比较字典中某个键的元素内部值等于下一个键

Question

d = {
    0:{1,2,3},
    1:{567},
    2:{2,3,5,8},
    3:{4,5,7,9},
    4:{6,7,8}
    }

我想将第一个kv对的值与下一个kv对的键进行比较。

示例：要检查{1,2,3}是否存在1或{567}是否存在2 ，如果确实存在，那么我想删除值中存在的k。

输出应如下所示：

d = {
    0:{1,2,3},
    2:{2,3,5,8}
    }

我尝试使用具有各种排列和组合的字典迭代器，但没有结果。 达到结果的最佳方法是什么？

Answer 1

Python字典不是按顺序排列的，因此我不确定您在这里真的可以说“上一个”的“下一个”。 使用pandas系列会更合适。

但是，您仍然可以遍历键并将其定义为订单。

previous = {}
dict_items = list(d.items())
for k,v in dict_items:
    if k in previous:
        del d[k]
    previous = v

编辑：确保键的顺序正确，更改dict_items为：

dict_items = sorted(d.items(),key=lambda x:x[0])

会做到的

Answer 2

根据您的示例和要求进行猜测，您使用的是Python 3.6+，这些字典会保留插入顺序。 你可以做：

In [57]: d = { 
    ...:     0:{1,2,3}, 
    ...:     1:{567}, 
    ...:     2:{2,3,5,8}, 
    ...:     3:{4,5,7,9}, 
    ...:     4:{6,7,8} 
    ...:     }                                                                                                                                                                                              

# Get an iterator from dict.keys
In [58]: keys_iter = iter(d.keys())                                                                                                                                                                         

# Get the first key
In [59]: first_key = next(keys_iter)                                                                                                                                                                        

# Populate output dict with first key-value
In [60]: out = {first_key: d[first_key]}                                                                                                                                                                    

# Populate out dict with key-values based on condition by
# looping over the `zip`-ed key iterator and dict values 
In [61]: out.update({k: d[k] for k, v in zip(keys_iter, d.values())
                     if k not in v})                                                                                                                         

In [62]: out                                                                                                                                                                                                
Out[62]: {0: {1, 2, 3}, 2: {2, 3, 5, 8}}

Answer 3

我花了一段时间才弄明白你想要什么。 在下面，我重新措辞了您的示例：

    EXAMPLE:

    Suppose the input dictionary is as follows:
       0:{1,2,3},    
       1:{567},      
       2:{2,3,5,8},  
       3:{4,5,7,9},  
       4:{6,7,8}    

    We have...

          key of 0:{1,2,3}  is 0
        value of 0:{1,2,3} is {1,2,3}

          key of 2:{2,3,5,8} is 2
        value of 2:{2,3,5,8} is {2,3,5,8}

    We execute code similar to the following:

    if key of 1:{567} in value of 0:{1,2,3}: 
    # if 1 in {1,2,3}:    
        delete 1:{567}

    if key of 2:{2,3,5,8} in value of 1:{567}:
    # if 2 in {567}: 
        delete 2:{2,3,5,8}

    and so on...    
    """

以下代码应该可以实现您的目标：

def cleanup_dict(in_dict):  

    # sort the dictionary keys
    keys = sorted(indict.keys())

    # keys_to_delete will tell us whether to delete
    # an entry or not
    keys_to_delete = list()

    try:
        while True:
            prv_key = next(keys)
            nxt_key = next(keys)
            prv_val = in_dict[prv_key]
            if (nxt_key in prv_val):
                keys_to_delete.append(nxt_key)
    except StopIteration:
        pass 

    for key in keys_to_delete:
        del in_dict[key]            
    return