Question

我是python新手，有以下问题。我有一个嵌套字典，形式为

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

，并希望找到所有具有相同值的键。输出应与此类似。

1 : [a,b]
2 : [a,c]

..

在此先感谢您的帮助！

Answer 1

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
output = {}

for key, value in dict.items():
    for v in value:
        if v in output.keys():
            output[v].append(key)
        else:
            output[v] = [ key ]

print(output)

输出将是

{'2': ['a', 'c'], '1': ['a', 'b'], '5': ['b'], '3': ['c']}

Answer 2

您可以使用defaultdict轻松构建输出（如果您希望按排序顺序对键进行排序）：

from collections import defaultdict

d = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

out  = defaultdict(list)

for key, values in d.items():
    for value in values:
        out[value].append(key)

# for a sorted output (dicts are ordered since Python 3.7):        
sorted_out = dict((k, out[k]) for k in sorted(out))
print(sorted_out)

#{'1': ['a', 'b'], '2': ['a', 'c'], '3': ['c'], '5': ['b']}

Answer 3

在我们寻求解决方案之前，莱姆告诉您一些事情。您所拥有的不是嵌套字典，而是字典中的设置。

一些python术语可以清除这一点：

数组：[1，2]

数组用大括号括起来，并用逗号分隔。
字典：{“ a”：1，“ b”：2}

字典用大括号括起来，并用逗号分隔“ key”：value对。这里，“ a”和“ b”是键，而1和2分别是它们的值。
设置：{1，2}

集合用大括号括起来，并用逗号分隔。

dict = {'a'：{'1'，'2'}，'b'：{'5'，'1'}，'c'：{'3'，'2'}}

在这里， {'1'，'2'} 是字典中的键为'a'的集合。因此，您所拥有的实际上是在词典中设置的，而不是嵌套词典中的。

解决方案

进入解决方案，集合不是可迭代的，这意味着您不能一一遍解它们。因此，您必须将它们转换为列表，然后对其进行迭代。

# Initialize the dictionary to be processed
data = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}

# Create dictionary to store solution
sol = {} # dictionary to store element as a key & sets containing that element as an array
# Eg., sol = { "1" : [ "a" , "b" ] }
# This shows that the value 1 is present in the sets contained in keys a & b.

# Record all elements & list every set containing those elements
for key in data. keys (): # iterate all keys in the dictionary
    l = list ( data [ key ] ) # convert set to list
    for elem in l: # iterate every element in the list
        if elem in sol. keys (): # check if elem already exists in solution as a key
            sol [ elem ]. append ( key ) # record that key contains elem
        else:
            sol [ elem ] = [ key ] # create a new list with elem as key & store that key contains elem

# At this time, sol would be
# {
#     "1" : [ "a" , "b" ] ,
#     "2" : [ "a" , "C" ] ,
#     "3" : [ "c" ] ,
#     "5" : [ "b" ]
# }

# Since, you want only the ones that are present in more than 1 sets, let's remove them
for key in sol : # iterate all keys in sol
    if sol [ key ]. length < 2 : # Only keys in at least 2 sets will be retained
        del sol [ key ] # remove the unrequired element

# Now, you have your required output in sol
print ( sol )
# Prints:
# {
#     "1" : [ "a" , "b" ] ,
#     "2" : [ "a" , "c" ]
# }

希望对您有帮助...

Answer 4

您可以反转dict中的键值，创建一个值-键dict，如果您只想要重复的值（find all the keys that have the same values），则可以filter：

from collections import defaultdict
def get_duplicates(dict1):
    dict2 = defaultdict(list)
    for k, v in dict1.items():
        for c in v:
            dict2[c].append(k)
    # if you want to all values, just return dict2
    # return dict2
    return dict(filter(lambda x: len(x[1]) > 1, dict2.items()))

输出：

{'1': ['a', 'b'], '2': ['a', 'c']}

Answer 5

使用defaultdict中的collections可以轻松完成此操作，

>>> d = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
>>> from collections import defaultdict
>>> dd = defaultdict(list)
>>> for key,vals in d.items():
...   for val in vals:
...     dd[val].append(key)
... 
>>>>>> dict(dd)
{'1': ['a', 'b'], '3': ['c'], '2': ['a', 'c'], '5': ['b']}

Answer 6

这可以通过两个内部for循环轻松实现：

dict = {'a': {'1','2'}, 'b':{'5','1'}, 'c':{'3','2'}}
out = {}
for key in dict:
    for value in dict[key]:
        if value not in out:
            out[value]= [key]
        else:
            out[value]+= [key]

print out  # {'1': ['a', 'b'], '3': ['c'], '2': ['a', 'c'], '5': ['b']}

从嵌套字典中获取重复项

6 个答案:

解决方案