Access an arbitrary element in a dictionary in Python

On Python 3, non-destructively and iteratively:

next(iter(mydict.values()))

On Python 2, non-destructively and iteratively:

mydict.itervalues().next()

If you want it to work in both Python 2 and 3, you can use the six package:

six.next(six.itervalues(mydict))

though at this point it is quite cryptic and I'd rather prefer your code.

If you want to remove any item, do:

key, value = mydict.popitem()

Note that "first" may not be an appropriate term here because dict is not an ordered type in Python < 3.6. Python 3.6+ dicts are ordered.

If you only need to access one element (being the first by chance, since dicts do not guarantee ordering) you can simply do this in Python 2:

my_dict.keys()[0]     -> key of "first" elementmy_dict.values()[0]   -> value of "first" elementmy_dict.items()[0]    -> (key, value) tuple of "first" element

Please note that (at best of my knowledge) Python does not guarantee that 2 successive calls to any of these methods will return list with the same ordering. This is not supported with Python3.

in Python 3:

list(my_dict.keys())[0]     -> key of "first" elementlist(my_dict.values())[0]   -> value of "first" elementlist(my_dict.items())[0]    -> (key, value) tuple of "first" element

In python3, The way :

dict.keys() 

return a value in type : dict_keys(), we'll got an error when got 1st member of keys of dict by this way:

dict.keys()[0]TypeError: 'dict_keys' object does not support indexing

Finally, I convert dict.keys() to list @1st, and got 1st member by list splice method:

list(dict.keys())[0]