获取两个字典中具有相同值的键

Question

我想用Python比较两个字典，主要是找出那些键相同或者说值相同的键。我有两个目标：

1. 找出两个字典中值相同的键，
2. 找出在另一个字典中没有对应键的键。

我的字典大概长这样：

dict_1={
'3471': ['AY219713', 'AJ504663'], '3460': ['AJ621556', 'AJ575744'], '0': ['AM158981', 'AM158980', 'AM158982', 'AM158979', 'AY594216', 'AY594215', 'EU053207', 'AM392286', 'L26168', 'L37584']}

dict_2 = {'478': ['AY219713', 'AJ504663'], '43': ['AJ575744', 'AJ621556'], '2321': ['AM158979', 'L37584', 'AM158982', 'EU053207', 'AY594215', 'AM392286', 'AY594216', 'L26168', 'AM158981', 'AM158980', 'CP000758', 'AY776289', 'AJ242581', 'AM422370', 'U70978', 'AY457038', 'FR668302', 'AM422371', 'AM490632', 'AM490617', 'AJ242584']}

所以我想得到的结果大概是这样的：

Matching keys from Dictionary 1 and Dictionary 2:
3471 = 478, 2 values in common
3460 = 43, 2 values in common

Keys with no equal match:
Dictionary 1 = 0, etc.
Dictionary 2 = 2321, etc.

那些值不同但键相同的情况不需要考虑。

Answer 1

这里有一点有趣的列表推导，使用了 集合 而不是 排序列表。

for x in [i[0] for i in filter(None,[["%s = %s" %(key1,key2) for key1 in dict_1.keys() if set(dict_1[key1]) == set(dict_2[key2])] for key2 in dict_2.keys()])]:
    print x

Answer 2

这个方法是通过反转字典，然后遍历一组键（也就是原字典中的值）来实现的。

dict_1 = {
'3471': ['AY219713', 'AJ504663'],
'3460': ['AJ621556', 'AJ575744'],
'0': ['AM158981', 'AM158980', 'AM158982']}

dict_2 = {
'478': ['AY219713', 'AJ504663'],
'43': ['AJ575744', 'AJ621556'],
'2321': ['AM158979', 'L37584', 'AM158982']}

def compare_dicts(D1, D2):
    dict_1 = {str(sorted(v)): k for k, v in D1.iteritems()}
    dict_2 = {str(sorted(v)): k for k, v in D2.iteritems()}
    KEYS  = set(dict_1.keys() + dict_2.keys())
    both = []
    just = {0:[], 1:[]}
    for k in KEYS:
        Vals = dict_1.get(k, False), dict_2.get(k, False)
        if all(map(bool, Vals)):
            both.append('{} = {}'.format(*map(str, Vals)))
        else:
            i = 0 if Vals[0] else 1
            just[i].append(Vals[i])

    print 'Matching keys from Dictionary 1 and Dictionary 2:'
    print '\n'.join(both)
    print
    print 'Keys with no equal match:'
    print 'Dictionary 1 = {}'.format(','.join(map(str, just[0])))
    print 'Dictionary 2 = {}'.format(','.join(map(str, just[1])))


compare_dicts(dict_1, dict_2) 

这样就能得到我们想要的结果：

Matching keys from Dictionary 1 and Dictionary 2:
3460 = 43
3471 = 478

Keys with no equal match:
Dictionary 1 = 0
Dictionary 2 = 2321

Answer 3

使用以下代码：

dict_1={
'3471': ['AY219713', 'AJ504663'], '3460': ['AJ621556', 'AJ575744'], '0': ['AM158981', 'AM158980', 'AM158982', 'AM158979', 'AY594216', 'AY594215', 'EU053207', 'AM392286', 'L26168', 'L37584']}

dict_2 = {'478': ['AY219713', 'AJ504663'], '43': ['AJ575744', 'AJ621556'], '2321': ['AM158979', 'L37584', 'AM158982', 'EU053207', 'AY594215', 'AM392286', 'AY594216', 'L26168', 'AM158981', 'AM158980', 'CP000758', 'AY776289', 'AJ242581', 'AM422370', 'U70978', 'AY457038', 'FR668302', 'AM42237
for k in dict_1:
    for i in dict_2:
        if sorted(dict_1[k]) == sorted(dict_2[i]):
            matches.append((k, i))

for k in matches:
    print '%s = %s' %(k[0], k[1])

这段代码运行的结果是：

bash-3.2$ python sortkey.py
3471 = 478
3460 = 43
bash-3.2$

Answer 4

这里是对aj8uppal回答的一个小改动——这个版本在匹配和不匹配的定义上更清晰，我觉得结果更接近你想要的。承认这部分是受到aj8uppal的启发；注意我把dict_1和dict_2缩短了，以便更易读，而且我只对它们进行了一次排序（当你有5000个键的时候，这一点很重要）：

dict_1 = {
'3471': ['AY219713', 'AJ504663'],
'3460': ['AJ621556', 'AJ575744'],
'0': ['AM158981', 'AM158980', 'AM158982']}

dict_2 = {
'478': ['AY219713', 'AJ504663'],
'43': ['AJ575744', 'AJ621556'],
'2321': ['AM158979', 'L37584', 'AM158982']}

match1 = {}
match2 = {}
# sort once: makes a difference when you have lots of elements
for k in dict_1:
  dict_1[k] = sorted(dict_1[k])
for j in dict_2:
  dict_2[j] = sorted(dict_2[j])
for k in dict_1:
  for j in dict_2:
    if (dict_1[k] == dict_2[j]):
      match1[k]=1
      match2[j]=1
      break;

print "matching keys:"
print list(match1)
print list(match2)
print "\nnon matching keys:"
print list(set(dict_1.keys()) - set(match1))
print list(set(dict_2.keys()) - set(match2))

这将产生以下输出：

matching keys:
['3471', '3460']
['478', '43']

non matching keys:
['0']
['2321']