cartesian-product

问题 I'm looking for an example of how, in Ruby, a C like language, or pseudo code, to create the Cartesian product of a variable number of arrays of integers, each of differing length, and step through the results in a particular order: So given, [1,2,3],[1,2,3],[1,2,3]: [1, 1, 1] [2, 1, 1] [1, 2, 1] [1, 1, 2] [2, 2, 1] [1, 2, 2] [2, 1, 2] [2, 2, 2] [3, 1, 1] [1, 3, 1] etc. Instead of the typical result I've seen (including the example I give below): [1, 1, 1] [2, 1, 1] [3, 1, 1] [1, 2, 1] [2, 2,

问题 I'm looking for an example of how, in Ruby, a C like language, or pseudo code, to create the Cartesian product of a variable number of arrays of integers, each of differing length, and step through the results in a particular order: So given, [1,2,3],[1,2,3],[1,2,3]: [1, 1, 1] [2, 1, 1] [1, 2, 1] [1, 1, 2] [2, 2, 1] [1, 2, 2] [2, 1, 2] [2, 2, 2] [3, 1, 1] [1, 3, 1] etc. Instead of the typical result I've seen (including the example I give below): [1, 1, 1] [2, 1, 1] [3, 1, 1] [1, 2, 1] [2, 2,

问题 Using pythons itertools , I'd like to create an iterator over the outer product of all permutations of a bunch of lists. An explicit example: import itertools A = [1,2,3] B = [4,5] C = [6,7] for x in itertools.product(itertools.permutations(A),itertools.permutations(B),itertools.permutations(C)): print x While this works, I'd like to generalize it to an arbitrary list of lists. I tried: for x in itertools.product(map(itertools.permutations,[A,B,C])): print x but it did not do what I intended.

问题 Using pythons itertools , I'd like to create an iterator over the outer product of all permutations of a bunch of lists. An explicit example: import itertools A = [1,2,3] B = [4,5] C = [6,7] for x in itertools.product(itertools.permutations(A),itertools.permutations(B),itertools.permutations(C)): print x While this works, I'd like to generalize it to an arbitrary list of lists. I tried: for x in itertools.product(map(itertools.permutations,[A,B,C])): print x but it did not do what I intended.

问题 I am really confused about this. I searched a lot of tutorials but i couldnot find a clear answer. A B B D 1 X x 5 2 x y 6 x 4 I want to cross this two tables.A , B, B,d are attributes. A B B D 1 X x 5 2 x x 5 1 X y 6 2 x y 6 1 X x 4 2 x x 4 This should be answer normally according to rule of cartesian. Cross all rows. But i am confused about same column B. Same column will seem twice? 回答1: Some relational query language/algebra relations have ordered column names . There is a way to

问题 I hope someone is able to help me with what is, at least to me, quite a tricky algorithm. The Problem I have a List ( 1 <= size <= 5 , but size unknown until run-time) of Lists ( 1 <= size <= 2 ) that I need to combine. Here is an example of what I am looking at:- ListOfLists = { {1}, {2,3}, {2,3}, {4}, {2,3} } So, there are 2 stages to what I need to do:- (1). I need to combine the inner lists in such a way that any combination has exactly ONE item from each list, that is, the possible

问题 I'm trying to construct multiple cross products of traversables of different (but each homogeneous) types. The desired return type is a traversable of a tuple with the type matching the types in the input traversables. For example: List(1, 2, 3) cross Seq("a", "b") cross Set(0.5, 7.3) This should give a Traversable[(Int, String, Double)] with all possible combinations from the three sources. The case of combining only two sources was nicely answered here. The given idea is: implicit class

问题 I have an unknown number of buckets(collections), and each bucket having an unknown number of entities I need to produce a cartesian product of all the entities, so that I endup with a single COLLECTION that has ARRAYS of entities and in each array, there is 1 representetive from EVERY bucket. So that if I have 5 buckets (B1..B5), and buckets B1, B2 have 1 item each, and bucket B3, B4 and B5 have 4, 8 and 10 items each, I'll have a collection of 320 arrays, and each array will have 5 items.

问题 Say I have a list, and I want to produce a list of all unique pairs of elements without considering the order. One way to do this is: mylist = ['W','X','Y','Z'] for i in xrange(len(mylist)): for j in xrange(i+1,len(mylist)): print mylist[i],mylist[j] W X W Y W Z X Y X Z Y Z I want to do this with iterators, I thought of the following, even though it doesn't have brevity: import copy it1 = iter(mylist) for a in it1: it2 = copy.copy(it1) for b in it2: print a,b But this doesn't even work. What

问题 I am trying to understand what will be the result of performing a natural join between two relations R and S, where they have no common attributes. By following the below definition, I thought the answer might be an empty set: Natural Join definition. My line of thought was because the condition in the 'Select' symbol is not met, the projection of all of the attributes won't take place. When I asked my lecturer about this, he said that the output will be the same as doing a cartezian product