If I have a dictionary like:
{ 'a': 1, 'b': 2, 'c': 3 }
How can I convert it to this?
[ ('a', 1), ('b', 2), ('c', 3) ]
And how can I convert it to this?
[ (1, 'a'), (2, 'b'), (3, 'c') ]
>>> d = { 'a': 1, 'b': 2, 'c': 3 }
>>> d.items()
[('a', 1), ('c', 3), ('b', 2)]
>>> [(v, k) for k, v in d.iteritems()]
[(1, 'a'), (3, 'c'), (2, 'b')]
It's not in the order you want, but dicts don't have any specific order anyway.1 Sort it or organize it as necessary.
See: items(), iteritems()
In Python 3.x, you would not use iteritems
(which no longer exists), but instead use items
, which now returns a "view" into the dictionary items. See the What's New document for Python 3.0, and the new documentation on views.
1: Insertion-order preservation for dicts was added in Python 3.7
You can use list comprehensions.
[(k,v) for k,v in a.iteritems()]
will get you [ ('a', 1), ('b', 2), ('c', 3) ]
and
[(v,k) for k,v in a.iteritems()]
the other example.
Read more about list comprehensions if you like, it's very interesting what you can do with them.
Create a list of namedtuples
It can often be very handy to use namedtuple. For example, you have a dictionary of 'name' as keys and 'score' as values like:
d = {'John':5, 'Alex':10, 'Richard': 7}
You can list the items as tuples, sorted if you like, and get the name and score of, let's say the player with the highest score (index=0) very Pythonically like this:
>>> player = best[0]
>>> player.name
'Alex'
>>> player.score
10
How to do this:
list in random order or keeping order of collections.OrderedDict:
import collections
Player = collections.namedtuple('Player', 'name score')
players = list(Player(*item) for item in d.items())
in order, sorted by value ('score'):
import collections
Player = collections.namedtuple('Player', 'score name')
sorted with lowest score first:
worst = sorted(Player(v,k) for (k,v) in d.items())
sorted with highest score first:
best = sorted([Player(v,k) for (k,v) in d.items()], reverse=True)
What you want is dict
's items()
and iteritems()
methods. items
returns a list of (key,value) tuples. Since tuples are immutable, they can't be reversed. Thus, you have to iterate the items and create new tuples to get the reversed (value,key) tuples. For iteration, iteritems
is preferable since it uses a generator to produce the (key,value) tuples rather than having to keep the entire list in memory.
Python 2.5.1 (r251:54863, Jan 13 2009, 10:26:13)
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> a = { 'a': 1, 'b': 2, 'c': 3 }
>>> a.items()
[('a', 1), ('c', 3), ('b', 2)]
>>> [(v,k) for (k,v) in a.iteritems()]
[(1, 'a'), (3, 'c'), (2, 'b')]
>>>
Python3 dict.values()
not return a list. This is the example
mydict = {
"a": {"a1": 1, "a2": 2},
"b": {"b1": 11, "b2": 22}
}
print(mydict.values())
> output: dict_values([{'a1': 1, 'a2': 2}, {'b1': 11, 'b2': 22}])
print(type(mydict.values()))
> output: <class 'dict_values'>
print(list(mydict.values()))
> output: [{'a1': 1, 'a2': 2}, {'b1': 11, 'b2': 22}]
print(type(list(mydict.values())))
> output: <class 'list'>
[tuple(reversed(x)) for x in d.items()]
- garejx
is already atuple
in your code (it's the nature ofitems
to produce an iterable oftuple
s), it would be simpler/faster to just do[x[::-1] for x in d.items()]
; the reversing slice directly constructs a reversedtuple
of the proper size rather than having thetuple
constructor iteratively populate (with overallocation and resizing at the end) atuple
from areversed
iterator. - ShadowRangerk, v
pattern in such cases. - garejcompatibility
tag because of py2.x // py3.x differences - dreftymac