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R中圆圈上等距n点的坐标?

[英]Coordinates of equally distanced n points on a circle in R?

I want to get the coordinates of the equally distanced n points on a circle in R. 我想得到R中圆圈上等距 n点的坐标。

Mathematically the solution is : exp((2*pi * i)*(k/n)) where 0 <= k < n 数学上解决方案是 :exp((2 * pi * i)*(k / n))其中0 <= k <n

There are many SOF questions to handle this problem. 有许多SOF问题可以解决这个问题。 All the solutions are in non-R environments: 所有解决方案都在非R环境中:

Evenly distributing n points on a sphere (java, python solutions presented) 均匀分布球体上的n个点 (java,python解决方案)

Generating points on a circle (non-R solution) 在圆上生成点 (非R解)

calculate pixel coordinates for 8 equidistant points on a circle (python solution) 计算圆上8个等距点的像素坐标 (python解)

drawing points evenly distributed on a circle (non-R solution) 绘图点均匀分布在圆上 (非R解)

How to plot points around a circle in R (no equally distancing) 如何在R中围绕圆圈绘制点 (没有相同的距离)

Coordinates of every point on a circle's circumference (non-R solution) 圆周上每个点的坐标 (非R解)

Coordinates of points dividing circle into n equal halves in Pebble 在Pebble中将圆分为n等分的坐标

How to efficiently draw exactly N points on screen? 如何在屏幕上有效地绘制N个点? (python solution) (python解决方案)

Approximate position on circle for n points (non-R solution) n点圆的近似位置 (非R解)

Determining Vector points on a circle 确定圆上的矢量点

What I did for solution: 我为解决方案做了什么:

# For 4 points, 0<=k<4    
exp((2*pi*sqrt(-1))*(0/4)); exp((2*pi*sqrt(-1))*(1/4)); exp((2*pi*sqrt(-1))*(2/4)); exp((2*pi*sqrt(-1))*(3/4)) 

Complex number i is not defined in R. There is no such constant as opposite to pi (3.14). 复数i在R中没有定义。没有与pi相反的常数(3.14)。 The trick sqrt(-1) to similate i does not work; 使用sqrt(-1)来模拟我不起作用; the error: 错误:

[1] NaN 
Warning message: In sqrt(-1) : NaNs produced

we can use complex numbers to achieve this quite simply, but you need to use the correct syntax. 我们可以使用复数来实现这一点,但是您需要使用正确的语法。 in general, complex numbers can be written as ai + b (eg 3i + 2 ). 通常,复数可以写为ai + b (例如3i + 2 )。 If there is only an imaginary component, we can write just ai . 如果只有一个虚构的组件,我们可以只写一个ai So, imaginary one is simply 1i . 所以,想象一个就是1i

Npoints = 10
points = exp(pi * 1i * seq(0, 2, length.out = Npoints+1)[-1])
plot(points)

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If, for any reason, you need to translate from a complex to a Cartesian plane, you can extract the real and imaginary components using Re() and Im() . 如果由于任何原因,您需要从复数平面转换为笛卡尔平面,则可以使用Re()Im()提取实部和虚部。

points.Cartesian = data.frame(x=Re(points), y=Im(points))
f <- function(x){
  i <- sqrt(as.complex(-1))
  exp(2*pi*i*x)
}

> f(0/4)
[1] 1+0i
> f(1/4)
[1] 0+1i
> f(2/4)
[1] -1+0i
> f(3/4)
[1] 0-1i

Having said that, couldn't you find equally spaced points on a circle without resorting to complex numbers? 话虽如此,你不能在不借助复杂数字的情况下在圆上找到等间距的点吗?

eq_spacing <- function(n, r = 1){
  polypoints <- seq(0, 2*pi, length.out=n+1)
  polypoints <- polypoints[-length(polypoints)]
  circx <- r * sin(polypoints)
  circy <- r * cos(polypoints)
  data.frame(x=circx, y=circy)
}

eq_spacing(4)
               x             y
 1  0.000000e+00  1.000000e+00
 2  1.000000e+00  6.123032e-17
 3  1.224606e-16 -1.000000e+00
 4 -1.000000e+00 -1.836910e-16

plot(eq_spacing(20), asp = 1)

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Yo can try this too (and avoid complex arithmetic) to have points on the unit circle on the real plane: 你也可以尝试这个(并避免复杂的算术)在真实平面上的单位圆上有点:

n <- 50 # number of points you want on the unit circle
pts.circle <- t(sapply(1:n,function(r)c(cos(2*r*pi/n),sin(2*r*pi/n))))
plot(pts.circle, col='red', pch=19, xlab='x', ylab='y')

在此输入图像描述

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