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我正在尝试设计一个圆形的二维码。我必须用python对其进行编码，但我对此没有太多经验。经过大量的努力，我创建了形状。现在我需要在这些形状之间填充颜色，但我做不到。我的输出图形现在看起来像这样：

当前输出截图

我需要用颜色填充那些空的形状。下面给出了我为创建此图形而实现的代码。我不确定我是否使用了正确的库。我看到了很多关于 matplotlib.pyplot 的例子，但我认为它对于创建图表更有用。因此，任何贡献都会受到赞赏。我真的被困住了。

from graphics import *
import math

def rect(r, theta):
    """theta in degrees

    returns tuple; (float, float); (x,y)
    """
    x = r * math.cos(math.radians(theta))
    y = r * math.sin(math.radians(theta))
    return x, y


def polar(x, y):
    """returns r, theta(degrees)
    """
    r = (x ** 2 + y ** 2) ** .5
    if y == 0:
        theta = 180 if x < 0 else 0
    elif x == 0:
        theta = 90 if y > 0 else 270
    else:
        theta = math.degrees(math.atan(float(y) / x))
    return r, theta


def main():
    win = GraphWin('Canvas', 640, 480)  # give title and dimensions
    win.setCoords(-320, -240, 320, 240)

    seventh = Circle(Point(0, 0), 90)  # set center and radius
    # seventh.setFill("yellow")
    seventh.draw(win)

    sixth = Circle(Point(0, 0), 80)  # set center and radius
    # sixth.setFill("yellow")
    sixth.draw(win)

    fifth = Circle(Point(0, 0), 70)  # set center and radius
    # fifth.setFill("yellow")
    fifth.draw(win)

    fourth = Circle(Point(0, 0), 60)  # set center and radius
    # fourth.setFill("yellow")
    fourth.draw(win)

    third = Circle(Point(0, 0), 50)  # set center and radius
    # third.setFill("yellow")
    third.draw(win)

    second = Circle(Point(0, 0), 40)  # set center and radius
    # second.setFill("yellow")
    second.draw(win)

    first = Circle(Point(0, 0), 30)  # set center and radius
    # first.setFill("yellow")
    first.draw(win)

    line = Line(Point(0, 0), Point(rect(90, 0)[0], rect(90, 0)[1]))
    line.draw(win)

    line2 = Line(Point(0, 0), Point(rect(90, 30)[0], rect(90, 30)[1]))
    line2.draw(win)

    line3 = Line(Point(0, 0), Point(rect(90, 60)[0], rect(90, 60)[1]))
    line3.draw(win)

    line4 = Line(Point(0, 0), Point(rect(90, 90)[0], rect(90, 90)[1]))
    line4.draw(win)

    line5 = Line(Point(0, 0), Point(rect(90, 120)[0], rect(90, 120)[1]))
    line5.draw(win)

    line6 = Line(Point(0, 0), Point(rect(90, 150)[0], rect(90, 150)[1]))
    line6.draw(win)

    line7 = Line(Point(0, 0), Point(rect(90, 180)[0], rect(90, 180)[1]))
    line7.draw(win)

    line8 = Line(Point(0, 0), Point(rect(90, 210)[0], rect(90, 210)[1]))
    line8.draw(win)

    line9 = Line(Point(0, 0), Point(rect(90, 240)[0], rect(90, 240)[1]))
    line9.draw(win)

    line10 = Line(Point(0, 0), Point(rect(90, 270)[0], rect(90, 270)[1]))
    line10.draw(win)

    line11 = Line(Point(0, 0), Point(rect(90, 300)[0], rect(90, 300)[1]))
    line11.draw(win)

    line12 = Line(Point(0, 0), Point(rect(90, 330)[0], rect(90, 330)[1]))
    line12.draw(win)

    line13 = Line(Point(0, 0), Point(rect(90, 360)[0], rect(90, 360)[1]))
    line13.draw(win)

    line14 = Line(Point(0, 0), Point(rect(90, 210)[0], rect(90, 210)[1]))
    line14.draw(win)

    first.setFill('black')


    message = Text(Point(0, 200), 'Click anywhere to quit.')
    message.draw(win)
    win.getMouse()

    win.close()

main()

如果我能做到，我想画一张看起来像这样的图片（在 Photoshop 中绘制）：

使用 Photoshop 创建的所需图像的屏幕截图

经过一些有用的评论后，我决定寻找另一个合适的 python 模块。但我找不到合适的。如果有人了解我的问题可以建议适当的 python 模块？

4

1 回答 1

1

我相信同样基于 tkinter 构建的 Python turtle 有足够的能力用更直观的逻辑来模拟你的循环二维码：

from math import pi, sin, cos
from turtle import Turtle, Screen

TWO_PI = pi * 2

SEGMENTS = 10  # pie slices
SECTORS = 8  # concentric circles
MAXIMUM_RADIUS = 100
MINIMUM_RADIUS = 10
DELTA_RADIUS = (MAXIMUM_RADIUS - MINIMUM_RADIUS) // SECTORS
SEGMENT_EXTENT = TWO_PI / SEGMENTS
DEFAULT_LINE_WIDTH = 1

DATA = [  # data to match OP's example chart
    (2, 0), (2, 3), (2, 5), (2, 6), (2, 9),
    (3, 0), (3, 1), (3, 3), (3, 4), (3, 6),
    (4, 1), (4, 2), (4, 7), (4, 8),
    (5, 0), (5, 1), (5, 4), (5, 5), (5, 6), (5, 7),
    (6, 3), (6, 5), (6, 8), (6, 9),
    (7, 0), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 8),
    ]

def centered_circle(turtle, radius, extent=TWO_PI):
    """ draw a circle centered about the current turtle location """

    position = turtle.position()
    heading = turtle.heading()

    turtle.penup()
    turtle.setx(turtle.xcor() + cos(heading) * radius)
    turtle.sety(turtle.ycor() - sin(heading) * radius)
    turtle.pendown()

    turtle.circle(radius, extent, steps=60)

    turtle.penup()
    turtle.setposition(position)  # return turtle to starting position
    turtle.pendown()

def draw_blank_chart(turtle):
    """ draw the empty chart """

    turtle.dot(MINIMUM_RADIUS * 2)  # center of chart

    for radius in range(MINIMUM_RADIUS + DELTA_RADIUS, MAXIMUM_RADIUS + 1, DELTA_RADIUS):
        centered_circle(turtle, radius)  # divide chart into sectors

    turtle.width(2 * DELTA_RADIUS / 3)  # outer edge of chart
    centered_circle(turtle, MAXIMUM_RADIUS + DELTA_RADIUS)
    turtle.width(DEFAULT_LINE_WIDTH)

    for _ in range(SEGMENTS):  # divide chart into segments
        turtle.forward(MAXIMUM_RADIUS + DELTA_RADIUS)
        turtle.backward(MAXIMUM_RADIUS + DELTA_RADIUS)
        turtle.left(SEGMENT_EXTENT)

def set_bit(turtle, sector, segment):
    """ turn on (blacken) one sector:segment of chart """

    turtle.setheading(SEGMENT_EXTENT * segment - pi/2)
    turtle.width(DELTA_RADIUS)
    centered_circle(turtle, (MINIMUM_RADIUS + DELTA_RADIUS * sector + DELTA_RADIUS/2), -SEGMENT_EXTENT)

screen = Screen()
screen.mode('logo')  # force zero degrees to be straight up

yertle = Turtle(visible=False)
yertle.width(DEFAULT_LINE_WIDTH)
yertle.speed('fastest')
yertle.radians()  # to make working with sin and cos easier

draw_blank_chart(yertle)

for sector, segment in DATA:
    set_bit(yertle, sector, segment)

screen.exitonclick()

在此处输入图像描述

于 2018-03-28T08:10:43.377 回答