如何手动计算字典的大小(它在内存中占用的字节数).我读到最初它是280字节,第6个键增加,然后是第86个等等.我想计算当我有超过10000个键时它将占用的大小.
小智 22
不适sys.getsizeof
用于嵌套字典,如下例所示。
>>> import sys
>>> d = { "onj1": {"name":"object 01", "id": "123"},"onj2": {"name":"object 02", "id": "124"}}
>>> d0 = {}
>>> sys.getsizeof(d0)
240
>>> sys.getsizeof(d)
240
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所以找到的解决方案就是这个网站提供的功能:post OR github
遵循函数:
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
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sys.getsizeof
在这种情况下将有所帮助:
from sys import getsizeof
dct = {'a': 5, 'b': 7}
print(getsizeof(dct))
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特别是对于字典,大小将取决于您的python版本(最近实现已更改).
创建N
大小字典的快速方法是:
from itertools import zip_longest
dct = dict(zip_longest(range(N), (), fillvalue=None))
# {0: None, 1: None, 2: None, ....}
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这应该有助于测试您对特定python版本的假设.
这个问题可能有关系.
您可以使用以下命令进行快速检查sys.getsizeof()
(它将返回对象的大小(以字节为单位)):
>>> import sys, itertools
>>> sys.getsizeof(dict(zip(range(1), itertools.cycle([1]))))
280
>>> sys.getsizeof(dict(zip(range(5), itertools.cycle([1]))))
280
>>> sys.getsizeof(dict(zip(range(6), itertools.cycle([1]))))
1048
>>> sys.getsizeof(dict(zip(range(85), itertools.cycle([1]))))
3352
>>> sys.getsizeof(dict(zip(range(86), itertools.cycle([1]))))
12568
>>> sys.getsizeof(dict(zip(range(87), itertools.cycle([1]))))
12568
>>> sys.getsizeof(dict(zip(range(10000), itertools.cycle([1]))))
786712
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如果您对 Python 字典的实际内部工作原理感兴趣,这dictobject.c
是权威的资源(此处为最新的 Python 3.6 分支)。另外,请看一下dictnotes.txt
.