Accessing elements of Python dictionary by index

Question

Consider a dict like

mydict = {
  \'Apple\': {\'American\':\'16\', \'Mexican\':10, \'Chinese\':5},
  \'Grapes\':{\'Arabian\':\'25\',\'Indian\':\'20\'} }


        

Answer 1

With the following small function, digging into a tree-shaped dictionary becomes quite easy:

def dig(tree, path):
    for key in path.split("."):
        if isinstance(tree, dict) and tree.get(key):
            tree = tree[key]
        else:
            return None
    return tree

Now, dig(mydict, "Apple.Mexican") returns 10, while dig(mydict, "Grape") yields the subtree {'Arabian':'25','Indian':'20'}. If a key is not contained in the dictionary, dig returns None.

Note that you can easily change (or even parameterize) the separator char from '.' to '/', '|' etc.

Answer 2

I know this is 8 years old, but no one seems to have actually read and answered the question.

You can call .values() on a dict to get a list of the inner dicts and thus access them by index.

>>> mydict = {
...  'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
...  'Grapes':{'Arabian':'25','Indian':'20'} }

>>>mylist = list(mydict.values())
>>>mylist[0]
{'American':'16', 'Mexican':10, 'Chinese':5},
>>>mylist[1]
{'Arabian':'25','Indian':'20'}

>>>myInnerList1 = list(mylist[0].values())
>>>myInnerList1
['16', 10, 5]
>>>myInnerList2 = list(mylist[1].values())
>>>myInnerList2
['25', '20']

Answer 3

As I noticed your description, you just know that your parser will give you a dictionary that its values are dictionary too like this:

sampleDict = {
              "key1": {"key10": "value10", "key11": "value11"},
              "key2": {"key20": "value20", "key21": "value21"}
              }

So you have to iterate over your parent dictionary. If you want to print out or access all first dictionary keys in sampleDict.values() list, you may use something like this:

for key, value in sampleDict.items():
    print value.keys()[0]

If you want to just access first key of the first item in sampleDict.values(), this may be useful:

print sampleDict.values()[0].keys()[0]

If you use the example you gave in the question, I mean:

sampleDict = {
              'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
              'Grapes':{'Arabian':'25','Indian':'20'}
              }

The output for the first code is:

American
Indian

And the output for the second code is:

American

EDIT 1:

Above code examples does not work for version 3 and above of python; since from version 3, python changed the type of output of methods keys and values from list to dict_values. Type dict_values is not accepting indexing, but it is iterable. So you need to change above codes as below:

First One:

for key, value in sampleDict.items():
    print(list(value.keys())[0])

Second One:

print(list(list(sampleDict.values())[0].keys())[0])

Answer 4

You can't rely on order on dictionaries. But you may try this:

dict['Apple'].items()[0][0]

If you want the order to be preserved you may want to use this: http://www.python.org/dev/peps/pep-0372/#ordered-dict-api