Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
238 views
in Technique[技术] by (71.8m points)

python - Why do two identical lists have a different memory footprint?

I created two lists l1 and l2, but each one with a different creation method:

import sys

l1 = [None] * 10
l2 = [None for _ in range(10)]

print('Size of l1 =', sys.getsizeof(l1))
print('Size of l2 =', sys.getsizeof(l2))

But the output surprised me:

Size of l1 = 144
Size of l2 = 192

The list created with a list comprehension is a bigger size in memory, but the two lists are identical in Python otherwise.

Why is that? Is this some CPython internal thing, or some other explanation?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

When you write [None] * 10, Python knows that it will need a list of exactly 10 objects, so it allocates exactly that.

When you use a list comprehension, Python doesn't know how much it will need. So it gradually grows the list as elements are added. For each reallocation it allocates more room than is immediately needed, so that it doesn't have to reallocate for each element. The resulting list is likely to be somewhat bigger than needed.

You can see this behavior when comparing lists created with similar sizes:

>>> sys.getsizeof([None]*15)
184
>>> sys.getsizeof([None]*16)
192
>>> sys.getsizeof([None for _ in range(15)])
192
>>> sys.getsizeof([None for _ in range(16)])
192
>>> sys.getsizeof([None for _ in range(17)])
264

You can see that the first method allocates just what is needed, while the second one grows periodically. In this example, it allocates enough for 16 elements, and had to reallocate when reaching the 17th.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...