Iterating Backwards through the List sounds like a better approach, because if you remove an element and other elements “fall into the gap”, that does not matter because you have already looked at those. Also, you do not have to worry about your counter variable becoming larger than the .Count. List<int> test = new List<int>(); … Read more
Iterative – you don’t finish a feature in one go. You are in a code >> get feedback >> code >> … cycle. You keep iterating till done. Incremental – you build as much as you need right now. You don’t over-engineer or add flexibility unless the need is proven. When the need arises, you … Read more
Why don’t you just do it the idiomatic C way? This ought to be bullet-proof, but it won’t be fast. I’m pretty sure indexing into a list in Python walks the linked list, so this is a “Shlemiel the Painter” algorithm. But I tend not to worry about optimization until it becomes clear that a … Read more
It’s clear you’re using numpy. With numpy you can just do: for cell in self.cells.flat: do_somethin(cell)
The reason the set iteration order changes from run-to-run appears to be because Python uses hash seed randomization by default. (See command option -R.) Thus set iteration is not only arbitrary (because of hashing), but also non-deterministic (because of the random seed). You can override the random seed with a fixed value by setting the … Read more
There’s nothing wrong with direct, “ad-hoc” solution. It can be clean enough too. Just notice that spiral is built from segments. And you can get next segment from current one rotating it by 90 degrees. And each two rotations, length of segment grows by 1. edit Illustration, those segments numbered … 11 10 7 7 … Read more
You need to set the target compiler option to es6 or higher for NodeListOf<T> to be iterable.
I don’t know about cleaner, but there’s another alternative: for (op, code) in zip(s[0::2], s[1::2]): print op, code A no-copy version: from itertools import izip, islice for (op, code) in izip(islice(s, 0, None, 2), islice(s, 1, None, 2)): print op, code
If you want to divide a list into slices you can use this trick: list_of_slices = zip(*(iter(the_list),) * slice_size) For example >>> zip(*(iter(range(10)),) * 3) [(0, 1, 2), (3, 4, 5), (6, 7, 8)] If the number of items is not dividable by the slice size and you want to pad the list with None … Read more
Some quick timing runs seem to give the 2nd option using range() a slight edge over enumerate(): timeit a = [f(n) for n, _ in enumerate(mlist)] 10000 loops, best of 3: 118 us per loop timeit a = [f(n) for n in range(len(mlist))] 10000 loops, best of 3: 102 us per loop and just for … Read more