如何在Python中重采样数组

Question

我刚开始学习Python，有个关于数组/矩阵的问题。

这是我得到的矩阵：

A =

[[85 77 83 ..., 59 58 59]

[80 83 80 ..., 57 60 58]

[75 76 81 ..., 59 58 60]]

我想对这个矩阵进行重新采样（我不知道这个词用得对不对），希望它变成：

B =

[[ 85 85 85 85 77 77 77 77 83 83 83 83 ....... 59 59 59 59 58 58 58 58 59 59 59 59]

[ 85 85 85 85 77 77 77 77 83 83 83 83 ....... 59 59 59 59 58 58 58 58 59 59 59 59]

[ 85 85 85 85 77 77 77 77 83 83 83 83 ....... 59 59 59 59 58 58 58 58 59 59 59 59]

[ 85 85 85 85 77 77 77 77 83 83 83 83 ....... 59 59 59 59 58 58 58 58 59 59 59 59]

[ 80 80 80 80 83 83 83 83 80 80 80 80 ....... 57 57 57 57 60 60 60 60 58 58 58 58]

[ 80 80 80 80 83 83 83 83 80 80 80 80 ....... 57 57 57 57 60 60 60 60 58 58 58 58]

[ 80 80 80 80 83 83 83 83 80 80 80 80 ....... 57 57 57 57 60 60 60 60 58 58 58 58]

[ 80 80 80 80 83 83 83 83 80 80 80 80 ....... 57 57 57 57 60 60 60 60 58 58 58 58]

[ 75 75 75 75 76 76 76 76 81 81 81 81 ....... 59 59 59 59 58 58 58 58 60 60 60 60]

[ 75 75 75 75 76 76 76 76 81 81 81 81 ....... 59 59 59 59 58 58 58 58 60 60 60 60]

[ 75 75 75 75 76 76 76 76 81 81 81 81 ....... 59 59 59 59 58 58 58 58 60 60 60 60]]

我在网上搜索了很多帖子，但还是不知道该怎么做。

所以请教教我该怎么做，我会非常感激。

Answer 1

我不太明白你的算法或者你想做什么，不过：

a=[[1,2],[3,4]]
# grow horizontally, 5 times
b=[[c for d in zip(x,x,x,x,x) for c in d] for x in a]
# grow vertically, 5 times
c= [z[:] for x in ((y[:],y[:],y[:],y[:],y[:]) for y in b) for z in x]

请注意，这个算法可以处理任何类型的数组，因为它只使用了基本的语言元素。

Answer 2

肯定要参考一下这个链接里的信息，特别是评论部分。

不过我想自己试试，所以得出了以下结果：

这可能是有史以来最糟糕的做法：

>>> import pprint
>>> a = [[85, 77, 99],
...      [11, 22, 33],
...      [44, 55, 66]]
>>> 
>>> def transform(n,matrix):
...     return [item for sublist in [[[item for sublist in [[element]*n for element in row] for item in sublist] for _ in range(n)] for row in matrix] for item in sublist]
... 
>>> pprint.pprint(transform(3,a))
[[85, 85, 85, 77, 77, 77, 99, 99, 99],
 [85, 85, 85, 77, 77, 77, 99, 99, 99],
 [85, 85, 85, 77, 77, 77, 99, 99, 99],
 [11, 11, 11, 22, 22, 22, 33, 33, 33],
 [11, 11, 11, 22, 22, 22, 33, 33, 33],
 [11, 11, 11, 22, 22, 22, 33, 33, 33],
 [44, 44, 44, 55, 55, 55, 66, 66, 66],
 [44, 44, 44, 55, 55, 55, 66, 66, 66],
 [44, 44, 44, 55, 55, 55, 66, 66, 66]]
>>> pprint.pprint(transform(4,a))
[[85, 85, 85, 85, 77, 77, 77, 77, 99, 99, 99, 99],
 [85, 85, 85, 85, 77, 77, 77, 77, 99, 99, 99, 99],
 [85, 85, 85, 85, 77, 77, 77, 77, 99, 99, 99, 99],
 [85, 85, 85, 85, 77, 77, 77, 77, 99, 99, 99, 99],
 [11, 11, 11, 11, 22, 22, 22, 22, 33, 33, 33, 33],
 [11, 11, 11, 11, 22, 22, 22, 22, 33, 33, 33, 33],
 [11, 11, 11, 11, 22, 22, 22, 22, 33, 33, 33, 33],
 [11, 11, 11, 11, 22, 22, 22, 22, 33, 33, 33, 33],
 [44, 44, 44, 44, 55, 55, 55, 55, 66, 66, 66, 66],
 [44, 44, 44, 44, 55, 55, 55, 55, 66, 66, 66, 66],
 [44, 44, 44, 44, 55, 55, 55, 55, 66, 66, 66, 66],
 [44, 44, 44, 44, 55, 55, 55, 55, 66, 66, 66, 66]]
>>> pprint.pprint(transform(5,a))
[[85, 85, 85, 85, 85, 77, 77, 77, 77, 77, 99, 99, 99, 99, 99],
 [85, 85, 85, 85, 85, 77, 77, 77, 77, 77, 99, 99, 99, 99, 99],
 [85, 85, 85, 85, 85, 77, 77, 77, 77, 77, 99, 99, 99, 99, 99],
 [85, 85, 85, 85, 85, 77, 77, 77, 77, 77, 99, 99, 99, 99, 99],
 [85, 85, 85, 85, 85, 77, 77, 77, 77, 77, 99, 99, 99, 99, 99],
 [11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33],
 [11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33],
 [11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33],
 [11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33],
 [11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33],
 [44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66],
 [44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66],
 [44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66],
 [44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66],
 [44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66]]
>>> 

Answer 3

下面这个实现方法可以递归地对一个包含数字的矩阵或列表（任何可迭代的容器类型）进行重新采样（也就是复制）。这个方法运行速度快，而且比其他方法更容易理解。它可以处理任意层级嵌套的列表，每个子列表都会被正确地深度复制。

import itertools

def resample(obj, n):
    try:
        return list(itertools.chain.from_iterable((resample(row, n) for c in xrange(n)) for row in obj))
    except TypeError:
        return obj

使用方法：

>>> l = [1, 2, 3, 4]
>>> resample(l, 4)
[1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]

>>> m = [[1, 2, 3, 4], [5, 6, 7, 8]]
>>> resample(m, 4)
[[1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4],
 [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4],
 [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4],
 [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4],
 [5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8],
 [5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8],
 [5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8],
 [5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8]]