Now someone told me that this is not very good programming because I use the return statement inside a loop and this would cause garbage collection to malfunction. That’s incorrect, and suggests you should treat other advice from that person with a degree of skepticism. The mantra of “only have one return statement” (or more … Read more
even shorter if you can lose the yearStart value: var yearStart = 2000; var yearEnd = 2040; var arr = []; while(yearStart < yearEnd+1){ arr.push(yearStart++); } UPDATE: If you can use the ES6 syntax you can do it the way proposed here: let yearStart = 2000; let yearEnd = 2040; let years = Array(yearEnd-yearStart+1) .fill() … Read more
Try for (a <- 0 until max by 2)
You can use random.shuffle() to, well, shuffle a list: import random r = list(range(1000)) random.shuffle(r) for i in r: # do something with i By the way, in many cases where you’d use a for loop over a range of integers in other programming languages, you can directly describe the “thing” you want to iterate … Read more
Here is an easier (and more Swifty) approach. for i in (0 ..< 5).reversed() { print(i) // 4,3,2,1,0 } let array = [“a”, “b”, “c”, “d”, “e”] for element in array.reversed() { print(element) // e,d,c,b,a } array.reversed().forEach { print($0) } // e,d,c,b,a print(Array(array.reversed())) // e,d,c,b,a
Because std::vector<bool> is not a container ! std::vector<T>‘s iterators usually dereference to a T&, which you can bind to your own auto&. std::vector<bool>, however, packs its bools together inside integers, so you need a proxy to do the bit-masking when accessing them. Thus, its iterators return a Proxy. And since the returned Proxy is an … Read more
Direct initialization using a constructor vs copy initialization These are exactly identical for ints and will generate identical code. Use whichever one you prefer to read or what your code policies are, etc. != could be faster in hardware than < The generated code won’t actually be i < n vs i != n, it’ll … Read more
If you have some given list, and want to iterate over its items and indices, you can use enumerate(): for index, item in enumerate(my_list): print index, item If you only need the indices, you can use range(): for i in range(len(my_list)): print i
I thought it would be useful to actually time some of the solutions presented here on a larger sample input. For this input and on my machine, I find Cardstdani’s approach to be the fastest, followed by the numpy isin() approach. Setup 1 import random list_1 = [random.randint(1, 10_000) for i in range(100_000)] list_2 = … Read more
Is it possible to find the present index in an enhanced for loop? No. If you need the index, I suggest you use an ordinary for-loop. However, you don’t actually seem to need the index in this situation. Unless you’re dealing with some really strange type of list, you could go with an Iterator and … Read more