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Guamokolatokint
Asked:2024-10-02 18:53:11 +0000 UTC2024-10-02 18:53:11 +0000 UTC

Python通过锯齿形将列表转换为矩阵

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请告诉我，如何使用锯齿形将列表转换为矩阵？相反，有代码以锯齿形方式将矩阵转换为列表：

def zigzag(a):

    m, n = a.shape
    solution = [[] for i in range(m + n - 1)]

    for i in range(m):
        for j in range(n):
            sum = i + j
            value = a[i][j]
            index = (i, j)
            if sum % 2 == 0:
                solution[sum].insert(0, (value, index))
            else:
                solution[sum].append((value, index))

    output_vector = []
    for i in solution:
        for j in i:
            output_vector.append(j[0])

    return output_vector
python

1 个回答

  1. Best Answer

    是的，我什至不明白用哪种方法来解决这个问题...从 zigzag 函数返回的列表中，您需要获取一个矩阵

    这里有一个问题：您返回了一个列表，但是您建议如何仅使用一个列表来了解构建矩阵的维度？

    假设我们有一个包含9 个元素的列表。这是一个3x3还是9x1矩阵？！

    如果zigzag()也返回原始矩阵的维度或者我们自己设置它们来创建矩阵，那么问题就完全解决了：

    import numpy as np

def zigzag(a):
    m, n = a.shape  
    solution = [[] for i in range(m + n - 1)]

    for i in range(m):
        for j in range(n):
            sum = i + j
            value = a[i][j]
            if sum % 2 == 0:
                solution[sum].insert(0, value)
            else:
                solution[sum].append(value)

    output_vector = []
    for i in solution:
        output_vector.extend(i)

    return output_vector, m, n  
    
a = np.array([[1, 2, 3],
              [4, 5, 6],
              [7, 8, 9]])

result_lst, m, n = zigzag(a)
print(result_lst) 
print(m, n) # Размеры матрицы

def list_zigzag_to_matrix(lst, m, n):
    a = np.zeros((m, n), dtype=int)
    solution = [[] for i in range(m + n - 1)]
    
    for i in range(m):
        for j in range(n):
            sum = i + j
            if sum % 2 == 0:
                solution[sum].insert(0, (i, j))
            else:
                solution[sum].append((i, j))
                
    index = 0  
    for d in solution:
        for i, j in d:
            a[i][j] = lst[index]
            index += 1
    
    return a

result_matrix = list_zigzag_to_matrix(result_lst, m, n)
print(result_matrix)

    结论：

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

[Program finished]
    • 1

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