Have a look at the Except method, which you use like this: var resultingList = listOfOriginalItems.Except(listOfItemsToLeaveOut, equalityComparer) You’ll want to use the overload I’ve linked to, which lets you specify a custom IEqualityComparer. That way you can specify how items match based on your composite key. (If you’ve already overridden Equals, though, you shouldn’t need … Read more
Here is a demo of possible solution. Tested with Xcode 12b. List { ForEach(0..<3) { _ in VStack { Text(“Hello, World!”).padding(.leading) } .frame(maxWidth: .infinity, maxHeight: .infinity, alignment: .leading) .listRowInsets(EdgeInsets()) .background(Color(UIColor.systemBackground))) } }
How about something simple like: >>> [x for x in sents if not x.startswith(‘@$\t’) and not x.startswith(‘#’)] [‘this doesnt’, ‘this shouldnt’, ‘this isnt’, ‘this musnt’]
[:] creates a slice, usually used to get just a part of a list. Without any minimum/maximum index given, it creates a copy of the entire list. Here’s a Python session demonstrating it: >>> a = [1,2,3] >>> b1 = a >>> b2 = a[:] >>> b1.append(50) >>> b2.append(51) >>> a [1, 2, 3, 50] … Read more
You should use the HashSet<T> class, which is specifically designed for what you’re doing.
With Convert a list of characters into a string you can just do ”.join(map(str,new))
Use itertools.cycle to cycle around to the beginning of L2: from itertools import cycle dict(zip(L1, cycle(L2))) # {‘A’: ‘1’, ‘B’: ‘2’, ‘C’: ‘3’, ‘D’: ‘1’, ‘E’: ‘2’} In your case, concatenating L2 with itself also works. # dict(zip(L1, L2 * 2)) dict(zip(L1, L2 + L2)) # {‘A’: ‘1’, ‘B’: ‘2’, ‘C’: ‘3’, ‘D’: ‘1’, ‘E’: … Read more
You can use bitshifting: out = 0 for bit in bitlist: out = (out << 1) | bit This easily beats the “int cast” method proposed by A. R. S., or the modified cast with lookup proposed by Steven Rumbalski: >>> def intcaststr(bitlist): … return int(“”.join(str(i) for i in bitlist), 2) … >>> def intcastlookup(bitlist): … Read more
How about: var most = list.GroupBy(i=>i).OrderByDescending(grp=>grp.Count()) .Select(grp=>grp.Key).First(); or in query syntax: var most = (from i in list group i by i into grp orderby grp.Count() descending select grp.Key).First(); Of course, if you will use this repeatedly, you could add an extension method: public static T MostCommon<T>(this IEnumerable<T> list) { return … // previous code … Read more
If you’re worried about performance, the following code will stop as soon as it finds the first duplicate item – all the other solutions so far require the whole input to be iterated at least once. var hashset = new HashSet<string>(); if (myList.Where(s => s != null).Any(s => !hashset.Add(s))) { // there are duplicates } … Read more