is it possible to do things like this:
def f():
d={}
d[1]='qwe'
d[2]='rty'
return d
a,b=f()[1:2]
not a,b=f()[1],f()[2]
and not
t=f()
a,b=t[1],t[2]
I wont avoid extra lines and call function only once.
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1Are the keys you use for indexing actually sequential integers like 1 and 2? (If yes, use lists).Lukas Graf– Lukas Graf2012-10-06 16:41:49 +00:00Commented Oct 6, 2012 at 16:41
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Is the order of the items relevant for you?Lukas Graf– Lukas Graf2012-10-06 16:42:43 +00:00Commented Oct 6, 2012 at 16:42
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3 Answers
It's unclear to me what it is exactly you want to do. You're indexing a dictionary by integers. Integers can be used as keys to a dictionary just fine, but it looks like a list would make more sense in your case. An important difference between dictionaries and lists is that dicts are not ordered. d[1] and d[2] will not necessarily be in first and second place in your example. So if you depend on the order, standard dictionaries won't work anyway (an OrderedDict would).
(The fact that your data structure is returned by a function f() doesn't make any difference here, so I'll ignore it for better readability)
If you got
lst = ['qwe', 'rty']
you can simply do
a, b = lst
to assign lst[0] to a, and lst[1] to b. Lists can also be sliced:
long_lst = range(1000)
a, b = long_lst[500:502]
If you actually do want do use a dictionary, you can access its components like this:
>>> d = {'a': 'AAA', 'b': 'BBB'}
>>> d.keys()
['a', 'b']
>>> d.values()
['AAA', 'BBB']
>>> d.items()
[('a', 'AAA'), ('b', 'BBB')]
But as stated above, even if they happen to be in the right order in this example, you can't rely on the order of a dictionary, it will change over time.
Comments
No; dictionary indexing doesn't support slicing, since it usually doesn't make any sense in dictionary keys.
You can roll it yourself, though, with something like
a, b = map(f().get, [1, 2])
or (thanks @DSM)
a, b = operator.itemgetter(1, 2)(f())
These are slightly different in that if the result of f() is missing 1 or 2 as keys, the first will default to assigning None and the second will raise KeyError. map(f().__getitem__, ...) should be equivalent to itemgetter, though probably slightly slower.
1 Comment
DSM
Alternatively,
itemgetter(1,2)(f()).Yes, just use method dict.values():
x,y = f().values()[0:2]
2 Comments
Danica
You can't rely on the order of
values though.
Dmytro Sirenko
@Dougal yes, you're right: Python documentation: docs.python.org/library/stdtypes.html#dict.items : "Keys and values are listed in an arbitrary order which is non-random, varies across Python implementations, and depends on the dictionary’s history of insertions and deletions" but "If items(), keys(), values(), iteritems(), iterkeys(), and itervalues() are called with no intervening modifications to the dictionary, the lists will directly correspond"