The for expression in your local value association-list is producing a list of list of lists of strings, but your references to it are treating it as a list of lists of strings. To get the flattened representation you wanted, you can use the flatten function, but because it would otherwise group everything into a … Read more
I agree with the answer currently proposed by fedorqui in the context of the question currently asked. The below is given only to provide some more general answers. One more general approach (for bash 4.0 or newer) is to store your pairs in an associative array: declare -A pairs=( [4_1]=4_2 [5_1]=5_2 [6_1]=6_2 [7_1]=7_2 [8_1]=8_2 ) … Read more
If you’re frequently iterating over a Cartesian product like in your example, you might want to investigate Python 2.6’s itertools.product — or write your own if you’re in an earlier Python. from itertools import product for y, x in product(range(3), repeat=2): do_something() for y1, x1 in product(range(3), repeat=2): do_something_else()
You can use two flatMap then a filter then you can pick the first one or if no result return null : C c1 = listOfAObjects.stream() .flatMap(a -> a.getList().stream()) .flatMap(b -> b.getPr().stream()) .filter(c -> c.getName().equalsIgnoreCase(name)) .findFirst() .orElse(null);
In Java you can use a label to specify which loop to break/continue: mainLoop: while (goal <= 100) { for (int i = 0; i < goal; i++) { if (points > 50) { break mainLoop; } points += i; } }
The recommended way in Python for breaking nested loops is… Exception class Found(Exception): pass try: for i in range(100): for j in range(1000): for k in range(10000): if i + j + k == 777: raise Found except Found: print i, j, k
The labels in JavaScript are used mainly with break, or continue in nested loops to be able to break the outer, or continue the outer loop from the code inside inner loop: outer: for (let i = 0; i < 10; i++) { let k = 5; for (let j = 0; j < 10; … Read more
The question does not contain a nested loop, just a single loop. But THIS nested version works, too: # for i in c d; do for j in a b; do echo $i $j; done; done c a c b d a d b
It’s an issue of memory cache. matrix[i][j] has better cache hits than matrix[j][i], since matrix[i][j] has more continuous memory accessing chances. For example, when we access matrix[i][0], the cache may load a continuous segment of memory containing matrix[i][0], thus, accessing matrix[i][1], matrix[i][2], …, will benefit from caching speed, since matrix[i][1], matrix[i][2], … are near to … Read more
Yep, it’s still O(n^2), it has a smaller constant factor, but that doesn’t affect O notation.