Probably a duplicate, but I can't find the answer by searching with these terms, at least.
Is there a quicker way to do this in Python?
level1 = {}
level2 = {}
level3 = {}
I've tried
level1 = level2 = level3 = {}
But that seems to create copies of the object, which isn't what I want. And
level1, level2, level3 = {}
throws an error.
level1 = level2 = level3 = {}
Doesn’t create copies. It lets reference level{1-3} to the same object. You can use a list comprehension instead:
level1, level2, level3 = [{} for dummy in range(3)]
or more readable:
level1, level2, level3 = {}, {}, {}
dummy is better indicated by a single underscore _
_ for gettext.Your variable naming is a possible sign that your design could be improved. It might be better to use a list instead of three separate variables:
levels = [{}, {}, {}]
levels = [{} for _ in range(3)]
levels = [] and add your dictionaries as you make themYou could do
level1, level2, level3 = {}, {}, {}
level1, level2, level3, = {}, {}, {}
I personally never change code to squeeze more stuff on one line. Have three different assignments is the way I would do it, one per line.
However, if there are documented performance improvements by making such a change, that's another thing. I just don't know of any performance improvements with these changes. Only obfuscation.
People have already answered the intent of your question, but...
Is there a quicker way to do this in Python?
What do you mean by quicker?
I think your real question should be "what is the pythonic way to do this?"
level1, level2, level3 = {}
From the Python documentation:
Tuples are constructed by the comma operator (not within square brackets), with or without enclosing parentheses, but an empty tuple must have the enclosing parentheses, such as a, b, c or (). A single item tuple must have a trailing comma, such as (d,).
a, b, c is implicitly (a, b, c). That's why you need to have a 3-tuple on the right side as suggested by other posters (nils, sdolan, gunner).
Shown explicitly they are suggesting this:
(level1, level2, level3) = ({}, {}, {},)
Python is interpreted which allows the definitions of things to be made at runtime.
def prefixed_range(prefix, start, stop=None, step=1):
" Like xrange() but results are prefixed numeric strings. """
prefix = str(prefix)
if stop is None:
start, stop = 0, start
for i in xrange(start, stop, step):
yield prefix + str(i)
def define_vars(names, value=None, namespace=None):
namespace = globals() if not namespace else namespace
factory = (lambda: value) if not callable(value) else value
for name in names:
namespace[name] = factory()
define_vars(prefixed_range('level', 1, 4), dict)
print 'level1: {}, level2: {}, level3: {}'.format(level1, level2, level3)
# level1: {}, level2: {}, level3: {}
Try this: level1,level2,level3=[{}]*3
level1 = level2 = level3 = {}does not create any copies of the dict.level1 = level2 = level3 = {}is in fact not creating copies. If it created a copy the behavior would be exactly what you want. Instead it is assigning all 3 to the same object, not copies of the object.