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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.

Mathematically the solution is: exp((2*pi * i)*(k/n)) where 0 <= k < n

There are many SOF questions to handle this problem. All the solutions are in non-R environments:

Evenly distributing n points on a sphere (java, python solutions presented)

Generating points on a circle (non-R solution)

calculate pixel coordinates for 8 equidistant points on a circle (python solution)

drawing points evenly distributed on a circle (non-R solution)

How to plot points around a circle in R (no equally distancing)

Coordinates of every point on a circle's circumference (non-R solution)

Coordinates of points dividing circle into n equal halves in Pebble

How to efficiently draw exactly N points on screen? (python solution)

Approximate position on circle for n points (non-R solution)

Determining Vector points on a circle

What I did for solution:

# 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). The trick sqrt(-1) to similate i does not work; the error:

[1] NaN 
Warning message: In sqrt(-1) : NaNs produced
like image 954
Erdogan CEVHER Avatar asked Dec 25 '22 00:12

Erdogan CEVHER


2 Answers

we can use complex numbers to achieve this quite simply, but you need to use the correct syntax. in general, complex numbers can be written as ai + b (e.g. 3i + 2). If there is only an imaginary component, we can write just ai. So, imaginary one is simply 1i.

Npoints = 20
points = exp(2i * pi * (1:Npoints)/Npoints)
plot(points)

enter image description here

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().

points.Cartesian = data.frame(x=Re(points), y=Im(points))
like image 165
dww Avatar answered Mar 09 '23 00:03

dww


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?

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)

enter image description here

like image 43
sebastian-c Avatar answered Mar 09 '23 00:03

sebastian-c