Question

我有一本字典：

{0: [1, 2, 3, 4, 5, 6, 7, 8, 9], 1: [0, 9, 4, 6, 7], 2: [8, 9, 3, 6, 0], 3: [8, 9, 2, 0], 4: [8, 0, 1, 6, 7], 5: [0, 9], 6: [8, 0, 2, 4, 1], 7: [8, 0, 4, 1], 8: [0, 2, 3, 4, 6, 7], 9: [0, 1, 2, 3, 5]}

然后我接受一个引用其中一个键的输入：

flag = 0
user = input("Enter an integer for a user ID: ")
if isinstance(user, float):
    flag = 1
if user.isdigit() and int(user) >= 0 and int(user) <= (len(network)-1) and flag == 0:
    return(user)
else:
    while user.isdigit() == False or int(user) < 0 or int(user) > (len(network)-1):
        user = input("Enter an integer for a user ID: ")

所以现在我需要帮助理解找到具有重叠键的值，例如，如果用户选择4将输出代码：

"For user with ID 4 we recommend the user with ID 2"
"That is because users 4 and 2 have 3 common friends and user 4 does not have more common friends with anyone else."

用户也无法相互了解。因此，尽管4与0有更多共同点，但他们彼此了解，因此无法工作。

谢谢！

Answer 1

方法：

接受matchMaker(id, d)和id

d

在d
检查密钥是否与id
检查身份证是否彼此了解
找到id和其他id之间的共同朋友;附加到列表l2
如果匹配，则查找全部并返回列表matches else []

代码

d={0: [1, 2, 3, 4, 5, 6, 7, 8, 9], 1: [0, 9, 4, 6, 7], 2: [8, 9, 3, 6, 0], 3: [8, 9, 2, 0], 4: [8, 0, 1, 6, 7], 5: [0, 9], 6: [8, 0, 2, 4, 1], 7: [8, 0, 4, 1], 8: [0, 2, 3, 4, 6, 7], 9: [0, 1, 2, 3, 5]}

def matchMaker(id, d):
    l2=[]
    for k,v in d.items():
        if k!=id:
            if id not in v:
                f = set(d[id]).intersection(v)
                l2.append([len(f), k])        # append f if you want the suggestions
    matches=[]
    if l2:
        match_count = max(l2)[0]
        matchid = max(l2)[1]
        matches = [matchid]

        for item in l2:
            if item[1] not in matches:
                if match_count == item[0]:
                    matches.append(item[1])
    return matches

id=5
all_matches = matchMaker(id,d)

print(all_matches)

要获取id的整数值，请使用all_matches[0]，all_matches[1]等。

输出

# matches for id = 5
[3, 1, 2]

# matches for id = 4
[2]

Answer 2

最简单的方法是将朋友列表转换为集合。您可以使用像这样的词典理解

来做到这一点

d  = {0: [1, 2, 3, 4, 5, 6, 7, 8, 9], 1: [0, 9, 4, 6, 7], 2: [8, 9, 3, 6, 0], 3: [8, 9, 2, 0], 4: [8, 0, 1, 6, 7], 5: [0, 9], 6: [8, 0, 2, 4, 1], 7: [8, 0, 4, 1], 8: [0, 2, 3, 4, 6, 7], 9: [0, 1, 2, 3, 5]}
ds = {k:set(v) for k, v in d.items()}

现在你可以通过交叉

找到共享的朋友，比如2和4

ds[2] & ds[4]

-> {0, 8, 6}

您可以使用len

计算它们

len(ds[2] & ds[4])

最后，你可以创建一个循环来找到最佳匹配（a是输入，b是输出）：

def best_match_for(a):
    b = None
    af = ds[a]
    best = -1
    for k, v in ds.items():
        if k!=a and not k in af:
            ovlp = len(af & v)
            if ovlp > best:
                best = ovlp
                b = k
    return b, best if best > 0 else None

如何检查字典中哪些值具有重叠键？

2 个答案: