Access an arbitrary element in a dictionary in Python
Question:
If a mydict is not empty, I access an arbitrary element as:
mydict[mydict.keys()[0]]
Is there any better way to do this?
Answers:
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.
As others mentioned, there is no “first item”, since dictionaries have no guaranteed order (they’re implemented as hash tables). If you want, for example, the value corresponding to the smallest key, thedict[min(thedict)] will do that. If you care about the order in which the keys were inserted, i.e., by “first” you mean “inserted earliest”, then in Python 3.1 you can use collections.OrderedDict, which is also in the forthcoming Python 2.7; for older versions of Python, download, install, and use the ordered dict backport (2.4 and later) which you can find here.
Python 3.7
Now dicts are insertion ordered.
Ignoring issues surrounding dict ordering, this might be better:
next(dict.itervalues())
This way we avoid item lookup and generating a list of keys that we don’t use.
Python3
next(iter(dict.values()))
You can always do:
for k in sorted(d.keys()):
print d[k]
This will give you a consistently sorted (with respect to builtin.hash() I guess) set of keys you can process on if the sorting has any meaning to you. That means for example numeric types are sorted consistently even if you expand the dictionary.
EXAMPLE
# lets create a simple dictionary
d = {1:1, 2:2, 3:3, 4:4, 10:10, 100:100}
print d.keys()
print sorted(d.keys())
# add some other stuff
d['peter'] = 'peter'
d['parker'] = 'parker'
print d.keys()
print sorted(d.keys())
# some more stuff, numeric of different type, this will "mess up" the keys set order
d[0.001] = 0.001
d[3.14] = 'pie'
d[2.71] = 'apple pie'
print d.keys()
print sorted(d.keys())
Note that the dictionary is sorted when printed. But the key set is essentially a hashmap!
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" element
my_dict.values()[0] # value of "first" element
my_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" element
list(my_dict.values())[0] # value of "first" element
list(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]
For both Python 2 and 3:
import six
six.next(six.itervalues(d))
In python3
list(dict.values())[0]
How about, this. Not mentioned here yet.
py 2 & 3
a = {"a":2,"b":3}
a[list(a)[0]] # the first element is here
>>> 2
to get a key
next(iter(mydict))
to get a value
next(iter(mydict.values()))
to get both
next(iter(mydict.items())) # or next(iter(mydict.viewitems())) in python 2
The first two are Python 2 and 3. The last two are lazy in Python 3, but not in Python 2.
No external libraries, works on both Python 2.7 and 3.x:
>>> list(set({"a":1, "b": 2}.values()))[0]
1
For aribtrary key just leave out .values()
>>> list(set({"a":1, "b": 2}))[0]
'a'
Subclassing dict is one method, though not efficient. Here if you supply an integer it will return d[list(d)[n]], otherwise access the dictionary as expected:
class mydict(dict):
def __getitem__(self, value):
if isinstance(value, int):
return self.get(list(self)[value])
else:
return self.get(value)
d = mydict({'a': 'hello', 'b': 'this', 'c': 'is', 'd': 'a',
'e': 'test', 'f': 'dictionary', 'g': 'testing'})
d[0] # 'hello'
d[1] # 'this'
d['c'] # 'is'
first_key, *rest_keys = mydict
Another way to do this in one line while keeping the dictionary intact is:
arbitrary_value = mydict.setdefault(*mydict.popitem())
popitem() returns a tuple of (key, value) for the last item that was added into the dictionary and this pair is passed into setdefault as positional arguments. The setdefault tries to insert key into mydict with value value if it doesn’t already exist, but does nothing if does exist; and then returns the value of that key to the caller. Because we already popped the (key, value) pair out of the dictionary, we insert it back into it via setdefault and then proceed to return value, which is what we want.
If a mydict is not empty, I access an arbitrary element as:
mydict[mydict.keys()[0]]
Is there any better way to do this?
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.
As others mentioned, there is no “first item”, since dictionaries have no guaranteed order (they’re implemented as hash tables). If you want, for example, the value corresponding to the smallest key, thedict[min(thedict)] will do that. If you care about the order in which the keys were inserted, i.e., by “first” you mean “inserted earliest”, then in Python 3.1 you can use collections.OrderedDict, which is also in the forthcoming Python 2.7; for older versions of Python, download, install, and use the ordered dict backport (2.4 and later) which you can find here.
Python 3.7
Now dicts are insertion ordered.
Ignoring issues surrounding dict ordering, this might be better:
next(dict.itervalues())
This way we avoid item lookup and generating a list of keys that we don’t use.
Python3
next(iter(dict.values()))
You can always do:
for k in sorted(d.keys()):
print d[k]
This will give you a consistently sorted (with respect to builtin.hash() I guess) set of keys you can process on if the sorting has any meaning to you. That means for example numeric types are sorted consistently even if you expand the dictionary.
EXAMPLE
# lets create a simple dictionary
d = {1:1, 2:2, 3:3, 4:4, 10:10, 100:100}
print d.keys()
print sorted(d.keys())
# add some other stuff
d['peter'] = 'peter'
d['parker'] = 'parker'
print d.keys()
print sorted(d.keys())
# some more stuff, numeric of different type, this will "mess up" the keys set order
d[0.001] = 0.001
d[3.14] = 'pie'
d[2.71] = 'apple pie'
print d.keys()
print sorted(d.keys())
Note that the dictionary is sorted when printed. But the key set is essentially a hashmap!
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" element
my_dict.values()[0] # value of "first" element
my_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" element
list(my_dict.values())[0] # value of "first" element
list(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]
For both Python 2 and 3:
import six
six.next(six.itervalues(d))
In python3
list(dict.values())[0]
How about, this. Not mentioned here yet.
py 2 & 3
a = {"a":2,"b":3}
a[list(a)[0]] # the first element is here
>>> 2
to get a key
next(iter(mydict))
to get a value
next(iter(mydict.values()))
to get both
next(iter(mydict.items())) # or next(iter(mydict.viewitems())) in python 2
The first two are Python 2 and 3. The last two are lazy in Python 3, but not in Python 2.
No external libraries, works on both Python 2.7 and 3.x:
>>> list(set({"a":1, "b": 2}.values()))[0]
1
For aribtrary key just leave out .values()
>>> list(set({"a":1, "b": 2}))[0]
'a'
Subclassing dict is one method, though not efficient. Here if you supply an integer it will return d[list(d)[n]], otherwise access the dictionary as expected:
class mydict(dict):
def __getitem__(self, value):
if isinstance(value, int):
return self.get(list(self)[value])
else:
return self.get(value)
d = mydict({'a': 'hello', 'b': 'this', 'c': 'is', 'd': 'a',
'e': 'test', 'f': 'dictionary', 'g': 'testing'})
d[0] # 'hello'
d[1] # 'this'
d['c'] # 'is'
first_key, *rest_keys = mydict
Another way to do this in one line while keeping the dictionary intact is:
arbitrary_value = mydict.setdefault(*mydict.popitem())
popitem() returns a tuple of (key, value) for the last item that was added into the dictionary and this pair is passed into setdefault as positional arguments. The setdefault tries to insert key into mydict with value value if it doesn’t already exist, but does nothing if does exist; and then returns the value of that key to the caller. Because we already popped the (key, value) pair out of the dictionary, we insert it back into it via setdefault and then proceed to return value, which is what we want.