如何在python中的数据数组/映射中绘制一个圆-HowcanIdrawacircleinadataarray/mapinpython

I have an array that will be 100 * 100, I can access any point like

我有一个100 * 100的数组,我可以访问任何点,如

map[x][y]

It kinda will look like this:

有点像这样:

for i in map:
    for ii in i:
        print ii,
    print '\n',

output:

. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .

I want to make a circle in it like:

我想在其中制作一个圆圈,如:

. . . . . # . . . . .
. . . # # . # # . . .
. . # . . . . . # . .
. # . . . . . . . # .
. # . . . . . . . # .
# . . . . . . . . . #
. # . . . . . . . # .
. # . . . . . . . # .
. . # . . . . . # . .
. . . # # . # # . . .
. . . . . # . . . . .

How can i do this?

我怎样才能做到这一点?

I want to try and make a triangulation system where i will find the point that 3 circles will overlap. Is there any other way I can achieve this.

我想尝试建立一个三角测量系统,我将找到3个圆圈重叠的点。有没有其他方法可以实现这一目标。

I just want to get the distance (dots from center) and the direction.

我只想得到距离(中心点)和方向。

2 个解决方案

(x - a)**2 + (y - b)**2 = r**2

width, height = 11, 11
a, b = 5, 5
r = 5
EPSILON = 2.2

map_ = [['.' for x in range(width)] for y in range(height)]

# draw the circle
for y in range(height):
    for x in range(width):
        # see if we're close to (x-a)**2 + (y-b)**2 == r**2
        if abs((x-a)**2 + (y-b)**2 - r**2) 

. . . # # # # # . . .
. . # . . . . . # . .
. # . . . . . . . # .
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
. # . . . . . . . # .
. . # . . . . . # . .
. . . # # # # # . . .

import math
# draw the circle
for angle in range(0, 360, 5):
    x = r * math.sin(math.radians(angle)) + a
    y = r * math.cos(math.radians(angle)) + b
    map_[int(round(y))][int(round(x))] = '#'

. . . # # # # # . . .
. # # . . . . . # # .
. # . . . . . . . # .
# . . . . . . . . # #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
. # . . . . . . . # .
. # # . . . . . # # .
. . . # # # # # . . .

(x-a)^2 + (y-b)^2 - r^2 = 0