Question

如果我想将列表初始化为5个零，那就非常简单了：

[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。我必须这样做：

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

感觉笨重并且使用变量，所以有时我甚至会这样做：

[[0] for _ in "     "]

但是如果我想要一份零列表，那就更加丑陋了：

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

所有这些而不是我想做的事情：

[[[0]]*5]*5

有没有人找到一种处理这个“问题”的优雅方式？

Answer 1

在考虑了一下之后，我提出了这个解决方案:(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，其中s的形状类似于(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......我只需将其建筑物打包成一个小型工厂功能，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！ - ）

Answer 3

另一个是扩展列表类：

import copy
class mlist(list):
  def __mul__(self, n):
    res = mlist()
    for _ in xrange(n):
      for l in self:
    res.append(copy.deepcopy(l))
  return res

然后：

>>> hey = mlist([mlist([0])])
>>> hey
[[0]]
>>> hey * 4
[[0], [0], [0], [0]]
>>> blah = hey * 4
>>> blah[0][0] = 9
>>> blah
[[9], [0], [0], [0]]

但初始化mlist很烦人。

Answer 4

一种解决方案是拥有辅助功能：

import copy
def r(i,n):
    return [copy.deepcopy(i) for _ in xrange(n)]

然后：

r([0],5)
r(r([0],5),5)

但这种语法很难看。

更优雅的方法来初始化Python中重复项的列表

4 个答案: