If you don't need the exact size of the object but roughly to know how big it is, one quick (and dirty) way is to let the program run, sleep for an extended period of time, and check the memory usage (ex: Mac's activity monitor) by this particular python process. This would be effective when you are trying to find the size of one single large object in a python process. For example, I recently wanted to check the memory usage of a new data structure and compare it with that of Python's set data structure. First I wrote the elements (words from a large public domain book) to a set, then checked the size of the process, and then did the same thing with the other data structure. I found out the Python process with a set is taking twice as much memory as the new data structure. Again, you wouldn't be able to exactly say the memory used by the process is equal to the size of the object. As the size of the object gets large, this becomes close as the memory consumed by the rest of the process becomes negligible compared to the size of the object you are trying to monitor.

对于numpy数组，getsizeof不起作用-对我来说，它总是出于某种原因返回40:

from pylab import *
from sys import getsizeof
A = rand(10)
B = rand(10000)

然后(在ipython中):

In [64]: getsizeof(A)
Out[64]: 40

In [65]: getsizeof(B)
Out[65]: 40

不过令人高兴的是,:

In [66]: A.nbytes
Out[66]: 80

In [67]: B.nbytes
Out[67]: 80000

这可能不是最相关的答案，但我只对对象存储和检索感兴趣。因此将对象转储为pickle并检查pickle的大小就足够了

我自己也遇到过很多次这个问题，我写了一个小函数(受到@aaron-hall的回答的启发)和测试，它完成了我所期望的sys。Getsizeof to do:

https://github.com/bosswissam/pysize

如果你对背景故事感兴趣，这就是

编辑:附上下面的代码以方便参考。要查看最新的代码，请检查github链接。

    import sys

    def get_size(obj, seen=None):
        """Recursively finds size of objects"""
        size = sys.getsizeof(obj)
        if seen is None:
            seen = set()
        obj_id = id(obj)
        if obj_id in seen:
            return 0
        # Important mark as seen *before* entering recursion to gracefully handle
        # self-referential objects
        seen.add(obj_id)
        if isinstance(obj, dict):
            size += sum([get_size(v, seen) for v in obj.values()])
            size += sum([get_size(k, seen) for k in obj.keys()])
        elif hasattr(obj, '__dict__'):
            size += get_size(obj.__dict__, seen)
        elif hasattr(obj, '__iter__') and not isinstance(obj, (str, bytes, bytearray)):
            size += sum([get_size(i, seen) for i in obj])
        return size

Python 3.8(2019年第一季度)将改变sys. js的一些结果。getsizeof, Raymond Hettinger在此宣布:

Python容器在64位版本上要小8个字节。

tuple ()  48 -> 40       
list  []  64 ->56
set()    224 -> 216
dict  {} 240 -> 232

这是在议题33597和稻田直树(甲烷)围绕紧凑型PyGC_Head和PR 7043的工作之后

这个想法将PyGC_Head大小减少到两个单词。 目前，PyGC_Head包含三个单词;Gc_prev, gc_next和gc_refcnt。 收集时使用Gc_refcnt，用于尝试删除。 Gc_prev用于跟踪和取消跟踪。 因此，如果我们可以在试删除时避免跟踪/取消跟踪，gc_prev和gc_refcnt可以共享相同的内存空间。

参见commit d5c875b:

从PyGC_Head中移除一个Py_ssize_t成员。 所有GC跟踪的对象(例如元组，列表，dict)大小减少4或8字节。

下面是我根据之前对所有变量的列表大小的回答编写的一个快速脚本

for i in dir():
    print (i, sys.getsizeof(eval(i)) )