任何人都知道如何从这个数组中检索部分：
a=[[1, 3, 2, 5, 7, 9, 4, 6, 8],
[4, 9, 8, 2, 6, 1, 3, 7, 5],
[7, 5, 6, 3, 8, 4, 2, 1, 9],
[6, 4, 3, 1, 5, 8, 7, 9, 2],
[5, 2, 1, 7, 9, 3, 8, 4, 6],
[9, 8, 7, 4, 2, 6, 5, 3, 1],
[2, 1, 4, 9, 3, 5, 6, 8, 7],
[3, 6, 5, 8, 1, 7, 9, 2, 4],
[8, 7, 9, 6, 4, 2, 1, 5, 3]]
我的意思是我想要检索第3X3节，例如，左上角是：
[[1,3,2],
[4,9,8],
[7,5,6]]
需要检索的部分是：
[[0:3,0:3]],[[3:6,0:3]],[[6:9,0:3]]
[[0:3,3:6]],[[3:6,3:6]],[[6:9,3:6]]
[[0:3,6:9]],[[3:6,6:9]],[[6:9,6:9]]
如何检索所有这些部分？ 是否有必要使用numpy？
答案 0 :(得分：3)
这是使用reshaping和置换维度的矢量化方法 -
a.reshape(3,3,3,3).transpose(2,0,1,3).reshape(9,3,3)
示例运行 -
In [197]: a
Out[197]:
array([[1, 3, 2, 5, 7, 9, 4, 6, 8],
[4, 9, 8, 2, 6, 1, 3, 7, 5],
[7, 5, 6, 3, 8, 4, 2, 1, 9],
[6, 4, 3, 1, 5, 8, 7, 9, 2],
[5, 2, 1, 7, 9, 3, 8, 4, 6],
[9, 8, 7, 4, 2, 6, 5, 3, 1],
[2, 1, 4, 9, 3, 5, 6, 8, 7],
[3, 6, 5, 8, 1, 7, 9, 2, 4],
[8, 7, 9, 6, 4, 2, 1, 5, 3]])
In [198]: a.reshape(3,3,3,3).transpose(2,0,1,3).reshape(9,3,3)
Out[198]:
array([[[1, 3, 2],
[4, 9, 8],
[7, 5, 6]],
[[6, 4, 3],
[5, 2, 1],
[9, 8, 7]],
[[2, 1, 4],
[3, 6, 5],
[8, 7, 9]], ....
如果您需要展平每个这样的部分/窗口，只需调整最后一次整形，就像这样 -
In [199]: a.reshape(3,3,3,3).transpose(2,0,1,3).reshape(9,9)
Out[199]:
array([[1, 3, 2, 4, 9, 8, 7, 5, 6],
[6, 4, 3, 5, 2, 1, 9, 8, 7],
[2, 1, 4, 3, 6, 5, 8, 7, 9],
[5, 7, 9, 2, 6, 1, 3, 8, 4],
[1, 5, 8, 7, 9, 3, 4, 2, 6],
[9, 3, 5, 8, 1, 7, 6, 4, 2],
[4, 6, 8, 3, 7, 5, 2, 1, 9],
[7, 9, 2, 8, 4, 6, 5, 3, 1],
[6, 8, 7, 9, 2, 4, 1, 5, 3]])
答案 1 :(得分：1)
以下是使用列表理解
的答案>>> [x[:3] for x in a[:3]]
[[1, 3, 2], [4, 9, 8], [7, 5, 6]]
左栏：
[x[0:3] for x in a[0:3]]
[x[0:3] for x in a[3:6]]
[x[0:3] for x in a[6:9]]
中段：
[x[3:6] for x in a[0:3]]
[x[3:6] for x in a[3:6]]
[x[3:6] for x in a[6:9]]
右图：
[x[6:9] for x in a[0:3]]
[x[6:9] for x in a[3:6]]
[x[6:9] for x in a[6:9]]
a[i:j]从索引i到j-1
对于所述行<{p>，x[i,j]将索引i的元素取为j-1
制作“扁平化”＆＃39;列表，使用pwnsauce评论的输入：
左栏：
[x for b in [x[0:3] for x in a[0:3]] for x in b]
[x for b in [x[0:3] for x in a[3:6]] for x in b]
[x for b in [x[0:3] for x in a[6:9]] for x in b]
中段：
[x for b in [x[3:6] for x in a[0:3]] for x in b]
[x for b in [x[3:6] for x in a[3:6]] for x in b]
[x for b in [x[3:6] for x in a[6:9]] for x in b]
右图：
[x for b in [x[6:9] for x in a[0:3]] for x in b]
[x for b in [x[6:9] for x in a[3:6]] for x in b]
[x for b in [x[6:9] for x in a[6:9]] for x in b]
答案 2 :(得分：0)
你可以使用类似的：
for i in range(0, len(a), 3):
left_section = []
middle_section = []
right_section = []
left_section.append(a[i][:3])
left_section.append(a[i+1][:3])
left_section.append(a[i+2][:3])
middle_section.append(a[i][3:6])
middle_section.append(a[i+1][3:6])
middle_section.append(a[i+2][3:6])
right_section.append(a[i][3:6])
right_section.append(a[i+1][3:6])
right_section.append(a[i+2][3:6])
print(left_section)
print(middle_section)
print(right_section)
OR
for i in range(0, len(a), 3):
left_section = []
middle_section = []
right_section = []
left_section.extend(a[i][:3])
left_section.extend(a[i+1][:3])
left_section.extend(a[i+2][:3])
middle_section.extend(a[i][3:6])
middle_section.extend(a[i+1][3:6])
middle_section.extend(a[i+2][3:6])
right_section.extend(a[i][3:6])
right_section.extend(a[i+1][3:6])
right_section.extend(a[i+2][3:6])
print(left_section)
print(middle_section)
print(right_section)
答案 3 :(得分：0)
纯粹的python方式，不使用numpy：
由于你需要扁平化的列表，我们使用这个功能（如前所述）：
def flatten_list(li):
return [el for sub_li in li for el in sub_li]
我们知道我们可以像这样检索第一部分：
flatten_list([a[row][0:3] for row in range(3)]) # retrieves the first section
>>> [1, 3, 2, 4, 9, 8, 7, 5, 6]
和所有left_sections如：
[flatten_list([a[row][0:3] for row in range(y*3, y*3+3)]) for y in range(3)]
>>> [[1, 3, 2, 4, 9, 8, 7, 5, 6], [6, 4, 3, 5, 2, 1, 9, 8, 7], [2, 1, 4, 3, 6, 5, 8, 7, 9]]
所有在一起：
[[flatten_list([a[row][x*3:x*3+3] for row in range(y*3, y*3+3)]) for y in range(3)] for x in range(3)]
>>> [[[1, 3, 2, 4, 9, 8, 7, 5, 6],
[6, 4, 3, 5, 2, 1, 9, 8, 7],
[2, 1, 4, 3, 6, 5, 8, 7, 9]],
[[5, 7, 9, 2, 6, 1, 3, 8, 4],
[1, 5, 8, 7, 9, 3, 4, 2, 6],
[9, 3, 5, 8, 1, 7, 6, 4, 2]],
[[4, 6, 8, 3, 7, 5, 2, 1, 9],
[7, 9, 2, 8, 4, 6, 5, 3, 1],
[6, 8, 7, 9, 2, 4, 1, 5, 3]]]