Hey all, thanks for all the responses. I think I get the gist of just about
everything. One thing to mention though is that I'm mainly working with scipy
so most of what I'm working with in numpy.ndarrays. I think most of what was
mentioned with respect to lists applies equally to numpy.ndarrays. The id()
function/sub-routine is something that'll be handy in the future as is knowing
the difference between 'is' and '= ='.
I have a few follow-up questions:
1) A lot of what I'm doing is image processing that involves slicing and
performing standard arithmetic and logic to arrays. So I think I've manage to
avoid this trap most of the time since at least shallow copies are made. Are
there any clear rules for when a shallow copy is made versus a deep copy?
2) Luke gave a tip on using the append rather than the += when creating
strings in a for-loop to be more memory-efficient. Does this only apply to
immutable objects such as strings?? I'm creating a lot of arrays by doing
something like this:
A=scipy.zeros(100,100) #initialize array
for ii in xrange(0,shape(A)[0]):
for jj in xrange(0,shape(A)[1]):
A[ii,jj]=3*ii+jj+sqrt(ii*jj)
I would think this is okay since the array is mutable, am I correct or is there
a better technique?
3) Now you all have me a little worried that I'm not being very memory
efficient with my programming style. A lot of the time, I'll perform a lot of
operations in sequence on an array. For example:
#B is a 1000x1000 pixel image that I've imported as an array. We'll call this
B1
B=(B-B.min())/(B.max()-B.min()) #image array scaled from 0 to 1. We'll call
this B2
B=B[B>0.5] #image with values less than 0.5 masked (set to zero). We'll call
this B3
So in this example, is memory allocated for B1, B2 and B3? Does every
operation force a new copy? Will this memory ever be 'recovered' by the
python memory manger (garbage collector?)?
4) The typical knee-jerk reaction to this 'oddity' is "what a pain, how
stupid" etc, but I'm sure there is a good explanation. Can someone explain why
python acts this way? faster processing? preserve memory? etc?
Thanks for all your help.
-Keith
----- Original Message ----
From: Luke Paireepinart <[EMAIL PROTECTED]>
To: Brett Wilkins <[EMAIL PROTECTED]>; Tutor <tutor@python.org>
Sent: Thursday, February 28, 2008 2:49:11 AM
Subject: Re: [Tutor] Python oddity
Brett
Wilkins
wrote:
>
As
everybody
else
has
told
you,
assigning
bb
=
aa
just
gives
bb
the
>
reference
to
the
same
object
that
aa
has.
Unless
I
missed
something,
>
then
nobody's
actually
mentioned
how
to
make
this
not
happen...
and
it's
>
actually
rather
easy...
instead
of
bb
=
aa,
do
this:
>
bb
=
aa[:]
>
Looks
like
a
splice,
and
indeed
it
is
a
splice,
without
bounds.
When
a
>
splice
is
used
like
this,
I
believe
it
is
known
as
the
copy
directive.
Yes,
Terry
explained
other
ways
as
well.
The
problem
with
using
slices
is
the
following:
>>>
a
=
[['a'],'b']
>>>
b
=
a
>>>
b
is
a
True
>>>
b
=
a[:]
>>>
b
is
a
False
#okay,
we're
fine
so
far.
>>>
b[0]
['a']
>>>
b[0].append('c')
>>>
b[0]
['a',
'c']
>>>
a[0]
['a',
'c']
and
here's
where
we
run
into
problems.
Turns
out
the
slice
notation
just
makes
a
shallow
copy,
so
if
you
have
any
lists
in
your
list,
they
will
still
refer
to
the
same
object.
(I'm
guessing
that
any
mutable
objects
will
still
refer
to
the
same
object)
copy.copy()
has
this
same
problem.
The
way
to
avoid
this,
as
I
mentioned
before,
is
to
use
a
deepcopy(),
but
as
I
said,
you
usually
don't
need
to
make
a
full
copy
of
a
list.
Oh,
and
if
you're
curious:
>>>
b
=
list(a)
>>>
b
is
a
False
>>>
b[0].append('d')
>>>
a[0]
['a',
'c',
'd']
So
a
list()
does
a
shallow
copy
as
well.
Here's
a
continuation,
with
deepcopy():
>>>
import
copy
>>>
b
=
copy.deepcopy(a)
>>>
b
is
a
False
>>>
b[0]
is
a[0]
False
>>>
b[0].append('e')
>>>
a[0]
['a',
'c',
'd']
HTH,
-Luke
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