R极坐标方程玫瑰曲线图问题 [英] R Plotly Polar Equation Rose Curve Graph Issue

问题描述

我试图用R绘制R中的极坐标方程.但是，当我绘制玫瑰曲线时，我没有在图中得到正确数量的花瓣.

I am trying to graph polar equations in R using plotly. However, when I graph a rose curve, I'm not getting the right amount of petals in the graph.

我的代码...

library(plotly)

f <- function(){
   output <- matrix(ncol=2, nrow = 361)
   for (i in 0:360){
        output[i,1] <- i
       output[i,2] <- 3 * cos(2 * (i * pi/180))
   }
   return(output)
 }
 mf <- f()
 df <- data.frame("theta" = mf[,1], "r"=mf[,2])

 p <- plot_ly(
     df,
     type = 'scatterpolar',
     mode = 'lines'
   ) %>%
   add_trace(
     r = ~r,
     theta = ~theta,
     name = 'Function',
     line = list(
        color = 'red'
    )
   ) %>%
   layout(
      title = 'Polar Graph',
      font = list(
        family = 'Arial',
        size = 12,
        color = '#000'
      ),
      showlegend = F
 )
 p

结果图...

图形应类似于...

有人可以告诉我我在做什么错吗，或者是否有更简单的方法可以在R中做到这一点?谢谢.

Can anyone tell me what I'm doing wrong or if there's an easier way to do this in R? Thanks.

推荐答案

对于常见的极坐标图，我习惯于半径为零的中心，但在您的情况下，图的中心的半径为-3.因此，使用plot_lt()时不会获得预期的结果. plot_lt()可能允许您在配置中进行更改，但我找不到它.

For common polar plot, I am used to the center having zero radius, but in your case the center of the plot has -3 as radius. Therefor you don’t get the expected result when using plot_lt(). it is possible that plot_lt() allow you to change this in the configuration, but I couldn’t find it.

可能的解决方案是移动角度和半径，以使半径始终大于零.这在下面的函数"shift_center_zero"中完成.对于每个负半径，乘以-1即可为正，对于这些行进行角度偏移，以使半径位于中心的另一侧.角度偏移是通过增加半个圆(180度)并采用完整圆的模量(360度)来限制角度，使其在单个圆内.

A possible solution is to shift the angels and radius such that the radius is always larger than zero. This is done in the function "shift_center_zero" below. For every negative radius, multiply by -1 to make positive, and for those rows make an angle shift such that the radius is on the other side of the center. The angle shift is done by adding half a round (180 degrees) and take modulus of a complete round (360 degrees) to restrict the angle such that it is within a single round.

shift_center_zero <- function(m){
      m_negative <- m[,2]<0 # get negative rows
      m[m_negative,1] <- (m[m_negative,1]+180)%%360 # angle shift
      m[m_negative,2] <- -1*m[m_negative,2] # radius shift
      return(m)
    }

其余的代码几乎相同，使用nrow = 360插入361，以删除NA，并使用新的"shift_center_zero"功能.

The rest of your code is pretty much the same, with nrow = 360 insted of 361 to remove NA's and the use of the new "shift_center_zero" function.

library(plotly)

f <- function(){
  output <- matrix(ncol=2, nrow = 360)
  for (i in 0:360){
    output[i,1] <- i
    output[i,2] <- 3 * cos(2 * (i * pi/180))
  }
  return(output)
}
mf <- f()

# make the shift
mf<-shift_center_zero(mf)

df <- data.frame("theta" = mf[,1], "r"=mf[,2])


p <- plot_ly(
  df,
  type = 'scatterpolar',
  mode = 'lines'
) %>%
  add_trace(
    r = ~r,
    theta = ~theta,
    name = 'Function',
    line = list(
      color = 'red'
    )
  ) %>%
  layout(
    title = 'Polar Graph',
    font = list(
      family = 'Arial',
      size = 12,
      color = '#000'
    ),
    showlegend = F
  )
p

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