In Perl, I can do this:
push(@{$h->[x]}, y);
Can I simplify the foll开发者_运维问答owing python codes according to above Perl example?
if x not in h:
h[x] = []
h[x].append(y)
I want to simplify this, because it goes many places in my code, (and I cannot initialize all possible x with []). I do not want to make it a function, because there is no 'inline' keyword.
Any ideas?
A very elegant way (since Python 2.5) is to use defaultdict
from the "collections" module:
>>> from collections import defaultdict
>>> h = defaultdict(list)
>>> h['a'].append('b')
>>> h
defaultdict(<type 'list'>, {'a': ['b']})
defaultdict
is like a dict, but provides a default value using whichever constructor you passed to it when you created it (in this example, a list).
I particularly like this over the setdefault
dict method, because 1) you define the variable as a defaultdict, and generally no other changes are required on the code (except perhaps to remove previous kludges for default values); and 2) setdefault is a terrible name :P
There are a couple of ways to do this with the dict methods:
h.setdefault(x, []).append(y)
or
h[x] = h.pop(x,[]).append(y)
You can use setdefault
h = {}
h.setdefault(x, []).append(y)
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