python sorting array of XYZ coordinates in table matrix

Question

I need to sort and array of XYZ coordinates in a table style matrix to export as an .csv file.

With the help from user Michael0x2a I managed to do it more or less. My problem now is if I have repeated X and Y it will return 0 for Z.

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)

if I run the example below it will return

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

so column headers are the y values, row headers are the x values.

for the 1st set of [1,1,20] it returns 1,1,0 because of the second set [1,1,11] has the same x and y values.

the end result should be:

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

I think it has something to do with this function:

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

can anyone help me out with this?

thanks so much

Francisco

What's the expected result? If you have two different values for the same coordinates then you can't fit them both into one place. Or... you can as a tuple - if that's what you want? — BartoszKP, Aug 29 '13 at 09:41
Please edit your question and include the expected output. That way you can format the matrix properly. — thefourtheye, Aug 29 '13 at 09:43
I edit the code, I was trying stuff out before I posted the question and I copied a code which wasn't working. now if you copy the code it should work, but still giving the initial error of putting a 0 instead of a 20 — user2725701, Aug 29 '13 at 09:54
May you use numpy, or does it have to be pure Python? General remark: As you need to "track" the indices of your coordinates, the solution probably includes the "key" paramater to the sorted function. — ojdo, Aug 29 '13 at 10:45
@ojdo it really needs to be in pure python because I'm using it in rhino python. it doesn't support numpy unfortunatly — user2725701, Aug 29 '13 at 11:14
I tried to give an answer, but you will have to clarify *exactly* how duplicate values in either dimension are to be handled before I can make it behave correctly. Why does the duplicate `[1,1]` spawn a new column, not a new row? What should happen if there are three identical coordinates? — ojdo, Aug 29 '13 at 13:01

score 0 · Accepted Answer · edited May 23 '17 at 12:21

Here is a suggestion. It uses sorted function on a range with the key parameter to get the indices needed to sort x and y later on (more details in the question 'How to get indices of a sorted array in Python'). This automatically deals with duplicate values.

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]]

However, the result of this script is not (yet) as desired. A looks like this in the end:

[[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]]

Note that the duplicate value 1 not only created a new column, but also a new row.

Assumption: only rows with identical index shall be merged. This can be accomplished using itertools groupby function and zip+sum to "merge" the rows by simply summing them column by column. The first column (the row index) has to be sliced:

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:])

The resulting list of lists AA looks like this:

[[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 sorting array of XYZ coordinates in table matrix

1 Answers1