set

问题 I have two arrays, say A and B with |A|=8 and |B|=4. I want to calculate the set difference A-B. How do I proceed? Please note that there are no repeated elements in either of the sets. Edit: Thank you so much everybody for a myriad of elegant solutions. Since I am in prototyping stage of my project, for now I implemented the simplest solution told by Brian and Owen. But I do appreciate the clever use of data structures as suggested here by the rest of you, even Though I am not a computer

问题 I was wondering whether I should have my data structure as a set or a list. Mostly I will do set operations, but in the end I will need to sort it. I wondered whether I should first make the set a list, and then use sorted(list(my_set)) , or just sort the set immediately sorted(my_set) . Arguably, I might consider a general "to list" phase, since having an ordered iterable at that point in time might make sense anyway. So I decided to test it, expecting a list to be quicker. Benchmarker:

问题 For any given set, for instance, val fruits = Set("apple", "grape", "pear", "banana") how to get a random element from fruits ? Many Thanks. 回答1: convert into Vector and get random element from it scala> val fruits = Set("apple", "grape", "pear", "banana") fruits: scala.collection.immutable.Set[String] = Set(apple, grape, pear, banana) scala> import scala.util.Random import scala.util.Random scala> val rnd=new Random rnd: scala.util.Random = scala.util.Random@31a9253 scala> fruits.toVector

问题 Given two sets, each containing integer values, how could one find a set containing all possible pair-wise ORs of the values of those two sets? E.g. (all numbers are binary) {1, 10} x {100, 1000} = {101, 1001, 110, 1010} {1, 10} x {11, 101} = {11, 101, 111} The first example results in full four combinations, while the second one only results in three, since both sets share some bits. Obviously the result can be calculated in O(m*n) , but is there a faster way of solving this, taking into

问题 I am trying to define (1,2,3) as a set of elements in coq. I can define it using list as (1 :: (2 :: (3 :: nil))). Is there any way to define set in coq without using list. 回答1: The are basically four possible choices to be made when defining sets in Coq depending on your constraints on the base type of the set and computation needs: If the base type doesn't have decidable equality, it is common to use: Definition Set A := A -> Prop Definition cup A B := fun x => A x /\ B x. ... basically,

问题 Say I have a struct , which could be anything: struct Cube { var x: Int var y: Int var z: Int var width: Int // ... } How do I then create a Set of these points, such that there will be no two objects with the same properties? let points: Set<Cube> = Set() // Type ‘Cube’ does not conform to protocol ‘Hashable’ But it’s not immediately obvious how to implement hashable. From what I read, I need to make a hash function, but that doesn’t look easily possible with the amount of properties I have

问题 I think this may be related to set being mutable . Basically, I can remove an element from a set using set.discard(element) . However, set.discard(element) itself returns None . But I'd like to get a copy of the updated set. For example, if I have a list of sets, how can I get the an updated copy conveniently using list comprehension operations? Sample code: test = [{'', 'a'}, {'b', ''}] print [x.discard('') for x in test] print test will return [None, None] [set(['a']), set(['b'])] 回答1: You

问题 Before I post this question, I found somehow similar question posted here. But the answer was based on a String. However, I have a different situation here. I am not trying to remove String but another object called AwardYearSource. This class has an int attribute called year. So I want to remove duplicates based on the year. i.e if there is year 2010 mentioned more than once, I want to remove that AwardYearSource object. How can I do that? 回答1: The simplest way to remove elements based on a

问题 Before I post this question, I found somehow similar question posted here. But the answer was based on a String. However, I have a different situation here. I am not trying to remove String but another object called AwardYearSource. This class has an int attribute called year. So I want to remove duplicates based on the year. i.e if there is year 2010 mentioned more than once, I want to remove that AwardYearSource object. How can I do that? 回答1: The simplest way to remove elements based on a

问题 I have been looking into the speeds of different Java collection types and have come across something weird. I am adding 1,000,000 objects from a static array to a different collection type and returning the time required. This part of the code works fine. Under further investigation I noticed that the TreeSet is not receiving all of the 1,000,000 objects, and is receiving a different amount each time. Below is the method to transfer the objects from an array to the TreeSet : public int