How can I host a large list in my 8G DDR3 RAM?

Question

I am new to python and just wondering how memory allocation works there. It turns out that one way to measure the size of a variable stored is to use sys.getsizeof(x) and it will return the number of bytes that are occupied by x in the memory, right? The following is an example code:

import struct
import sys

x = struct.pack('



which gives:

4
37


The variable x that I have just created is a 4-byte string and the first question rises here. Why is the memory allocated to a 4-byte string is 37 bytes? Is not that too much extra space?

The story gets more complicated when I start to create a list of 2 * 4-byte strings.
Bellow you will find another few lines:

import struct
import sys

k   = 2
rng = range(0, k)

x = [b''] * k

for i in rng:
    x[i] = struct.pack('


from which I get:

2
4
80
37


Another question is that why when I store two 4-byte strings in a list the total sum of the memory allocated to them is not equal to the sum of their solo sizes?! That is 37 + 37 != 80. What are those extra 6 bytes for?

Lets enlarge k to 10000, the previous code gives:

10000
4
80064
37


Here the difference rises dramatically when comparing the solo size to the whole: 37 * 10000 = 370000 != 80064. It looks like that each item in the list is now occupying 80064/10000 = 8.0064 bytes. Sounds feasible but I still cannot address previously shown conflicts.

After all, the main question of mine is that when I rise k to 0xffffffff and expect to get a list of size ~ 8 * 0xffffffff = 34359738360 I actually encounter an exception of MemoryError. Is there any way to eliminate non-critical memory spaces so that my 8G DDR3 RAM can host this variable x?

user_736937 · Answer

Why is the memory allocated to a 4-byte string is 37 bytes? Is not that too much extra space?

All objects in Python have some amount of "slop" on a per-object basis. Note that in the case of bytes and probably all immutable stdlib types, this padding (here 33 bytes) is independent of the length of the object:

from sys import getsizeof as gso
print(gso(b'x'*1000) - gso(b''))
# 1000

Note, this is not the same as:

print(gso([b'x']*1000) - gso(b''))
# 8031

In the former, you're making a bytes object of 1000 x's.

In the latter, you're making a list of 1000 bytes objects. The important distinction is that in the latter, you're (a) replicating the bytes object 1000 times, and incorporating the size of the list container. (The reason for the difference being only ~8,000 and not ~34,000 (i.e. 8 bytes per element instead of 34 bytes per element (=sizeof(b'x')) comes next.)

Lets talk about containers:

print(gso([b'x'*100,]) - gso([]))

Here we print the difference between the getsizeof of a one element list (of a 100 byte long byte object) and an empty list. We're effectively taring out the size of the container.

We might expect that this is equal to getsizeof(b'x' * 100).

It is not.

The result of print(gso([b'x'*100,]) - gso([])) is 8 bytes (on my machine) and is because the list contains just references/pointers to underlying objects and those 8 bytes are just that -- the pointer to the single element of the list.

That is 37 + 37 != 80. What are those extra 6 bytes for?

Lets do the same thing and look at the net size, by subtracting the size of the container:

x = [b'\xff\xff\xff\xff', b'\xff\xff\xff\xff']

print(gso(x[0]) - gso(b''))   # 4
print(gso(x)    - gso([]))    # 16

In the first, the 4 returned is just as the 1000 returned in the first example I provided, one per byte. (len(x[0]) is 4).

In the second, it's 8 bytes per reference-to-sublist. It has nothing to do with the contents of those sublists:

N = 1000
x = [b'x']    * N
y = [b'xxxx'] * N
print(gso(x) == gso(y))
# True

But while mutable containers don't seem to have a fixed "slop":

lst = []
for _ in range(100):
    lst.append('-')
    x = list(lst)

    slop = gso(x) - (8 * len(x))
    print({"len": len(x), "slop": slop})

Output:

{'len': 1, 'slop': 88}
{'len': 2, 'slop': 88}
{'len': 3, 'slop': 88}
{'len': 4, 'slop': 88}
{'len': 5, 'slop': 88}
{'len': 6, 'slop': 88}
{'len': 7, 'slop': 88}
{'len': 8, 'slop': 96}
{'len': 9, 'slop': 120}
{'len': 10, 'slop': 120}
{'len': 11, 'slop': 120}
{'len': 12, 'slop': 120}
{'len': 13, 'slop': 120}
{'len': 14, 'slop': 120}
{'len': 15, 'slop': 120}
{'len': 16, 'slop': 128}
{'len': 17, 'slop': 128}
{'len': 18, 'slop': 128}
{'len': 19, 'slop': 128}
{'len': 20, 'slop': 128}
{'len': 21, 'slop': 128}
{'len': 22, 'slop': 128}
{'len': 23, 'slop': 128}
{'len': 24, 'slop': 136}
...

...Immutable containers do:

lst = []
for _ in range(100):
    lst.append('-')
    x = tuple(lst)

    slop = gso(x) - (8 * len(x))
    print({"len": len(x), "slop": slop})

{'len': 1, 'slop': 48}
{'len': 2, 'slop': 48}
{'len': 3, 'slop': 48}
{'len': 4, 'slop': 48}
{'len': 5, 'slop': 48}
{'len': 6, 'slop': 48}
{'len': 7, 'slop': 48}
{'len': 8, 'slop': 48}
{'len': 9, 'slop': 48}
{'len': 10, 'slop': 48}
{'len': 11, 'slop': 48}
{'len': 12, 'slop': 48}
{'len': 13, 'slop': 48}
{'len': 14, 'slop': 48}
...

Is there any way to eliminate non-critical memory spaces so that my 8G DDR3 RAM can host this variable x?

First, recall that the sizeof a container will not reflect the entire amount of memory used by Python. The ~8 bytes per element is the size of the pointer, each of those elements will consume an additional 37 (or whatever) bytes (sans interning or similar optimization).

But the good news is that it's unlikely you probably don't need the entire list at the same time. If you're just building a list to iterate over, then generate it one element at a time, with a for loop or generator function.

Or generate it a chunk at a time, process it, and then continue, letting the garbage collector clean up the no-longer-used memory.

One other interesting thing to point out

N = 1000
x = [b'x' for _ in range(N)]
y = [b'x'] * N
print(x == y)              # True
print(gso(x) == gso(y))    # False

(This is likely due to the size of y being known a priori, while the size of x is not and has been resized as it grew).

How can I host a large list in my 8G DDR3 RAM?

1 Answers