set

问题 This question already has answers here : user defined object equality for a set in harmony (es6) (2 answers) Closed 4 years ago . I want to create a set of value objects in JavaScript. The problem is that in JavaScript equality is based on identity. Hence, two different objects with the same value will be treated as unequal: var objects = new Set; objects.add({ a: 1 }); objects.add({ a: 1 }); alert(objects.size); // expected 1, actual 2 How do you work around this problem? 回答1: Use JSON

问题 This question already has answers here : Converting a list to a set changes element order (10 answers) Does Python have an ordered set? (12 answers) Closed 6 years ago . In Python3 I wrote a simple one line code as follows : { 2*x for x in {1,2,3,4} } but I m getting answer like this (order changed). {8, 2, 4, 6} Why I am getting answer {8,2,4,6} instead of {2,4,6,8} ? 回答1: That's because sets don't have any order. They're unordered collection. help on set : Build an unordered collection of

问题 I've been following the tutorial on the django website and have been running into a problem regarding setting up Django: https://docs.djangoproject.com/en/dev/intro/tutorial01/ I've installed Django through their guide using setup.py, but when I run "django-admin.py startproject mysite" on the terminal, it outputs the error "-bash: django-admin.py: command not found". I've tried django-admin as per some google searches and some searches have told me to try the following commands cd /usr/local

问题 I have this dataframe: d = {'city':['Barcelona','Madrid','Rome','Torino','London','Liverpool','Manchester','Paris'], 'country': ['ES','ES','IT','IT','UK','UK','UK','FR'], 'revenue': [1,2,3,4,5,6,7,8], 'amount': [8,7,6,5,4,3,2,1] df = pd.DataFrame(d) I want to obtain this for each country: españa = {'city':['Barcelona','Madrid'] 'revenue':[1,2] 'amount':[8,7]} ES = pd.DataFrame(españa) So that in the end I will have 4 dataframes named ES,IT,UK and FR. I have tried this so far: a = set(df.loc[:

问题 I have this code: set<int>::iterator new_end = set_difference(set1.begin(), set1.end(), set2.begin(), set2.end(), set1.begin()); set1.erase(new_end, set1.end); It compiles and runs fine in visual studio. However, in a previous question, people stated that a set 's iterators are supposed to be const . I don't see anything like that in the standard. Can someone tell me where it says that, or if this is well-defined behavior? If it's not, please provide code that does what I need. Is there a way

问题 Is there any way to observe additions to and removals from ES6 Maps and Sets? Object.observe doesn't work because it is only applies to direct properties of the observed object. Hypothetically the size property could be observed, but no indication would be provided of exactly what has changed. Another idea would be to replace the object's set and get functions with proxified versions. Is there a better way? If not, I'm surprised that nobody thought of this when the proposals were being

问题 So what I know is that a HashSet has no real sorting capabilities like a SortedSet, however I stumbled upon this : When I run the following code : public static void main(String[] args) { Set<String> collection = new HashSet<String>(2000); String[] data = {"a", "c", "g", "f", "b", "f", "b", "d","q","r","d","m"}; for(String input: data) { collection.add(input); } System.out.println("Output: " + collection); } I get the following output : Output: [a, b, c, d, f, g, m, q, r] Which is

问题 The C++ draft says about std::lower_bound: § 25.4.3.1 lower_bound [lower.bound] template<class ForwardIterator, class T> ForwardIterator lower_bound(ForwardIterator first, ForwardIterator last, const T& value); template<class ForwardIterator, class T, class Compare> ForwardIterator lower_bound(ForwardIterator first, ForwardIterator last, const T& value, Compare comp); Requires: The elements e of [first,last) shall be partitioned with respect to the expression e < value or comp(e, value).

问题 I'm having trouble understanding whether it is possible to prove that two sets (in this case regular languages) are identical and thus interchangeable. From what I understand, sets can be equivalent even if they are not constructively equal. Regular languages are sets of strings, but I don't see how to say that r1 = r2 so that something like symmetry can be used in a proof. Here is my RegularLanguage declaration: Inductive RegularLanguage (A : Set) : Set := | EmptyLang : RegularLanguage A |

问题 I wonder if it's possible to define Set in Hibernate mapping in a such way, that element would specify not column in original ( FOO ) table, but in joined one ( BAR ). Let's say we have some FooContainer.hbm.xml , which contains Set of Foo objects: <set ...> <key column="COLUMN_FROM_BAR" /> <one-to-many class="xyz.Foo" /> </set> Here FOO has FK to BAR ( FOO.BAR_ID ), so joining is done through element in Foo.hbm.xml: <many-to-one class="xyz.Bar" fetch="join" column="BAR_ID" foreign-key="barId