python: find all keys with a value

问题

I have a dictionary like below:

mydict = {'a' : 'apple',
     'b' : 'bobb',
    'c' : {
         'd' : 'dog'
       },
      'e' : 'dog'
     }

Suppose I want to find all the keys with value 'dog'In case of nesting, the keys should be separated by a dot.

So the output should be a list ['e', 'c.d']

Now if I write below code in python 3, it only outputs 'e'.

print(list(mydict.keys())[list(mydict.values()).index('dog')])

How can I get the nested keys?

回答1:

You can use a recursion function like following:

def find_key(mydict, pre=tuple()):
    for key, value in mydict.items():
        if isinstance(value, dict):
            yield from find_key(value, pre=pre+(key,))
        elif value == 'dog':
            if pre:
                yield '.'.join(pre + (key,))
            else:
                yield key

test:

In [23]: list(find_key(mydict))
Out[23]: ['e', 'c.d']
In [26]: mydict = {'a' : 'apple',
     'b' : 'bobb',
    'c' : {
         'd' : 'dog'
       },
      'e' : 'dog',
     'k':{'f':{'c':{'x':'dog'}}}}

In [27]: 

In [27]: list(find_key(mydict))
Out[27]: ['k.f.c.x', 'e', 'c.d']

回答2:

You can use recursion (although @Kasramvd beat me to it, here is a non-generator version for good measure):

>>> def find_key(value, d, level=None, acc=None):
...     acc = acc if acc is not None else []
...     level = level if level is not None else []
...     for k,v in d.items():
...         if isinstance(v, dict):
...             find_key(value, v, level=[*level, k], acc=acc)
...         elif v == value:
...             acc.append('.'.join([*level,k]))
...     return acc
...
>>>
>>> find_key('dog', mydict)
['c.d', 'e']

And with Kasramvd's messier test-case:

>>> mydict = {'a' : 'apple',
...      'b' : 'bobb',
...     'c' : {
...          'd' : 'dog'
...        },
...       'e' : 'dog',
...      'k':{'f':{'c':{'x':'dog'}}}}
>>> find_key('dog', mydict)
['c.d', 'e', 'k.f.c.x']

回答3:

You can also use recursion with a list comprehension:

mydict1 = {'a' : 'apple',
    'b' : 'bobb',
    'c' : {
       'd' : 'dog'
      },
     'e' : 'dog'
     }

mydict = {'a' : 'apple',
 'b' : 'bobb',
'c' : {
     'd' : 'dog'
   },
  'e' : 'dog',
 'k':{'f':{'c':{'x':'dog'}}}}

s = [mydict1, mydict]
def get_val(d, target='dog'):
  return reduce(lambda x, y:([x] if isinstance(x, str) else x)+([y] if isinstance(y, str) else y), list(filter(None, [a if b == target else "{}.{}".format(a, get_val(b)) if isinstance(b, dict) else None for a, b in d.items()])))


new_s = list(map(get_val, s))

Output:

[['c.d', 'e'], ['c.d', 'e', 'k.f.c.x']]

回答4:

Here you go :

mydict = {'a' : 'apple',
     'b' : 'bobb',
    'c' : {
         'd' : 'dog'
       },
      'e' : 'dog',
     'k':{'f':{'c':{'x':'dog'}}},

      'p':{'r':{'h':{'m':{'n':{'o':{'a':{'b':{'t':{'z':{'a':{'b':{'c':{'d':{'e':{'f':{'g':'dog'}}}}}}}}}}}}}}}}







          }

your_word='dog'

result = []

for key,value in mydict.items():
    if value==your_word:
        result.append(key)
    elif isinstance(value,dict):

        def recursive_approach(my_dict, nested_dict=key):

            for i, j in my_dict.items():
                if isinstance(j, dict):

                    nested_dict = "{}.{}".format(nested_dict, i)


                    return recursive_approach(j, nested_dict)
                else:
                    if nested_dict == '':

                        if j == your_word:
                            result.append(i)
                    else:
                        if j == your_word:
                            result.append("{}.{}".format(nested_dict, i))


        recursive_approach(value)





print(result)

output:

['p.r.h.m.n.o.a.b.t.z.a.b.c.d.e.f.g', 'k.f.c.x', 'e', 'c.d']

来源：https://stackoverflow.com/questions/48276165/python-find-all-keys-with-a-value

标签

python

python-3.x

nested