python 如何在python中创建嵌套列表？
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How to create nested lists in python?
提问by Jader Dias
I know you can create easily nested lists in python like this:
我知道您可以像这样在 python 中轻松创建嵌套列表：
[[1,2],[3,4]]
But how to create a 3x3x3 matrix of zeroes?
但是如何创建一个 3x3x3 的零矩阵呢？
[[[0] * 3 for i in range(0, 3)] for j in range (0,3)]
or
或者
[[[0]*3]*3]*3
Doesn't seem right. There is no way to create it just passing a list of dimensions to a method? Ex:
好像不太对 没有办法只将维度列表传递给方法来创建它吗？前任：
CreateArray([3,3,3])
回答by
In case a matrix is actually what you are looking for, consider the numpy package.
如果矩阵确实是您要查找的内容，请考虑使用 numpy 包。
http://docs.scipy.org/doc/numpy/reference/generated/numpy.zeros.html#numpy.zeros
http://docs.scipy.org/doc/numpy/reference/generated/numpy.zeros.html#numpy.zeros
This will give you a 3x3x3 array of zeros:
这会给你一个 3x3x3 的零数组：
numpy.zeros((3,3,3))
You also benefit from the convenience features of a module built for scientific computing.
您还可以从为科学计算构建的模块的便利功能中受益。
回答by rob
List comprehensions are just syntactic sugar for adding expressiveness to list initialization; in your case, I would not use them at all, and go for a simple nested loop.
列表推导式只是为列表初始化增加表现力的语法糖；在您的情况下，我根本不会使用它们，而是使用简单的嵌套循环。
On a completely different level: do you think the n-dimensional arrayof NumPy could be a better approach?
Although you can use lists to implement multi-dimensional matrices, I think they are not the best tool for that goal.
在完全不同的层面上：您认为NumPy的n 维数组可能是更好的方法吗？
尽管您可以使用列表来实现多维矩阵，但我认为它们不是实现该目标的最佳工具。
回答by Jader Dias
NumPy addresses this problem
NumPy 解决了这个问题
http://www.scipy.org/Tentative_NumPy_Tutorial#head-d3f8e5fe9b903f3c3b2a5c0dfceb60d71602cf93
http://www.scipy.org/Tentative_NumPy_Tutorial#head-d3f8e5fe9b903f3c3b2a5c0dfceb60d71602cf93
>>> a = array( [2,3,4] )
>>> a
array([2, 3, 4])
>>> type(a)
<type 'numpy.ndarray'>
But if you want to use the Python native lists as a matrix the following helper methods can become handy:
但是如果你想使用 Python 原生列表作为矩阵，下面的辅助方法会变得很方便：
import copy
def Create(dimensions, item):
for dimension in dimensions:
item = map(copy.copy, [item] * dimension)
return item
def Get(matrix, position):
for index in position:
matrix = matrix[index]
return matrix
def Set(matrix, position, value):
for index in position[:-1]:
matrix = matrix[index]
matrix[position[-1]] = value
回答by Sander Evers
回答by Dario
Just nest the multiplication syntax:
只需嵌套乘法语法：
[[[0] * 3] * 3] * 3
It's therefore simple to express this operation using folds
因此使用折叠来表达这个操作很简单
def zeros(dimensions):
return reduce(lambda x, d: [x] * d, [0] + dimensions)
Or if you want to avoid reference replication, so altering one item won't affect any other you should instead use copies:
或者，如果您想避免引用复制，因此更改一项不会影响任何其他项目，您应该使用副本：
import copy
def zeros(dimensions):
item = 0
for dimension in dimensions:
item = map(copy.copy, [item] * dimension)
return item
