set

问题 I am working removing the duplicate Dictionaries in an array of Dictionaries in swift 3.0 The below is the let Dict1 : [String : String] = ["messageTo":"Madhu"] let Dict2 : [String : String] = ["messageTo":"Kiran"] let Dict3 : [String : String] = ["messageTo":"Raju"] var arrOfDict = [[String:String]]() arrOfDict.append(Dict1) arrOfDict.append(Dict2) arrOfDict.append(Dict1) arrOfDict.append(Dict3) arrOfDict.append(Dict2 print(arrOfDict) //prints [["messageTo": "Madhu"], ["messageTo": "Kiran"],

问题 I am trying to update field value in mongoose. { "_id" : ObjectId("5b62c772efedb6bd3f0c983a"), "projectID" : ObjectId("0000000050e62416d0d75837"), "__v" : 0, "clientID" : ObjectId("00000000996b902b7c3f5efa"), "inspection_data" : [ { "pdf" : null, "published" : "N", "submissionTime" : ISODate("2018-08-02T08:57:08.532Z"), "userID" : ObjectId("00000000cac68e3bc04643f7"), "insSummary" : "inspected areas", "insName" : "Infotech", "_id" : ObjectId("5b62c772fa02622a18655e7b"), "published_date" :

问题 I have the following code in C++ #include <iostream> #include <set> using namespace std; int main() { set<int> a; int n; for(int i=0;i<3;i++){ cin>>n; a.insert(n); } cout << *a.end(); return 0; } Why is it always printing "3" instead of the greatest element in the set? Replacing cout << *a.end(); with cout << *--a.end(); works fine. 回答1: Why is it always printing "3" instead of the greatest element in the set? Because a.end() is one past the end of your vector. It doesn't hold any valid data,

问题 public static void main(String[] args) { HashSet set = new HashSet(); set.add(new StringBuffer("abc")); set.add(new StringBuffer("abc")); set.add(new StringBuffer("abc")); set.add(new StringBuffer("abc")); System.out.println(set); } Output: [abc,abc,abc,abc] Here in above code I added object of StringBuffer("abc") many times and Set adds it but Set never adds duplicates. 回答1: StringBuffer does not override Object#equals() and Object#hashCode(), so identity of StringBuffer instances is based

问题 I have two lists A and B (List). How to determine if they are equal in the cheapest way? I can write something like '(A minus B) union (B minus A) = empty set' or join them together and count amount of elements, but it is rather expensive. Is there workaround? 回答1: Well, that depends on how you interpret your lists. If you consider them as tuples (so the order of elements in lists matters), then you can go with this code: public bool AreEqual<T>(IList<T> A, IList<T> B) { if (A.Count != B

问题 I'm using gcc 4.3.3 to try to compile the following code: struct testStruct { int x; int y; bool operator<(testStruct &other) { return x < other.x; } testStruct(int x_, int y_) { x = x_; y = y_; } }; int main() { multiset<testStruct> setti; setti.insert(testStruct(10,10)); return 0; } I get this error: /usr/include/c++/4.4/bits/stl_function.h|230|error: no match for ‘operator<’ in ‘__x < __y’ I suspect I'm not doing the the operator overloading as it should be done, but I just can't pinpoint

问题 I've been playing with sets in SQL Server 2000 and have the following table structure for one of my temp tables (#Periods): RestCTR HoursCTR Duration Rest ---------------------------------------- 1 337 2 0 2 337 46 1 3 337 2 0 4 337 46 1 5 338 1 0 6 338 46 1 7 338 2 0 8 338 46 1 9 338 1 0 10 339 46 1 ... What I'd like to do is to calculate the Sum of the 2 longest Rest periods for each HoursCTR, preferably using sets and temp tables (rather than cursors, or nested subqueries). Here's the

问题 I've been playing with sets in SQL Server 2000 and have the following table structure for one of my temp tables (#Periods): RestCTR HoursCTR Duration Rest ---------------------------------------- 1 337 2 0 2 337 46 1 3 337 2 0 4 337 46 1 5 338 1 0 6 338 46 1 7 338 2 0 8 338 46 1 9 338 1 0 10 339 46 1 ... What I'd like to do is to calculate the Sum of the 2 longest Rest periods for each HoursCTR, preferably using sets and temp tables (rather than cursors, or nested subqueries). Here's the

问题 I can understand that multiset has count(), for counting the number of occurrences of a value, because elements can be repeated in multiset. But what's the point of having count() in set, when all the values are already unique? 回答1: count is part of the associative container requirements (1) . Every associative container is required to provide it as part of its interface, even if the result is always zero or one as is the case with std::set . (1) This is a link to the SGI STL documentation

问题 Does the C++ standard library have an "ordered set" datastructure? By ordered set, I mean something that is exactly the same as the ordinary std::set but that remembers the order in which you added the items to it. If not, what is the best way to simulate one? I know you could do something like have a set of pairs with each pair storing the number it was added in and the actual value, but I dont want to jump through hoops if there is a simpler solution. 回答1: No single, homogeneous data