慕沐林林
简短版：使用pre_allocated_list = [None] * size预先分配一个列表（即，能够解决列表的'size'元素，而不是通过附加逐渐形成列表）。即使在大型列表中，此操作也非常快。分配稍后将分配给列表元素的新对象将花费更长时间，并且将成为程序中的瓶颈，性能方面。长版：我认为应该考虑初始化时间。因为在python中，一切都是引用，无论你是将每个元素设置为None还是一些字符串都无关紧要 - 无论哪种方式，它都只是一个引用。如果要为每个要引用的元素创建新对象，则需要更长的时间。对于Python 3.2：import timeimport copydef print_timing (func): def wrapper (*arg): t1 = time.time () res = func (*arg) t2 = time.time () print ("{} took {} ms".format (func.__name__, (t2 - t1) * 1000.0)) return res return wrapper@print_timingdef prealloc_array (size, init = None, cp = True, cpmethod=copy.deepcopy, cpargs=(), use_num = False): result = [None] * size if init is not None: if cp: for i in range (size): result[i] = init else: if use_num: for i in range (size): result[i] = cpmethod (i) else: for i in range (size): result[i] = cpmethod (cpargs) return result@print_timingdef prealloc_array_by_appending (size): result = [] for i in range (size): result.append (None) return result@print_timingdef prealloc_array_by_extending (size): result = [] none_list = [None] for i in range (size): result.extend (none_list) return resultdef main (): n = 1000000 x = prealloc_array_by_appending(n) y = prealloc_array_by_extending(n) a = prealloc_array(n, None) b = prealloc_array(n, "content", True) c = prealloc_array(n, "content", False, "some object {}".format, ("blah"), False) d = prealloc_array(n, "content", False, "some object {}".format, None, True) e = prealloc_array(n, "content", False, copy.deepcopy, "a", False) f = prealloc_array(n, "content", False, copy.deepcopy, (), False) g = prealloc_array(n, "content", False, copy.deepcopy, [], False) print ("x[5] = {}".format (x[5])) print ("y[5] = {}".format (y[5])) print ("a[5] = {}".format (a[5])) print ("b[5] = {}".format (b[5])) print ("c[5] = {}".format (c[5])) print ("d[5] = {}".format (d[5])) print ("e[5] = {}".format (e[5])) print ("f[5] = {}".format (f[5])) print ("g[5] = {}".format (g[5]))if __name__ == '__main__': main()评价：prealloc_array_by_appending took 118.00003051757812 msprealloc_array_by_extending took 102.99992561340332 msprealloc_array took 3.000020980834961 msprealloc_array took 49.00002479553223 msprealloc_array took 316.9999122619629 msprealloc_array took 473.00004959106445 msprealloc_array took 1677.9999732971191 msprealloc_array took 2729.999780654907 msprealloc_array took 3001.999855041504 msx[5] = Noney[5] = Nonea[5] = Noneb[5] = contentc[5] = some object blahd[5] = some object 5e[5] = af[5] = []g[5] = ()正如您所看到的，只需创建一个对同一None对象的引用的大列表，只需要很少的时间。前置或延长需要更长的时间（我没有做任何平均值，但是在运行几次后我可以告诉你，延伸和追加大致需要相同的时间）。为每个元素分配新对象 - 这是花费最多时间的。而S.Lott的答案就是这样 - 每次都会格式化一个新的字符串。这不是严格要求的 - 如果您想预先分配一些空间，只需创建一个None列表，然后随意将数据分配给列表元素。无论哪种方式，生成数据都需要花费更多时间而不是追加/扩展列表，无论是在创建列表时生成还是在生成列表之后生成数据。但是如果你想要一个人口稀少的列表，那么从None列表开始肯定会更快。