字典中的Python数组交集 [英] Python intersection of arrays in dictionary

问题描述

我有数组字典，例如:

y_dict= {1: np.array([5, 124, 169, 111, 122, 184]),
         2: np.array([1, 2, 3, 4, 5, 6, 111, 184]), 
         3: np.array([169, 5, 111, 152]), 
         4: np.array([0, 567, 5, 78, 90, 111]),
         5: np.array([]),
         6: np.array([])}

我需要在我的字典中找到数组的拦截:y_dict. 第一步，我从空数组中清除了字典，就像

I need to find interception of arrays in my dictionary: y_dict. As a first step I cleared dictionary from empty arrays, as like

dic = {i:j for i,j in y_dict.items() if np.array(j).size != 0}

因此，dic具有以下视图:

dic = { 1: np.array([5, 124, 169, 111, 122, 184]),
        2: np.array([1, 2, 3, 4, 5, 6, 111, 184]), 
        3: np.array([169, 5, 111, 152]), 
        4: np.array([0, 567, 5, 78, 90, 111])}

要找到拦截，我尝试使用元组方法，例如:

To find interception I tried to use tuple approach as like:

result_dic = list(set.intersection(*({tuple(p) for p in v} for v in dic.values())))

实际结果为空列表:[];

Actual result is empty list: [];

预期结果应为:[5, 111]

您能帮我在字典中找到数组的交集吗?谢谢

Could you please help me to find intersection of arrays in dictionary? Thanks

推荐答案

您发布的代码过于复杂且错误，因为还需要进行一次额外的内部迭代.您想这样做:

The code you posted is overcomplex and wrong because there's one extra inner iteration that needs to go. You want to do:

result_dic = list(set.intersection(*(set(v) for v in dic.values())))

或带有map且没有for循环:

result_dic = list(set.intersection(*(map(set,dic.values()))))

结果

[5, 111]

• 迭代值(忽略键)
• 将每个numpy数组转换为set(也可以转换为tuple，但是intersection仍会将其转换为集合)
• 将参数传递给intersection，并将其拆包
• iterate on the values (ignore the keys)
• convert each numpy array to a set (converting to tuple also works, but intersection would convert those to sets anyway)
• pass the lot to intersection with argument unpacking

我们甚至可以通过在每个数组上创建集合并使用filter过滤掉空数组来摆脱步骤1:

We can even get rid of step 1 by creating sets on every array and filtering out the empty ones using filter:

result_dic = list(set.intersection(*(filter(None,map(set,y_dict.values())))))

这是出于单行的缘故，但在现实生活中，表达式可能会被分解，因此更易读&；值得评论.这种分解还可以帮助我们避免在不传递任何参数(因为没有非空集)时发生崩溃，这会挫败与集相交的聪明方法(首先在

That's for the sake of a one-liner, but in real life, expressions may be decomposed so they're more readable & commentable. That decomposition may also help us to avoid the crash which occurs when passed no arguments (because there were no non-empty sets) which defeats the smart way to intersect sets (first described in Best way to find the intersection of multiple sets?).

只需预先创建列表，并且仅当列表不为空时才调用交集.如果为空，则只需创建一个空集即可:

Just create the list beforehand, and call intersection only if the list is not empty. If empty, just create an empty set instead:

non_empty_sets = [set(x) for x in y_dict.values() if x.size]
result_dic = list(set.intersection(*non_empty_sets)) if non_empty_sets else set()

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