我想创建一个可以容纳10个元素的空列表(或者是最好的方法)。
之后我想在该列表中分配值,例如,这应该显示0到9:
s1 = list();
for i in range(0,9):
s1[i] = i
print s1
但是,当我运行此代码时,它会生成错误,或者在另一种情况下,它只显示[]
(空)。
有人可以解释原因吗?
发布于 2019-02-28 13:23:48
varunl目前接受的答案
>>> l = [None] * 10
>>> l
[None, None, None, None, None, None, None, None, None, None]
适用于非引用类型,如数字。不幸的是,如果你想创建一个列表列表,你将遇到引用错误。Python 2.7.6中的示例:
>>> a = [[]]*10
>>> a
[[], [], [], [], [], [], [], [], [], []]
>>> a[0].append(0)
>>> a
[[0], [0], [0], [0], [0], [0], [0], [0], [0], [0]]
>>>
如您所见,每个元素都指向同一个列表对象。要解决此问题,您可以创建一个方法,将每个位置初始化为不同的对象引用。
def init_list_of_objects(size):
list_of_objects = list()
for i in range(0,size):
list_of_objects.append( list() ) #different object reference each time
return list_of_objects
>>> a = init_list_of_objects(10)
>>> a
[[], [], [], [], [], [], [], [], [], []]
>>> a[0].append(0)
>>> a
[[0], [], [], [], [], [], [], [], [], []]
>>>
有一种默认的内置python方式(不是编写函数),但我不确定它是什么。很乐意纠正!
编辑:是的 [ [] for _ in range(10)]
示例:
>>> [ [random.random() for _ in range(2) ] for _ in range(5)]
>>> [[0.7528051908943816, 0.4325669600055032], [0.510983236521753, 0.7789949902294716], [0.09475179523690558, 0.30216475640534635], [0.3996890132468158, 0.6374322093017013], [0.3374204010027543, 0.4514925173253973]]
发布于 2019-02-28 14:13:09
有两种“快速”方法:
x = length_of_your_list
a = [None]*x
# or
a = [None for _ in xrange(x)]
似乎[None]*x
更快:
>>> from timeit import timeit
>>> timeit("[None]*100",number=10000)
0.023542165756225586
>>> timeit("[None for _ in xrange(100)]",number=10000)
0.07616496086120605
但是如果你对范围(例如[0,1,2,3,...,x-1]
)没问题,那么range(x)
可能是最快的:
>>> timeit("range(100)",number=10000)
0.012513160705566406
发布于 2020-06-28 17:04:00
用这个嵌套的可以,不仅可以做二维的,多次嵌套可以做三维的
a = [[[] for _ in range(100)] for _ in range(100)]()
type(a[0][0])
list
https://stackoverflow.com/questions/-100006395
复制相似问题