Use range. In Python 2, it returns a list directly: >>> range(11, 17) [11, 12, 13, 14, 15, 16] In Python 3, range is an iterator. To convert it to a list: >>> list(range(11, 17)) [11, 12, 13, 14, 15, 16] Note: The second number in range(start, stop) is exclusive. So, stop = 16+1 = … Read more
NumPy’s arrays are more compact than Python lists — a list of lists as you describe, in Python, would take at least 20 MB or so, while a NumPy 3D array with single-precision floats in the cells would fit in 4 MB. Access in reading and writing items is also faster with NumPy. Maybe you … Read more
Python has a language feature called List Comprehensions that is perfectly suited to making this sort of thing extremely easy. The following statement does exactly what you want and stores the result in l3: l3 = [x for x in l1 if x not in l2] l3 will contain [1, 6].
If you are exposing your class through a library that others will use, you generally want to expose it via interfaces rather than concrete implementations. This will help if you decide to change the implementation of your class later to use a different concrete class. In that case the users of your library won’t need … Read more
You can use a list comprehension with enumerate: indices = [i for i, x in enumerate(my_list) if x == “whatever”] The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down … Read more
dict.values returns a view of the dictionary’s values, so you have to wrap it in list: list(d.values())
Iterate your list in reverse with a for loop: for (int i = safePendingList.Count – 1; i >= 0; i–) { // some code // safePendingList.RemoveAt(i); } Example: var list = new List<int>(Enumerable.Range(1, 10)); for (int i = list.Count – 1; i >= 0; i–) { if (list[i] > 5) list.RemoveAt(i); } list.ForEach(i => Console.WriteLine(i)); … Read more
Shortest Code [x for _, x in sorted(zip(Y, X))] Example: X = [“a”, “b”, “c”, “d”, “e”, “f”, “g”, “h”, “i”] Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1] Z = [x for _,x in sorted(zip(Y,X))] print(Z) # [“a”, “d”, “h”, “b”, “c”, “e”, “i”, “f”, “g”] Generally Speaking [x for … Read more
If you don’t want duplicates in a Collection, you should consider why you’re using a Collection that allows duplicates. The easiest way to remove repeated elements is to add the contents to a Set (which will not allow duplicates) and then add the Set back to the ArrayList: Set<String> set = new HashSet<>(yourList); yourList.clear(); yourList.addAll(set); … Read more
Use str.join: >>> words = [‘this’, ‘is’, ‘a’, ‘sentence’] >>> ‘-‘.join(words) ‘this-is-a-sentence’ >>> ‘ ‘.join(words) ‘this is a sentence’