避免将重复项插入到Python列表中，并具有理解

问题描述：

我有一个字典：

XY_dict = {1: [(12, 55),(13, 55)],
2: [(14, 55),(15, 57)],
3: [(14, 55),(15, 58)],
4: [(14, 55),(16, 55)]}

我想知道哪些键有值元组其中唯一（不存在任何其他键值）。从示例字典中，键1是唯一的，因为（12,55）或（13，55）在任何其他字典的关键。通过获取具有共享值的密钥列表，我可以稍后反转结果，并获得唯一的密钥。

I want to find out which keys have values tuples of which are unique (don't present in any other key's value). From the sample dictionary, key 1 is unique because neither (12, 55) nor (13, 55) is present in any other dictionary's key. By getting the list of keys with shared values, I can invert the result later on and get the keys that are unique.

我正在使用列表解析获取密钥共享值：

I am using a list comprehension for getting keys with shared values:

keys_shared_values = [k1 for k1,v1 in XY_dict.iteritems()
                       for k,v in XY_dict.iteritems()
                       for XY_pair in v
                       if XY_pair in v1 and k != k1 and k1 not in keys_shared_values]

因此，我得到 [2，2，3，3，4，4] 但我预计重复不会被插入（因为我正在评估键值是否在结果列表中）。我可以通过运行列表（set（shared_values））来修复，但是想了解我的代码有什么问题。

As a result, I am getting [2, 2, 3, 3, 4, 4] yet I expect duplicates not to be inserted (since I am evaluating whether the key value is in the result list). I can fix that by running the list(set(shared_values)), but would like to understand what is wrong with my code.

答

问题在于，您完成理解之前， keys_shared_values 为空，所以您的 k1不在keys_shared_values 将始终返回 True 。你不能参考当前的理解。你最好的选择是转换为设置，如您所建议的那样。

The problem is that keys_shared_values is empty until you complete the comprehension, so your k1 not in keys_shared_values will always return True. You cannot refer to the current comprehension. Your best bet is to convert to set as you already suggested.

你应该将代码更改为循环你想要这个功能：

You should change your code to a loop if you want that functionality:

keys_shared_values = []
for k, v in XY_dict.iteritems():
    for k1, v1 in XY_dict.iteritems():
        for XY_pair in v:
            if XY_pair in v1 and k != k1 and k1 not in keys_shared_values:
                keys_shared_values.append(k1)
print keys_shared_values

结果：

[3, 4, 2]

避免将重复项插入到Python列表中，并具有理解

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