List behaviour

[barberomarcelo]

Maybe I'm missing something but the latter is not the behaviour I'm
expecting:

a = [[1,2,3,4], [5,6,7,8]]
b = a[:]
b [[1, 2, 3, 4], [5, 6, 7, 8]] a == b True a is b False for i in range(len(b)): .... for x in range(4):
.... b[i][x] = b[i][x] + 10
.... b [[11, 12, 13, 14], [15, 16, 17, 18]] a [[11, 12, 13, 14], [15, 16, 17, 18]]

ActivePython 2.4.2 Build 248 (ActiveState Corp.) based on
Python 2.4.2 (#67, Oct 30 2005, 16:11:18) [MSC v.1310 32 bit (Intel)]
on win32
Windows XP

Marcelo

[Diez B. Roggisch]

ba************@gmail.com wrote:

Maybe I'm missing something but the latter is not the behaviour I'm
expecting:

a = [[1,2,3,4], [5,6,7,8]]
b = a[:]
b [[1, 2, 3, 4], [5, 6, 7, 8]] a == b True a is b False for i in range(len(b)): ... for x in range(4):
... b[i][x] = b[i][x] + 10
... b [[11, 12, 13, 14], [15, 16, 17, 18]] a [[11, 12, 13, 14], [15, 16, 17, 18]]
b = a[:]

does clone a, but doesn't make a deepcopy of its contents, which you are
manipulating!

So do
[GCC 4.0.2 20050808 (prerelease) (Ubuntu 4.0.1-4ubuntu8)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
poWelcome to rlcompleter2 0.96
for nice experiences hit <tab> multiple times import copy
a = [[10]]
b = copy.deepcopy(a)
b[0][0] = 20
a [[10]] b [[20]]

HTH,

Diez

[Heiko Wundram]

Am Mittwoch 17 Mai 2006 17:06 schrieb ba************@gmail.com:

Maybe I'm missing something but the latter is not the behaviour I'm

expecting:

a = [[1,2,3,4], [5,6,7,8]]
b = a[:]
b
[[1, 2, 3, 4], [5, 6, 7, 8]]
a == b
True
a is b
False

Try an:

a[0] is b[0]
and
a[1] is b[1]

here, and you'll see, that [:] only creates a shallow copy. Thus, the lists in
the lists aren't copied, they are shared by the distinct lists a and b.

Hope this clears it up.

--- Heiko.

[Mike Kent]

When you did:
b = a[:]

b was then a copy of a, rather than just a reference to the same a.
But what does a contain? It contains two sublists -- that is, it
contains references to two sublists. So b, which is now a copy of a,
contains copies of the two references to the same two sublists.

What you need to do instead is:
b = copy.deepcopy(a)

to get you what you actually want.

[barberomarcelo]

Thank you very much. It was clear.