch_size=batch_size, seed=1) genX2 = gen.flow(X2, y, batch_size=batch_size, seed=1) genX3 … Read more
Simply call list on the generator. lst = list(gen) lst Be aware that this affects the generator which will not return any further items. You also cannot directly call list in IPython, as it conflicts with a command for listing lines of code. Tested on this file: def gen(): yield 1 yield 2 yield 3 … Read more