Question

我需要在表格样式矩阵中对XYZ坐标进行排序和排列，以导出为.csv文件。

在用户Michael0x2a的帮助下，我设法或多或少地做到了。我现在的问题是，如果我重复了X和Y，它将为Z返回0。

def find_x_and_y(array):
    """Step 1: Get unique x and y coordinates and the width and height of the matrix"""

    x = sorted(list(set([i[0] for i in array])))
    y = sorted(list([i[1] for i in array]))


    height = len(x) + 1
    width = len(y) + 1

    return x, y, width, height

def construct_initial_matrix(array):
    """Step 2: Make the initial matrix (filled with zeros)"""
    x, y, width, height = find_x_and_y(array)

    matrix = []
    for i in range(height):
        matrix.append([0] * width)

    return matrix

def add_edging(array, matrix):
    """Step 3: Add the x and y coordinates to the edges"""
    x, y, width, height = find_x_and_y(array)

    for coord, position in zip(x, range(1, height)):
        matrix[position][0] = coord

    for coord, position in zip(y, range(1, width)):
        matrix[0][position] = coord

    return matrix

def add_z_coordinates(array, matrix):
    """Step 4: Map the coordinates in the array to the position in the matrix"""
    x, y, width, height = find_x_and_y(array)

    x_to_pos = dict(zip(x, range(1, height)))
    y_to_pos = dict(zip(y, range(1, width)))

    for x, y, z in array:
        matrix[x_to_pos[x]][y_to_pos[y]] = z
    return matrix

def make_csv(matrix):
    """Step 5: Printing"""
    return '\n'.join(', '.join(str(i) for i in row) for row in matrix)


def main(array):
    matrix = construct_initial_matrix(array)
    matrix = add_edging(array, matrix)
    matrix = add_z_coordinates(array, matrix)

    print make_csv(matrix)

如果我运行下面的示例，它将返回

example = [[1, 1, 20], [1, 1, 11], [2, 3, 12.1], [2, 5, 13], [5,4,10], [3,6,15]]
main(example)

0, 1, 1, 3, 4, 5, 6
1, 0, 11, 0, 0, 0, 0
2, 0, 0, 12.1, 0, 13, 0
3, 0, 0, 0, 0, 0, 15
5, 0, 0, 0, 10, 0, 0

因此列标题是y值，行标题是x值。

对于第一组[1,1,20]它返回1,1,0因为第二组[1,1,11]具有相同的x和y值。

最终结果应为：

0, 1, 1, 3, 4, 5, 6
1, 20, 11, 0, 0, 0, 0
2, 0, 0, 12.1, 0, 13, 0
3, 0, 0, 0, 0, 0, 15
5, 0, 0, 0, 10, 0, 0

我认为这与此功能有关：

    x_to_pos = dict(zip(x, range(1, height)))
    y_to_pos = dict(zip(y, range(1, width)))

任何人都可以帮我解决这个问题吗？

非常感谢

旧金山

Answer 1

这是一个建议。它使用sorted上的range函数和key参数来获取稍后对x和y进行排序所需的索引（问题中的更多详细信息{ {3}}）。这会自动处理重复值。

example = [[1, 1, 20], [1, 1, 11], [2, 3, 12.1], [2, 5, 13], [5,4,10], [3,6,15]]
x = [el[0] for el in example]
y = [el[1] for el in example]
z = [el[2] for el in example]

# indices for x,y to get them in sorted order later
# duplicates in both dimensions are preserved
x_idx = sorted(range(len(x)), key=lambda k:x[k])
y_idx = sorted(range(len(y)), key=lambda k:y[k])

# initialize A with 0
A = [[0 for _ in range(len(y)+1)] for _ in range(len(x)+1)]

# and fill it with values
for k, val in enumerate(z):
    A[x_idx[k]+1][y_idx[k]+1] = val
    A[k+1][0] = x[x_idx[k]]
    A[0][k+1] = y[y_idx[k]]

但是，此脚本的结果不是（还）。 A最终看起来像这样：

[[0, 1, 1, 3, 4, 5, 6],
 [1, 20, 0, 0, 0, 0, 0],
 [1, 0, 11, 0, 0, 0, 0],
 [2, 0, 0, 12.1, 0, 0, 0],
 [2, 0, 0, 0, 0, 13, 0],
 [3, 0, 0, 0, 0, 0, 15],
 [5, 0, 0, 0, 10, 0, 0]]

请注意，重复值1不仅创建了新列，还创建了新行。

假设：仅合并具有相同索引的行。这可以使用itertools groupby函数和zip + sum来完成，通过简单地将它们逐列相加来“合并”行。第一列（行索引）必须切片：

AA = []
for row_index, rows_to_be_merged in itertools.groupby(A, lambda x: x[0]):
    AA.append([row_index] + 
              [sum(rows) for rows in zip(*rows_to_be_merged)][1:])

结果列表AA如下所示：

[[0, 1, 1, 3, 4, 5, 6],
 [1, 20, 11, 0, 0, 0, 0],
 [2, 0, 0, 12.1, 0, 13, 0],
 [3, 0, 0, 0, 0, 0, 15],
 [5, 0, 0, 0, 10, 0, 0]]

python排序表矩阵中的XYZ坐标数组

1 个答案: