set

问题 I have a list of items, and each item has a quantity. var items = { 1: 12, // we have 12 x item1 2: 1, // we have 1 x item2 3: 1, 4: 7, 5: 2, 6: 2 }; Alternatively this could be viewed as: var items = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6]; How would you go about obtaining a list of every combination of these items, keeping in mind that the order is entirely unimportant (and hence [1,2,3] == [3,2,1] ), and that not every item has to exist in the result. I

问题 I have a list of items, and each item has a quantity. var items = { 1: 12, // we have 12 x item1 2: 1, // we have 1 x item2 3: 1, 4: 7, 5: 2, 6: 2 }; Alternatively this could be viewed as: var items = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6]; How would you go about obtaining a list of every combination of these items, keeping in mind that the order is entirely unimportant (and hence [1,2,3] == [3,2,1] ), and that not every item has to exist in the result. I

问题 I am creating a GAMS model to solve a simple maximization problem. I have a set J with 3 elements (1,2,3) and a variable x(J) that encompasses all the elements. I am wondering if there is a way in GAMS to set a lower bound of 0 and upper bound of 3 to each element in the set without having to set each element bound individually and without using the positive variable keyword for the lower bound. I have tried using x.lo =e= 0 and x.up =e= 3 but none of these are working. I am guessing I am not

问题 I am creating a GAMS model to solve a simple maximization problem. I have a set J with 3 elements (1,2,3) and a variable x(J) that encompasses all the elements. I am wondering if there is a way in GAMS to set a lower bound of 0 and upper bound of 3 to each element in the set without having to set each element bound individually and without using the positive variable keyword for the lower bound. I have tried using x.lo =e= 0 and x.up =e= 3 but none of these are working. I am guessing I am not

问题 So I've generated a list of combinations, using itertools and I'm getting a result that looks like this: nums = [-5,5,4,-3,0,0,4,-2] x = [x for x in set(itertools.combinations(nums, 4)) if sum(x)==target] >>> x = [(-5, 5, 0, 4), (-5, 5, 4, 0), (5, 4, -3, -2), (5, -3, 4, -2)] What is the most time-complexity wise efficient way of removing unordered duplicates, such as x[0] and x[1] are the duplicates. Is there anything built in to handle this? My general approach would be to create a counter

问题 This question already has answers here : Does Python have an ordered set? (12 answers) Closed 5 years ago . EDIT: Thanks. I assumed sets were ordered since the output is almost in alphabetical order. But yes I want an ordered set. def create_table(secret): sec = set(secret) alpha = set("abcdefghiklmnopqrstuvwxyz") bet = alpha - sec newSecret = secret & bet print newSecret OUTPUT: set(['A', 'C', 'B', 'E', 'D', 'G', 'F', 'I', 'H', 'K', 'M', 'L', 'O', 'N', 'Q', 'P', 'S', 'R', 'U', 'T', 'W', 'V',

问题 This question already has answers here : Does Python have an ordered set? (12 answers) Closed 5 years ago . EDIT: Thanks. I assumed sets were ordered since the output is almost in alphabetical order. But yes I want an ordered set. def create_table(secret): sec = set(secret) alpha = set("abcdefghiklmnopqrstuvwxyz") bet = alpha - sec newSecret = secret & bet print newSecret OUTPUT: set(['A', 'C', 'B', 'E', 'D', 'G', 'F', 'I', 'H', 'K', 'M', 'L', 'O', 'N', 'Q', 'P', 'S', 'R', 'U', 'T', 'W', 'V',

问题 This question already has answers here : Does Python have an ordered set? (12 answers) Closed 5 years ago . EDIT: Thanks. I assumed sets were ordered since the output is almost in alphabetical order. But yes I want an ordered set. def create_table(secret): sec = set(secret) alpha = set("abcdefghiklmnopqrstuvwxyz") bet = alpha - sec newSecret = secret & bet print newSecret OUTPUT: set(['A', 'C', 'B', 'E', 'D', 'G', 'F', 'I', 'H', 'K', 'M', 'L', 'O', 'N', 'Q', 'P', 'S', 'R', 'U', 'T', 'W', 'V',

问题 JavaDoc defines set as : A collection that contains no duplicate elements. More formally, sets contain no pair of elements e1 and e2 such that e1.equals(e2) To verify the same, i created a very simple program: import java.util.HashSet; public class CheckHashSet { public static void main(String[] args) { HashSet<Employee> set = new HashSet<Employee>(); set.add(new Employee(10)); set.add(new Employee(10)); System.out.println(set.size()); System.out.println(new Employee(10).equals(new Employee

问题 The bounty expires in 7 days . Answers to this question are eligible for a +250 reputation bounty. Leo Dabus is looking for an answer from a reputable source . I am trying to implement a generic Mutable Ordered Set type and it needs to conform to many protocols to behave the same way as an Array and a Set does in Swift. First of all to accomplish that the generic type element needs to conform to Hashable and the generic struct needs to conform to RandomAccessCollection, SetAlgebra,