Question

例如

d1 == {'inds': [9, 9, 1, 9, 8, 1], 
       'vals': [0.28, 0.84, 0.71, 0.03, 0.04, 0.75]}

d2 == {'inds': [0, 9, 9, 1, 3, 3, 9], 
       'vals': [0.26, 0.06, 0.46, 0.58, 0.42, 0.21, 0.53, 0.76]}

如何通过下面的函数返回公共索引？

find_common_inds(d1, d2) == [1, 9]

这是我的代码，但似乎不起作用

intersect = []
for key in d1.keys():
    if key in d2.keys():
           intersect.append(key)
print(intersect)

Answer 1

您可以在平整的值列表上使用设置交集：

from itertools import chain

d1 = {'inds': [9, 9, 1, 9, 8, 1], 'vals': [0.28, 0.84, 0.71, 0.03, 0.04, 0.75]}

d2 = {'inds': [0, 9, 9, 1, 3, 3, 9], 'vals': [0.26, 0.06, 0.46, 0.58, 0.42, 0.21, 0.53, 0.76]}

print(set(chain.from_iterable(d1.values())).intersection(chain.from_iterable(d2.values())))
# {9, 1}

参考：

set.intersection()。
使用chain.from_iterable()对列表进行修饰。

Answer 2

您可以将功能更改为：

def find_common_values_in_identical_keys(d1,d2,**kwargs):
    if 'keyname' in kwargs: 
        kn = kwargs['keyname']   # this is hard  :P and it would still allow a "key"
                                 # that's not a hashable so it will crash donw below :/
    if kn: 
        try:
            return list(set(d1.get(kn, set())).intersection(d2.get(kn, [])))
        except TypeError as e:
            print("Bad keyname:", e)
    else:
        return [(k,find_common_values_in_identical_keys(d1,d2,keyname = k)) 
                for k in set(d1.keys()).union(d2)]  



d1 = {'inds': [9, 9, 1, 9, 8, 1], 
      'vals': [0.28, 0.84, 0.71, 0.03, 0.04, 0.75]}

d2 = {'inds': [0, 9, 9, 1, 3, 3, 9], 
      'vals': [0.26, 0.06, 0.46, 0.58, 0.42, 0.21, 0.53, 0.76]}


print(find_common_values_in_identical_keys(d1,d2,"inds"))

wich然后打印：

[9,1] # the order is random as sets are inheritently unordered

这也可以处理所有键的情况：

print(find_common_values_in_identical_keys(d1,d2))

输出：

[('vals', []), ('inds', [9, 1])]

返回（键名，[相同值的列表]）的元组列表

您可能仍要防止d1和d2不是dict的使用isinstance(...)的人，并抛出ValueError使其更加安全。< / p>

返回给定两个压缩向量共同的索引列表

2 个答案: