Change id of copied value from a list

Question

I've copied one element of a list but the id is still the same.

t=['bancorp', 'bancorp','bancorp ba','bancorp ali', 'gas eu', 'gas', 'karl', 'bancorp','bancorp america','uni','gas for','gas tr']

n=t[2]

id(n)
124661664

id(t[2])
124661664

I´ve tried this but the ID is still the same:

n=cp.deepcopy(t[2])

How Can I copy the value but with different ID?

EDIT:

When I slice:n=t[2:3][0]. The id is different.

But when I slice: n=t[0:1][0]. The ID is the same. Why did it happen?

Strings are immutable, so it's never useful to copy them. What are you trying to do? — RemcoGerlich, Apr 07 '14 at 14:46
I want an independent copy of the element of a list, so the ID should be different. Check my edit. — CreamStat, Apr 07 '14 at 14:57
@CreamStat an "independent copy" isn't relevant here, because strings are immutable; you cannot change them in-place. — jonrsharpe, Apr 07 '14 at 15:21
@Creamstat but why? There is no advantage to having two copies of the same string. — RemcoGerlich, Apr 07 '14 at 16:33
It makes as much sense as wanting several copies of the number 2. — RemcoGerlich, Apr 07 '14 at 16:46

score 3 · Answer 1 · edited May 23 '17 at 10:33

3

In python, smaller strings are interned. This allows for efficient memory use since strings are immutable. In short, for small identical strings you will get the same id value.

As to why t[2:3][0] was not interned, and t[0:1][0] was, it looks like Cython does not intern strings with whitespace.

edited May 23 '17 at 10:33

Community

1
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answered Apr 07 '14 at 14:45

C.B.

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So, Is there no way I can get an independent copy of the first value f my list? – CreamStat Apr 07 '14 at 15:16

jonrsharpe · Answer 2 · 2014-04-07T16:41:03.037

If you're really determined to create a separate copy of a string, you could do:

>>> s = 'foo'
>>> t = ''.join(s)
>>> s
'foo'
>>> t
'foo'
>>> id(s) == id(t)
False

However, as strings are immutable (i.e. can't be changed in-place) you don't really get any benefit from this. For example:

>>> a = list('foo')
>>> a
['f', 'o', 'o']
>>> b = a[:] # only a shallow copy
>>> id(a[0]) == id(b[0])
True # share the same object
>>> a[0] += 'l' # 'modify' one of them
>>> a
['fl', 'o', 'o'] # changes a
>>> b
['f', 'o', 'o'] # doesn't change b

MONTYHS · Answer 3 · 2014-04-07T14:57:58.917

-1

Craete a empty list and try to copy

t=['bancorp', 'bancorp','bancorp ba','bancorp ali', 'gas eu', 'gas', 'karl', 'bancorp','bancorp america','uni','gas for','gas tr']
n = []
n = t[2:3]

print id(n)

print id(t[2])

edited Apr 07 '14 at 14:57

answered Apr 07 '14 at 14:52

MONTYHS

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Not very useful - here `n` is an empty list, which (although it does not have the same id as `t[2]`) isn't what the OP wants. `t[2:3]` would give a list with the single element `t[2]`, but `id(t[2:3][0]) == id(t[2])`. – jonrsharpe Apr 07 '14 at 14:54
Just `print n` and you will see what I mean. – jonrsharpe Apr 07 '14 at 14:57
Why it doesn't work when n=t[0:1][0] ? In that case, the ID is still the same! – CreamStat Apr 07 '14 at 15:03
@MONTYHS `id(n) != id(t[2])` simply because `['bancorp ba'] != 'bancorp ba'` Note that `id(n[0]) == id(t[2])`. – jonrsharpe Apr 07 '14 at 15:20

Change id of copied value from a list

3 Answers3