Esmail schrieb:
Diez B. Roggisch wrote:
Esmail schrieb:
Could someone help confirm/clarify the semantics of the [:] operator
in Python?
a = range(51,55)
############# 1 ##################
b = a[:] # b receives a copy of a, but they are independent
>
# The following two are equivalent
############# 2 ##################
c = []
c = a[:] # c receives a copy of a, but they are independent
No, the both above are equivalent. Both just bind a name (b or c) to a
list. This list is in both cases a shallow copy of a.
############# 3 ##################
d = []
d[:] = a # d receives a copy of a, but they are independent
This is a totally different beast. It modifies d in place, no
rebinding a name. So whover had a refernce to d before, now has a
changed object,
I follow all of this up to here, the next sentence is giving me pause
whereas in the two cases above, the original lists aren't touched.
The original list 'a', isn't changed in any of these cases right? And
modifying b, c or d would not change 'a' either - or am I not
understanding this correctly?
None of your operations changes a. But I talked about the lists you
bound b and c to before. Those aren't changed as well - they simply are
not pointed to anymore. In your example, that means the will be
garbage-collected, in other scenarios, such as this, the stay:
a = []
foo = []
bar = foo
assert bar is foo
bar = a[:]
assert bar is not foo
Diez
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