尺寸为x的1的二维阵列菱形[重复]

Question

嘿，我正在尝试制作一个由 1 和 0 组成的二维数组，其中 1 形成菱形。钻石的大小应为 x：

尺寸 x = 3 的菱形看起来像这样：

[[0 0 1 0 0]
 [0 1 0 1 0]
 [1 0 0 0 1]
 [0 1 0 1 0]
 [0 0 1 0 0]]

尺寸 x = 1 的钻石形状如下所示：

[[1]]

有人知道如何实现吗？我认为numpy.eye 和concatenate 可能有用。但是，我找不到解决方案

Answer 1

我将利用numpy.diagflat 与numpy.flip 结合以下方式：

import numpy as np
arr = np.diagflat([1,1,1],2)  # now we have 1s in upper-right part
arr = np.maximum(arr,np.flip(arr,1))  # now we have 1s in upper part
arr = np.maximum(arr,np.flip(arr,0))  # now we have 1s everywhere
print(arr)

输出：

[[0 0 1 0 0]
 [0 1 0 1 0]
 [1 0 0 0 1]
 [0 1 0 1 0]
 [0 0 1 0 0]]

Answer 2

这里是np.eye的解决方案

import numpy as np

def diamond(n):
    a, b = np.eye(n, dtype=int), np.eye(n, dtype=int)[:,::-1]

    c, d = np.hstack((b,a[:,1:])), np.hstack((a,b[:,1:]))

    return np.vstack((c, d[1:,:]))

输出：

>>> print(diamond(3))

[[0 0 1 0 0]
 [0 1 0 1 0]
 [1 0 0 0 1]
 [0 1 0 1 0]
 [0 0 1 0 0]]

Answer 3

我的条目使用np.pad(... mode = 'reflect')：

def diamond(n):
    return np.pad(np.eye(n), ((n-1, 0), (0, n-1)), mode = 'reflect')
    

diamond(3)
Out: 
array([[0., 0., 1., 0., 0.],
       [0., 1., 0., 1., 0.],
       [1., 0., 0., 0., 1.],
       [0., 1., 0., 1., 0.],
       [0., 0., 1., 0., 0.]])

diamond(1)
Out: array([[1.]])

diamond(5)
Out: 
array([[0., 0., 0., 0., 1., 0., 0., 0., 0.],
       [0., 0., 0., 1., 0., 1., 0., 0., 0.],
       [0., 0., 1., 0., 0., 0., 1., 0., 0.],
       [0., 1., 0., 0., 0., 0., 0., 1., 0.],
       [1., 0., 0., 0., 0., 0., 0., 0., 1.],
       [0., 1., 0., 0., 0., 0., 0., 1., 0.],
       [0., 0., 1., 0., 0., 0., 1., 0., 0.],
       [0., 0., 0., 1., 0., 1., 0., 0., 0.],
       [0., 0., 0., 0., 1., 0., 0., 0., 0.]])

在我的时间里，这比@Daweo 的答案快 40%，而且代码也更简单

def diamond_Daweo(n):
    arr = np.diagflat(np.ones(n), n-1)  # now we have 1s in upper-right part
    arr = np.maximum(arr,np.flip(arr,1))  # now we have 1s in upper part
    return np.maximum(arr,np.flip(arr,0))  # now we have 1s everywhere

%timeit diamond_Daweo(100)
120 µs ± 1.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%timeit diamond(100)
75.8 µs ± 2.86 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 4

以下解决方案使用numpy.eye 和numpy.flip 来定义正确大小的数组，其中填充了代表绘制菱形所需线条的数组（一个对角线左上（lu）和一个对角线左下（ ld)。

numpy.zeros 然后用于创建一个正确尺寸的数组，以存储每条边上都有size 的钻石（d）。每个维度都是size*2-1，以允许沿每个轴的两条线的全宽，-1 占顶点/角。

然后，numpy.where 用于有效地将上面生成的两种类型的线映射到更正 d 上的位置以“绘制”菱形。这是通过首先对d 进行切片来实现的，以便在使用lu/ld 生成应用于相关选择以填充这些选择之前，每行都从正确的索引开始。

import numpy as np

def diamond(size):
    lu = np.eye(size) # Define left-upward edge
    ld = np.flip(lu, axis=0) # Define left-downward edge
    d = np.zeros((lu.shape[0]*2-1, lu.shape[1]*2-1)) # Create empty array of correct size filled with zeros
    mid = size-1 # Store mid-point index of each axis
    d[np.where(ld==1)] = 1 # Write upper left-centre edge
    d[:,mid:][np.where(lu==1)] = 1 # Write upper centre-right edge
    d[mid:,mid:][np.where(ld==1)] = 1 # Write lower right-centre edge
    d[mid:,:][np.where(lu==1)] = 1 # Write lower left-centre edge
    return d.astype(int) # Return diamond shaped array with values int instead of float

print(diamond(1),'\n')
print(diamond(2),'\n')
print(diamond(3),'\n')
print(diamond(4),'\n')

输出

[[1]]

[[0 0 1 0 0]
 [0 1 0 1 0]
 [1 0 0 0 1]
 [0 1 0 1 0]
 [0 0 1 0 0]]

[[0 0 1 0 0]
 [0 1 0 1 0]
 [1 0 0 0 1]
 [0 1 0 1 0]
 [0 0 1 0 0]]

[[0 0 0 1 0 0 0]
 [0 0 1 0 1 0 0]
 [0 1 0 0 0 1 0]
 [1 0 0 0 0 0 1]
 [0 1 0 0 0 1 0]
 [0 0 1 0 1 0 0]
 [0 0 0 1 0 0 0]]

Answer 5

如果你想使用 numpy.eye，这行得通。虽然有点复杂。

dim = 8
A = np.maximum(np.maximum(np.maximum(np.eye(dim, k=-(dim//2)), np.eye(dim, k=(dim//2))), np.fliplr(np.eye(dim, k=(dim//2)))), np.fliplr(np.eye(dim, k=-(dim//2))))

Answer 6

生成任何所需大小的菱形矩阵的通用代码。 使用单位矩阵和切片技术。

import numpy as np

x = 5  # assuming, this will always be odd

result_sz = 2*x-1
mid_pt = int(np.floor(result_sz/2.0))

result = np.zeros((result_sz, result_sz), dtype=np.int)
tile = np.eye(x)

# we have four regular graph quadrants
# first quadrant
result[0:x, mid_pt:] = tile    # direct identity
# second quadrant
result[mid_pt:, mid_pt:] = np.flip(tile, axis=1)     # identity flipped about y-axis
# third quadrant
result[mid_pt:, 0:x] = tile   # direct identity
# fourth quadrant
result[0:x, 0:x] = np.flip(tile, axis=0)     # identity flipped about x-axis

# The diamond matrix
print(result)

Answer 7

def diamond(m):
    n = 2 * m - 1
    res = [[0] * n for _ in range(n)]
    # make first m lines
    for x in range(m):
        res[x][m - x - 1] = 1
        res[x][n - (m - x)] = 1
    # copy other lines as they as symmetric with above line by line m - 1
    for x in range(m, n):
        res[x] = res[2 * (m - 1) - x][:]
    return res