sys.getsizeof gives you the amount of memory allocated to the list itself, but you also have 10...00 int objects that the list only contains a pointer to.

The memory assigned is not disproportional; you are creating 100,000 objects! As you can see, they take up roughly 34 megabytes of space:

>>> sys.getsizeof(Test())+sys.getsizeof(Test().__dict__)
344
>>> (sys.getsizeof(Test())+sys.getsizeof(Test().__dict__)) * 1000000 / 10**6
34.4 #megabytes

You can get a minor improvement with __slots__, but you will still need about 20MB of memory to store those 100,000 objects.

>>> sys.getsizeof(Test2())+sys.getsizeof(Test2().__slots__)
200
>>> sys.getsizeof(Test2())+sys.getsizeof(Test2().__slots__) * 1000000 / 10**6
20.0 #megabytes

(With credit to mensi's answer, sys.getsizeof is not taking into account references. You can autocomplete to see most of the attributes of an object.)

See SO answer: Usage of __slots__? http://docs.python.org/release/2.5.2/ref/slots.html

To use __slots__:

class Test2():
    __slots__ = ['a','b','c','d','e']

    def __init__(self):
        ...

If you check the size of a list, it will be provide the size of the list data structure, including the pointers to its constituent elements. It won't consider the size of elements.

str1_size = sys.getsizeof(['a' for i in xrange(0, 1024)])
str2_size = sys.getsizeof(['abc' for i in xrange(0, 1024)])
int_size = sys.getsizeof([123 for i in xrange(0, 1024)])
none_size = sys.getsizeof([None for i in xrange(0, 1024)])
str1_size == str2_size == int_size == none_size

The size of empty list: sys.getsizeof([]) == 72
Add an element: sys.getsizeof([1]) == 80
Add another element: sys.getsizeof([1, 1]) == 88
So each element adds 4 bytes.
To get 1024 bytes, we need (1024 - 72) / 8 = 119 elements.

The size of the list with 119 elements: sys.getsizeof([None for i in xrange(0, 119)]) == 1080.
This is because a list maintains an extra buffer for inserting more items, so that it doesn't have to resize every time. (The size comes out to be same as 1080 for number of elements between 107 and 126).

So what we need is an immutable data structure, which doesn't need to keep this buffer - tuple.

empty_tuple_size = sys.getsizeof(())                     # 56
single_element_size = sys.getsizeof((1,))                # 64
pointer_size = single_element_size - empty_tuple_size    # 8
n_1mb = (1024 - empty_tuple_size) / pointer_size         # (1024 - 56) / 8 = 121
tuple_1mb = (1,) * n_1mb
sys.getsizeof(tuple_1mb) == 1024

So this is your answer to get a 1MB data structure: (1,)*121

But note that this is only the size of tuple and the constituent pointers. For the total size, you actually need to add up the size of individual elements.

Alternate:

sys.getsizeof('') == 37
sys.getsizeof('1') == 38     # each character adds 1 byte

For 1 MB, we need 987 characters:

sys.getsizeof('1'*987) == 1024

And this is the actual size, not just the size of pointers.

They are not the same thing at all.

len() queries for the number of items contained in a container. For a string that's the number of characters:

Return the length (the number of items) of an object. The argument may be a sequence (string, tuple or list) or a mapping (dictionary).

sys.getsizeof() on the other hand returns the memory size of the object:

Return the size of an object in bytes. The object can be any type of object. All built-in objects will return correct results, but this does not have to hold true for third-party extensions as it is implementation specific.

Python string objects are not simple sequences of characters, 1 byte per character.

Specifically, the sys.getsizeof() function includes the garbage collector overhead if any:

getsizeof() calls the object’s __sizeof__ method and adds an additional garbage collector overhead if the object is managed by the garbage collector.

String objects do not need to be tracked (they cannot create circular references), but string objects do need more memory than just the bytes per character. In Python 2, __sizeof__ method returns (in C code):

Py_ssize_t res;
res = PyStringObject_SIZE + PyString_GET_SIZE(v) * Py_TYPE(v)->tp_itemsize;
return PyInt_FromSsize_t(res);

where PyStringObject_SIZE is the C struct header size for the type, PyString_GET_SIZE basically is the same as len() and Py_TYPE(v)->tp_itemsize is the per-character size. In Python 2.7, for byte strings, the size per character is 1, but it's PyStringObject_SIZE that is confusing you; on my Mac that size is 37 bytes:

>>> sys.getsizeof('')
37

For unicode strings the per-character size goes up to 2 or 4 (depending on compilation options). On Python 3.3 and newer, Unicode strings take up between 1 and 4 bytes per character, depending on the contents of the string.

For containers such as dictionaries or lists that reference other objects, the memory size given covers only the memory used by the container and the pointer values used to reference those other objects. There is no straightforward method of including the memory size of the ‘contained’ objects because those same objects could have many more references elsewhere and are not necessarily owned by a single container.

The documentation states it like this:

Only the memory consumption directly attributed to the object is accounted for, not the memory consumption of objects it refers to.

If you need to calculate the memory footprint of a container and anything referenced by that container you’ll have to use some method of traversing to those contained objects and get their size; the documentation points to a recursive recipe.