Question

我希望从一维列表中以最pythonic的方式填充3x3矩阵。

所以从第一个转换到第二个

[1,2,3,4,5,6,7,8,9]

[ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]

到目前为止，我可以填写第一个：

l = [1,2,3,4,5,6,7,8,9]
m = [[l[y] for y in range(3)] for x in range(3)]

但这给出了

[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]

Answer 1

您可以使用list comprehension执行此类操作：

a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [a[k:k+3] for k in range(0,len(a),3)]

输出：

print(b)
>>> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

您还可以为此案例创建一个方法来处理数组的大小调整：

def reshape(a = list, r = 1):
    return [a[k:k+r] for k in range(0,len(a),r)]

a = [1, 2, 3, 4, 5, 6, 7, 8, 9]

输出：

print(reshape(a, 3))
>>> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(reshape(a,2))
>>> [[1, 2], [3, 4], [5, 6], [7, 8], [9]]

Answer 2

假设：

l = [1,2,3,4,5,6,7,8,9]

您可以一起创建N iterators然后zip。在grouper下的文档的itertools Recipes部分中描述了此技术。

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

print(list(grouper(l, 3)))

输出：

[(1, 2, 3), (4, 5, 6), (7, 8, 9)]

用M矩阵填充M和1D list python

2 个答案: