我正在尝试，给定表示四边形的两个维度(宽度和高度(，将其划分为N部分，其中每个部分在比例上尽可能相似。
例如，想象一张纸。它由4个点A, B, C, D组成现在考虑一下，这张纸的尺寸为800 x 800，点为：
A: {0, 0}
B: {0, 800}
C: {800, 800}
D: {800, 0}
绘制将为您提供4个点，或使用折线图绘制3条线。添加一个额外的点E: {0, 0}以关闭单元格。
我已经用程序实现了这一点，对于N个细胞。不幸的是，由于某种原因，当我设置N=4时，我得到了8个单元格。。。我被炸了，我无法解决，所以我正在寻找三样东西：
- A(如何改进此代码以使其更具可读性
- B( 如何改进此代码以使其更具Python风格
- C( 你猜对了。我该如何解决
N=4问题
完整的代码：
import matplotlib.pyplot as plt
class QuadPartitioner:
@staticmethod
def get_factors(number):
'''
Takes a number and returns a list of factors
:param number: The number for which to find the factors
:return: a list of factors for the given number
'''
facts = []
for i in range(1, number + 1):
if number % i == 0:
facts.append(i)
return facts
@staticmethod
def get_partitions(N, quad_width, quad_height):
'''
Given a width and height, partition the area into N parts
:param N: The number of partitions to generate
:param quad_width: The width of the quadrilateral
:param quad_height: The height of the quadrilateral
:return: a list of a list of cells where each cell is defined as a list of 5 verticies
'''
# We reverse only because my brain feels more comfortable looking at a grid in this way
factors = list(reversed(QuadPartitioner.get_factors(N)))
# We need to find the middle of the factors so that we get cells
# with as close to equal width and heights as possible
split = int(len(factors)/2)
factors = factors[split-1:split+1]
# The width and height of an individual cell
cell_width = quad_width / factors[0]
cell_height = quad_height / factors[1]
number_of_cells_in_a_row = factors[0]
rows = factors[1]
row_of_cells = []
# We build just a single row of cells
# then for each additional row, we just duplicate this row and offset the cells
for n in range(0, number_of_cells_in_a_row):
cell_points = []
for i in range(0, 5):
cell_y = 0
cell_x = n * cell_width
if i == 2 or i == 3:
cell_x = n * cell_width + cell_width
if i == 1 or i == 2:
cell_y = cell_height
cell_points.append((cell_x, cell_y))
row_of_cells.append(cell_points)
rows_of_cells = [row_of_cells]
# With that 1 row of cells constructed, we can simply duplicate it and offset it
# by the height of a cell multiplied by the row number
for index in range(1, rows):
new_row_of_cells = [[ (point[0],point[1]+cell_height*index) for point in square] for square in row_of_cells]
rows_of_cells.append(new_row_of_cells)
return rows_of_cells
if __name__ == "__main__":
QP = QuadPartitioner()
partitions = QP.get_partitions(4, 800,800)
for row_of_cells in partitions:
for cell in row_of_cells:
x, y = zip(*cell)
plt.plot(x, y, marker='o')
plt.show()
我通过更改解决了这个问题
split = int(len(factors)/2)
factors = factors[split-1:split+1]
至：
factor_count = len(factors)
if factor_count % 2 == 0:
split = int(factor_count/2)
factors = factors[split-1:split+1]
else:
factors = []
split = ceil(factor_count/2)
factors.append(split)
factors.append(split)