<p>如果您要将数量可变的列表串联在一起,那么您的输入将是一个列表列表(或一些等效的集合)。性能测试需要考虑到这一点,因为您将无法执行类似list1+list2+list3的操作。你知道吗</p>
<p>以下是一些测试结果(1000次重复):</p>
<pre><code>option1 += loop 0.00097 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option2 itertools.chain 0.00138 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option3 functools.reduce 0.00174 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option4 comprehension 0.00188 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option5 extend loop 0.00127 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option6 deque 0.00180 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
</code></pre>
<p><em>这表明+=循环遍历列表列表是最快的方法</em></p>
<p>以及产生它们的来源:</p>
<pre><code>allLists = [ list(range(10)) for _ in range(5) ]
def option1():
result = allLists[0].copy()
for lst in allLists[1:]:
result += lst
return result
from itertools import chain
def option2(): return list(chain(*allLists))
from functools import reduce
def option3():
return list(reduce(lambda a,b:a+b,allLists))
def option4(): return [ e for l in allLists for e in l ]
def option5():
result = allLists[0].copy()
for lst in allLists[1:]:
result.extend(lst)
return result
from collections import deque
def option6():
result = deque()
for lst in allLists:
result.extend(lst)
return list(result)
from timeit import timeit
count = 1000
t = timeit(lambda:option1(), number = count)
print(f"option1 += loop {t:.5f}",option1()[:15])
t = timeit(lambda:option2(), number = count)
print(f"option2 itertools.chain {t:.5f}",option2()[:15])
t = timeit(lambda:option3(), number = count)
print(f"option3 functools.reduce {t:.5f}",option3()[:15])
t = timeit(lambda:option4(), number = count)
print(f"option4 comprehension {t:.5f}",option4()[:15])
t = timeit(lambda:option5(), number = count)
print(f"option5 extend loop {t:.5f}",option5()[:15])
t = timeit(lambda:option6(), number = count)
print(f"option6 deque {t:.5f}",option6()[:15])
</code></pre>