Question

我正在努力解决以下问题。想象一下，我有很多这样的数据：

one = {'A':'m','B':'n','C':'o'}
two = {'A':'m','B':'n','C':'p'}
three = {'A':'x','B':'n','C':'p'}

等等，不一定必须存储在dicts中。如何获得最常见条目的数据子集？

在上面的例子中，我想得到

one, two          with same A and B = m,n
two, three        with same B and C = n,p
one, two three    with same B       = n
one, two          with same A       = m

Answer 1

长字典的一种方式但效率不高是使用itertools.combinations查找词典之间的组合，然后遍历组合，然后循环设置并获得设置项之间的交集：

one = {'one':{'A':'m','B':'n','C':'o'}}
two ={'two':{'A':'m','B':'n','C':'p'}}
three = {'three':{'A':'x','B':'n','C':'p'}}

dict_list=[one,two,three]
v_item=[i.items() for i in dict_list]

from itertools import combinations
names=[]
items=[]
l=[combinations(v_item,i) for i in range(2,4)]
flat=[[[t[0] for t in k] for k in j] for j in l]  
"""this line is for flattening the combinations i don't know why but python puts every elements within a list :
>>> l
[[([('one', {'A': 'm', 'C': 'o', 'B': 'n'})], [('two', {'A': 'm', 'C': 'p', 'B': 'n'})]), 
([('one', {'A': 'm', 'C': 'o', 'B': 'n'})], [('three', {'A': 'x', 'C': 'p', 'B': 'n'})]), 
([('two', {'A': 'm', 'C': 'p', 'B': 'n'})], [('three', {'A': 'x', 'C': 'p', 'B': 'n'})])], 
[([('one', {'A': 'm', 'C': 'o', 'B': 'n'})], [('two', {'A': 'm', 'C': 'p', 'B': 'n'})], [('three', {'A': 'x', 'C': 'p', 'B': 'n'})])]]"""


for comb in flat :
   for pair in comb:
     names,items =zip(*pair)
     items=[i.viewitems() for i in items]
     print names,reduce(lambda x,y:x&y,items)

结果：

('one', 'two') set([('B', 'n'), ('A', 'm')])
('one', 'three') set([('B', 'n')])
('two', 'three') set([('B', 'n'), ('C', 'p')])
('one', 'two', 'three') set([('B', 'n')])

关于以下几行：

     items=[i.viewitems() for i in items]
     print names,reduce(lambda x,y:x&y,items)

您需要c reate a view object of your items作为set个对象，然后您可以使用&操作数计算项目的交集。使用reduce函数。

Answer 2

谢谢Kasra，这给了我最后的暗示:) 我改变了一些东西并将其转换为Python3（忘记提及......）但是作为你的代码，它在大型数据集（我确实拥有）上的速度非常慢并且内存不足。所以我必须寻找另一种方法：/。

这是我的最终代码：

from itertools import combinations
from functools import reduce

class Piece():
    def __init__(self,tag,A,B,C):
        self._tag = tag
        self.A = A
        self.B = B
        self.C = C

        self._dict = set(self.__dict__.items())  


pieces = []
pieces.append(Piece('one','m','n','o'))
pieces.append(Piece('two','m','n','p'))
pieces.append(Piece('three','x','n','p'))


l=[combinations(pieces,i) for i in range(2,4)]
flat =[]
for i in l:
    for k in i:
        flat.append(k)


for f in flat:
    print('-'*25)    
    print([j._tag for j in f])
    dicts = (i._dict for i in f)    
    matches = reduce(lambda x,y : x & y,dicts)    
    print(matches)

Python：具有最常见条目的数据子集

2 个答案: