You don't need to copy a Python string. They are immutable, and the copy module always returns the original in such cases, as do str(), the whole string slice, and concatenating with an empty string.
Moreover, your 'hello' string is interned (certain strings are). Python deliberately tries to keep just the one copy, as that makes dictionary lookups faster.
One way you could work around this is to actually create a new string, then slice that string back to the original content:
>>> a = 'hello'
>>> b = (a + '.')[:-1]
>>> id(a), id(b)
(4435312528, 4435312432)
But all you are doing now is waste memory. It is not as if you can mutate these string objects in any way, after all.
If all you wanted to know is how much memory a Python object requires, use sys.getsizeof(); it gives you the memory footprint of any Python object.
For containers this does not include the contents; you'd have to recurse into each container to calculate a total memory size:
>>> import sys
>>> a = 'hello'
>>> sys.getsizeof(a)
42
>>> b = {'foo': 'bar'}
>>> sys.getsizeof(b)
280
>>> sys.getsizeof(b) + sum(sys.getsizeof(k) + sys.getsizeof(v) for k, v in b.items())
360
You can then choose to use id() tracking to take an actual memory footprint or to estimate a maximum footprint if objects were not cached and reused.
I'm just starting some string manipulations and found this question. I was probably trying to do something like the OP, "usual me". The previous answers did not clear up my confusion, but after thinking a little about it I finally "got it".
As long as a, b, c, d, and e have the same value, they reference to the same place. Memory is saved. As soon as the variable start to have different values, they get start to have different references. My learning experience came from this code:
import copy
a = 'hello'
b = str(a)
c = a[:]
d = a + ''
e = copy.copy(a)
print map( id, [ a,b,c,d,e ] )
print a, b, c, d, e
e = a + 'something'
a = 'goodbye'
print map( id, [ a,b,c,d,e ] )
print a, b, c, d, e
The printed output is:
[4538504992, 4538504992, 4538504992, 4538504992, 4538504992]
hello hello hello hello hello
[6113502048, 4538504992, 4538504992, 4538504992, 5570935808]
goodbye hello hello hello hello something
To put it a different way "id()" is not what you care about. You want to know if the variable name can be modified without harming the source variable name.
>>> a = 'hello'
>>> b = a[:]
>>> c = a
>>> b += ' world'
>>> c += ', bye'
>>> a
'hello'
>>> b
'hello world'
>>> c
'hello, bye'
If you're used to C, then these are like pointer variables except you can't de-reference them to modify what they point at, but id() will tell you where they currently point.
The problem for python programmers comes when you consider deeper structures like lists or dicts:
>>> o={'a': 10}
>>> x=o
>>> y=o.copy()
>>> x['a'] = 20
>>> y['a'] = 30
>>> o
{'a': 20}
>>> x
{'a': 20}
>>> y
{'a': 30}
Here o and x refer to the same dict o['a'] and x['a'], and that dict is "mutable" in the sense that you can change the value for key 'a'. That's why "y" needs to be a copy and y['a'] can refer to something else.
As others have already explained, there's rarely an actual need for this, but nevertheless, here you go:
(works on Python 3, but there's probably something similar for Python 2)
import ctypes
copy = ctypes.pythonapi._PyUnicode_Copy
copy.argtypes = [ctypes.py_object]
copy.restype = ctypes.py_object
s1 = 'xxxxxxxxxxxxx'
s2 = copy(s1)
id(s1) == id(s2) # False
d[ 'hello' ] = e, wheree[ 'hi' ] = 'again'. To generate such a nested dictionary, I generated a singleedictionary and copied it multiple times. I noticed that the memory consumption was very low, which led to my question here. Now I understand that no string copies were created, hence the low memory consumption. - usual mebto be a modified version ofawithout modifyinga, just letbbe the result of whatever operation. e.g.b = a[2:-1]setsbto'll'andaremains 'hello'. - OJFord