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How to create nested lists in python?

开发者 https://www.devze.com 2022-12-19 05:08 出处:网络
I know you can create easily nested lists in python like this: [[1,2],[3,4]] But how to create a 3x3x3 matrix of zeroes?

I know you can create easily nested lists in python like this:

[[1,2],[3,4]]

But how to create a 3x3x3 matrix of zeroes?

[[[0] * 3 for i in range(0, 3)] for j in range (0,3)]

or

[[[0]*3]*3开发者_JS百科]*3

Doesn't seem right. There is no way to create it just passing a list of dimensions to a method? Ex:

CreateArray([3,3,3])


In case a matrix is actually what you are looking for, consider the numpy package.

http://docs.scipy.org/doc/numpy/reference/generated/numpy.zeros.html#numpy.zeros

This will give you a 3x3x3 array of zeros:

numpy.zeros((3,3,3)) 

You also benefit from the convenience features of a module built for scientific computing.


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.

On a completely different level: do you think the n-dimensional array of 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.


NumPy addresses this problem

Link

>>> 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:

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


Or use the nest function defined here, combined with repeat(0) from the itertools module:

nest(itertools.repeat(0),[3,3,3])


Just nest the multiplication syntax:

[[[0] * 3] * 3] * 3

It's therefore simple to express this operation using folds

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:

import copy
def zeros(dimensions):
    item = 0
    for dimension in dimensions:
        item = map(copy.copy, [item] * dimension)
   return item
0

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