在Python中初始化重复项列表的更优雅的方法
在Python中初始化重复项列表的更优雅的方法
提问于 2010-06-10 03:16:41
如果我想要一个列表初始化为5个零，这是非常好和容易的：
[0] * 5但是，如果我更改代码以放入更复杂的数据结构，比如一列零：
[[0]] * 5将不会像预期的那样工作，因为它将是相同列表的10个副本。我要做的是：
[[0] for i in xrange(5)]这感觉很笨重，而且使用了一个变量，所以有时我甚至会这样做：
[[0] for _ in " "]但是，如果我想要一个包含零的列表，它就会变得更加丑陋：
[[[0] for _ in " "] for _ in " "]所有这些，而不是我想要做的：
[[[0]]*5]*5有没有人找到一个优雅的方法来解决这个“问题”？
Stack Overflow用户
回答已采纳
发布于 2010-06-10 04:10:27
在考虑了一下之后，我想出了这个解决方案：(7行不导入)
# helper
def cl(n, func):
# return a lambda, that returns a list, where func(tion) is called
return (lambda: [func() for _ in range(n)])
def matrix(base, *ns):
# the grid lambda (at the start it returns the base-element)
grid = lambda: base
# traverse reversed, to handle the midmost values first
for n in reversed(ns):
# assign a new lambda with the last grid within (and call it)
grid = cl(n, grid)
return grid() # call the full grid (but the matrix calls you ^^)测试结果如下：
>>> from pprint import pprint as pp
>>>
>>> matrix(None, 2,3)
[[None, None, None], [None, None, None]]
>>>
>>> matrix(None, 4,3)
[[None, None, None], [None, None, None], [None, None, None], [None, None, None]]
>>>
>>> x = matrix(None, 3,5,2)
>>> pp(x)
[[[None, None], [None, None], [None, None], [None, None], [None, None]],
[[None, None], [None, None], [None, None], [None, None], [None, None]],
[[None, None], [None, None], [None, None], [None, None], [None, None]]]
>>> x[1][3][0] = "test"
>>> pp(x)
[[[None, None], [None, None], [None, None], [None, None], [None, None]],
[[None, None], [None, None], [None, None], ['test', None], [None, None]],
[[None, None], [None, None], [None, None], [None, None], [None, None]]]另一种解决方案，其优点是使用"[[]*5]*5"-syntax：
def uniq(base, l):
# function used to replace all values with the base
nl = []
for i in l:
if type(i) is list:
nl.append(uniq(base, i)) # recursion for deep lists
else:
nl.append(base)
return nl测试：
# first arg is the base, the 0 inside the [] is just a dummy
# (for what None is the best choice usually)
>>> x = uniq(0, [[[0]]*5]*5)
>>> x[0][3][0] = 5
>>> pp(x)
[[[0], [0], [0], [5], [0]],
[[0], [0], [0], [0], [0]],
[[0], [0], [0], [0], [0]],
[[0], [0], [0], [0], [0]],
[[0], [0], [0], [0], [0]]]顺便说一句。numpy库有一个np.zeros(s)-function，其中s是一个类似(3,4,5)的形状
>>> s = (2,2)
>>> np.zeros(s)
array([[ 0., 0.],
[ 0., 0.]])最后是性能测试：
# functions are already defined ...
import timeit
>>> # Alex Martelli's Code
>>> t1 = timeit.Timer( lambda: multi_dimension_list(None, 3,4,5) )
>>> # the two mentioned above
>>> t2 = timeit.Timer( lambda: matrix(None, 3,4,5) )
>>> t3 = timeit.Timer( lambda: uniq(None, [[[None]*5]*4]*3) )
>>>
>>> t1.timeit(10000)
2.1910018920898438
>>> t2.timeit(10000)
0.44953203201293945
>>> t3.timeit(10000)
0.48807907104492188我发现发现这个问题非常有趣。所以，谢谢你的问题:)
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原文链接：
https://stackoverflow.com/questions/3009091
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