在 Python 列表中交换元素的最快方法
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Fastest way to swap elements in Python list
提问by Gerald Senarclens de Grancy
采纳答案by Ignacio Vazquez-Abrams
Looks like the Python compiler optimizes out the temporary tuple with this construct:
看起来 Python 编译器使用以下构造优化了临时元组：
code:
代码：
import dis
def swap1():
a=5
b=4
a, b = b, a
def swap2():
a=5
b=4
c = a
a = b
b = c
print 'swap1():'
dis.dis(swap1)
print 'swap2():'
dis.dis(swap2)
output:
输出：
swap1():
6 0 LOAD_CONST 1 (5)
3 STORE_FAST 0 (a)
7 6 LOAD_CONST 2 (4)
9 STORE_FAST 1 (b)
8 12 LOAD_FAST 1 (b)
15 LOAD_FAST 0 (a)
18 ROT_TWO
19 STORE_FAST 0 (a)
22 STORE_FAST 1 (b)
25 LOAD_CONST 0 (None)
28 RETURN_VALUE
swap2():
11 0 LOAD_CONST 1 (5)
3 STORE_FAST 0 (a)
12 6 LOAD_CONST 2 (4)
9 STORE_FAST 1 (b)
13 12 LOAD_FAST 0 (a)
15 STORE_FAST 2 (c)
14 18 LOAD_FAST 1 (b)
21 STORE_FAST 0 (a)
15 24 LOAD_FAST 2 (c)
27 STORE_FAST 1 (b)
30 LOAD_CONST 0 (None)
33 RETURN_VALUE
Two loads, a ROT_TWO, and two saves, versus three loads and three saves. You are unlikely to find a faster mechanism.
两次加载，一次ROT_TWO和两次保存，对比三次加载和三次保存。您不太可能找到更快的机制。
回答by TryPyPy
If you could post a representative code sample, we could do a better job of benchmarking your options. FWIW, for the following dumb benchmark, I get about a 3x speedup with Shed Skinand a 10x speedup with PyPy.
如果您可以发布具有代表性的代码示例，我们可以更好地对您的选项进行基准测试。FWIW，对于以下愚蠢的基准测试，我使用Shed Skin获得了大约 3倍的加速，使用PyPy获得了大约10 倍的加速。
from time import time
def swap(L):
for i in xrange(1000000):
for b, a in enumerate(L):
L[a], L[b] = L[b], L[a]
def main():
start = time()
L = list(reversed(range(100)))
swap(L[:])
print time() - start
return L
if __name__ == "__main__":
print len(main())
# for shedskin:
# shedskin -b -r -e listswap.py && make
# python -c "import listswap; print len(listswap.main())"
回答by Ashish Mohapatra
I found this method as the fastest way to swap two numbers:
我发现这种方法是交换两个数字的最快方法：
mylist = [11,23,5,8,13,17];
first_el = mylist.pop(0)
last_el = mylist.pop(-1)
mylist.insert(0, last_el)
mylist.append(first_el)
