Accessing elements of Python dictionary by index

Question

Given that it is a dictionary you access it by using the keys. Getting the dictionary stored under "Apple", do the following:

>>> mydict["Apple"]
{'American': '16', 'Mexican': 10, 'Chinese': 5}

And getting how many of them are American (16), do like this:

>>> mydict["Apple"]["American"]
'16'

If the questions is, if I know that I have a dict of dicts that contains 'Apple' as a fruit and 'American' as a type of apple, I would use:

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


print myDict['Apple']['American']

as others suggested. If instead the questions is, you don't know whether 'Apple' as a fruit and 'American' as a type of 'Apple' exist when you read an arbitrary file into your dict of dict data structure, you could do something like:

print [ftype['American'] for f,ftype in myDict.iteritems() if f == 'Apple' and 'American' in ftype]

or better yet so you don't unnecessarily iterate over the entire dict of dicts if you know that only Apple has the type American:

if 'Apple' in myDict:
    if 'American' in myDict['Apple']:
        print myDict['Apple']['American']

In all of these cases it doesn't matter what order the dictionaries actually store the entries. If you are really concerned about the order, then you might consider using an OrderedDict:

http://docs.python.org/dev/library/collections.html#collections.OrderedDict

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])

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']