两个列表与一些不同的密钥字典，如何找到交集？

Question

我的数据是这样的：

buffer = [{"siteid": 1 , "distance": 2, "codes": "1|b|c", "urv": "545"}, {"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 3 , "distance": 2, "codes": "1|b|c", "urv": "546"}] 
layer = [{"siteid": 2 }, {"siteid": 4 }, {"siteid": 3 }] 

我希望能够返回所有缓冲区列表的成员，其中SITEID是一样的。

其结果将是：

[{"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 3 , "distance": 2, "codes": "1|b|c", "urv": "546"}] 

TIA， 克里斯

Answer 1

这样的事情？

buffer = [{"siteid": 1 , "distance": 2, "codes": "1|b|c", "urv": "545"}, {"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 2 , "distance": 2, "codes": "1|b|c", "urv": "55"}, {"siteid": 3 , "distance": 2, "codes": "1|b|c", "urv": "546"}] 
layer = [{"siteid": 2 }, {"siteid": 4 }, {"siteid": 3 }] 
ids = [l['siteid'] for l in layer] 
print [b for b in buffer if b['siteid'] in ids] 

Answer 2

我不知道，能做到这一点任何内置的，但你可以尝试编写自己的功能找到的交集你的两个词：

def find_intersection(buffer, layer): 
    siteids = [a['siteid'] for a in layer] 
    return [item for item in buffer if item['siteid'] in siteids] 

Answer 3

intersection = set(x['sideid'] for x in buffer) & set(x['siteid'] for x in layer) 
return [x for x in buffer if x['sideid'] in intersection]