通过递归地将用户输入与产品特征匹配来创建产品包

Question

我正在从事 Product Bundle 创建和推荐项目。 捆绑和推荐必须根据用户输入实时进行。

条件是 1. 产品包应尽可能涵盖用户输入。 2. 推荐项目中用户输入的重复应该更少。

user_input=['a','b','c']

d1=['a','c','d']
d2=['a','b','e','f']
d3=['a','c','b','f']
d4=['b']
d5=['g','e','a']
d6=['g']

expected output - d1 + d4, d3, d1+d2,d5+d4+d1

以下是我的代码，它给出了结果，但它显示了重复的结果，也没有显示所有的组合。 任何帮助表示赞赏。

dlist=[d1,d2,d3,d4,d5,d6]
diff_match=[]
# find match n diff of each product based on user input
for i in range(len(dlist)):
    match=set(user_input).intersection(set(dlist[i]))
    #print("match is",match)
    diff=set(user_input).difference(set(dlist[i]))
    #print("diff is",diff)
    temp={'match':match,'diff':diff}
    diff_match.append(temp)
    
for i in range(len(diff_match)):
    # if match is found, recommend the product alone
    diff_match_temp=diff_match[i]['match']
    print("diff match temp is",diff_match_temp)
    if diff_match_temp==user_input:
        print ("absolute match")
    #scenario where the user input is subset of product features, seperate from partial match
    elif (all(x in list(diff_match_temp) for x in list(user_input))):
        print("User input subset of product features")
        print("The parent list is",diff_match[i]['match'])
        print("the product is", dlist[i])
    else:
        '''else check if the difference between user input and the current product is fulfilled by other product, 
        if yes, these products are bundled together'''
        for j in range(len(diff_match)):
            temp_diff=diff_match[i]['diff']
            print("temp_diff is",temp_diff)
            # empty set should be explicitly checked to avoid wrong match
            if (temp_diff.intersection(diff_match[j]['match'])==temp_diff and len(temp_diff)!=0 and list(temp_diff) != user_input) :
            #if temp_diff==diff_match[j]['match'] and len(temp_diff)!=0 and list(temp_diff) != user_input :
                print("match found with another product")
                print("parent is",dlist[i])
                print("the combination is",dlist[j] )

Answer 1

要递归执行此操作，您可以创建一个 function 返回具有最大组件覆盖率的产品并递归以完成具有剩余组件的捆绑包：

def getBundles(C,P):
    cSet     = set(C) # use set operations,  largest to smallest coverage
    coverage = sorted(P.items(),key=lambda pc:-len(cSet.intersection(pc[1])))
    for i,(p,cs) in enumerate(coverage):
        if cSet.isdisjoint(cs):continue          # no coverage
        if cSet.issubset(cs): yield [p];continue # complete (stop recursion)
        remaining = cSet.difference(cs)   # remaining components to bundle
        unused    = dict(coverage[i+1:])  # products not already bundled
        yield from ([p]+rest for rest in getBundles(remaining,unused))

output：

prods = {"d1":['a','c','d'],
         "d2":['a','b','e','f'],
         "d3":['a','c','b','f'],
         "d4":['b'],
         "d5":['g','e','a'],
         "d6":['g']}

user_input=['a','b','c']

for bundle in getBundles(user_input,prods):
    print(bundle)

['d3']
['d1', 'd2']
['d1', 'd4']

请注意，排除了诸如 ['d5','d1','d2'] 之类的冗余组合，因为 d1+d2 涵盖了 d5 涵盖的所有内容以及更多内容，因此 d5 是多余的。 相同产品的排列也被排除在外（例如 d1+d2 与 d2+d1 相同）

[编辑]

如果您需要提供冗余组合（可能作为扩展选择选项），您可以编写一个略有不同的递归 function，它不会排除它们。 在呈现结果时，您还应该按照最接近的顺序对结果进行排序：

def getBundles(C,P,remain=None,bundle=None):
    cSet  = set(C)                               # use set operations
    if remain is None: remain,bundle = cSet,[]   # track coverage & budle
    prods = list(P.items())                      # remaining products
    for i,(p,cs) in enumerate(prods):
        if cSet.isdisjoint(cs):continue         # no coverage
        newBundle = bundle+[p]                  # add product to bundle
        if remain.issubset(cs): yield newBundle # full coverage bundle 
        toCover = remain.difference(cs)         # not yet covered 
        unused  = dict(prods[i+1:])             # remainin products
        yield from getBundles(C,unused,toCover,newBundle) # recurse for rest

output：

prods = {"d1":['a','c','d'],
         "d2":['a','b','e','f'],
         "d3":['a','c','b','f'],
         "d4":['b'],
         "d5":['g','e','a'],
         "d6":['g']}    
user_input=['a','b','c']

for bundle in sorted(getBundles(user_input,prods),
                     key=lambda b:sum(map(len,(prods[p] for p in b)))):
    print(bundle)

['d1', 'd4']
['d3']
['d3', 'd4']
['d1', 'd2']
['d1', 'd3']
['d1', 'd4', 'd5']
['d3', 'd5']
['d1', 'd2', 'd4']
['d1', 'd3', 'd4']
['d2', 'd3']
['d3', 'd4', 'd5']
['d2', 'd3', 'd4']
['d1', 'd2', 'd5']
['d1', 'd3', 'd5']
['d1', 'd2', 'd3']
['d1', 'd2', 'd4', 'd5']
['d1', 'd3', 'd4', 'd5']
['d2', 'd3', 'd5']
['d1', 'd2', 'd3', 'd4']
['d2', 'd3', 'd4', 'd5']
['d1', 'd2', 'd3', 'd5']
['d1', 'd2', 'd3', 'd4', 'd5']

Answer 2

我正在从事产品捆绑创建和推荐项目。 捆绑和推荐必须根据用户输入实时进行。

条件是 1.产品包应该尽可能地覆盖用户输入。 2. 推荐项目中用户输入的重复应该更少。

user_input=['a','b','c']

d1=['a','c','d']
d2=['a','b','e','f']
d3=['a','c','b','f']
d4=['b']
d5=['g','e','a']
d6=['g']

expected output - d1 + d4, d3, d1+d2,d5+d4+d1

以下是我的代码，它给出了结果，但它显示了重复的结果，也没有显示所有的组合。 任何帮助表示赞赏。

dlist=[d1,d2,d3,d4,d5,d6]
diff_match=[]
# find match n diff of each product based on user input
for i in range(len(dlist)):
    match=set(user_input).intersection(set(dlist[i]))
    #print("match is",match)
    diff=set(user_input).difference(set(dlist[i]))
    #print("diff is",diff)
    temp={'match':match,'diff':diff}
    diff_match.append(temp)
    
for i in range(len(diff_match)):
    # if match is found, recommend the product alone
    diff_match_temp=diff_match[i]['match']
    print("diff match temp is",diff_match_temp)
    if diff_match_temp==user_input:
        print ("absolute match")
    #scenario where the user input is subset of product features, seperate from partial match
    elif (all(x in list(diff_match_temp) for x in list(user_input))):
        print("User input subset of product features")
        print("The parent list is",diff_match[i]['match'])
        print("the product is", dlist[i])
    else:
        '''else check if the difference between user input and the current product is fulfilled by other product, 
        if yes, these products are bundled together'''
        for j in range(len(diff_match)):
            temp_diff=diff_match[i]['diff']
            print("temp_diff is",temp_diff)
            # empty set should be explicitly checked to avoid wrong match
            if (temp_diff.intersection(diff_match[j]['match'])==temp_diff and len(temp_diff)!=0 and list(temp_diff) != user_input) :
            #if temp_diff==diff_match[j]['match'] and len(temp_diff)!=0 and list(temp_diff) != user_input :
                print("match found with another product")
                print("parent is",dlist[i])
                print("the combination is",dlist[j] )