更优雅的方法来初始化Python中重复项的列表

Question

如果我想要一个初始化为5个零的列表,这非常好用且简单:

[0] * 5

但是,如果我更改我的代码以放置更复杂的数据结构,如零列表:

[[0]] * 5

将不会按预期工作,因为它将是同一列表的10个副本.我要做:

[[0] for i in xrange(5)]

感觉笨重并使用变量,所以有时我甚至做:

[[0] for _ in "     "]

但是,如果我想要一个零列表列表,它会变得更加丑陋:

[[[0] for _ in "     "] for _ in "     "]

所有这些而不是我想做的事情:

[[[0]]*5]*5

有没有人找到一种优雅的方式来处理这个"问题"？

Answer 1

在考虑了一下之后,我想出了这个解决方案:(没有导入的7行)

# 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 ^^)

测试给出以下结果:

>>> 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"[[[0]]*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)函数,s其形状就像(3,4,5)

>>> s = (2,2)
>>> np.zeros(s)
array([[ 0.,  0.],
       [ 0.,  0.]])

最后进行性能测试:

# 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真的很有趣.所以,谢谢你的问题:)

Answer 2

如果我经常要求列表列表......我只需将其构建包装成一个小工厂函数,例如:

import copy

def multi_dimension_list(baseitem, *dimensions):
  dimensions = list(dimensions)
  result = [baseitem] * dimensions.pop(-1)
  for d in reversed(dimensions):
    result = [copy.deepcopy(result) for _ in range(d)]
  return result

eg = multi_dimension_list(0, 3, 4, 5)
print(eg)
# and just to prove the parts are independent...:
eg[1][1][1] = 23
print(eg)

在实践中,我甚至都不打扰,因为我对这种多维列表的使用很少,所以内联列表理解就好了.然而,建立你自己的小实用函数模块的一般想法,你需要经常执行的简单任务(在你看来)并不是通过内联习语优雅地完成,这是唯一的方法! - )