计算运动矢量之间的角度

Question

我有一个如下所示的数据框：

    structure(list(K = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), T = c(1L, 2L, 3L, 4L, 5L, 6L, 
7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 
20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 
33L, 34L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 
13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 
26L, 27L, 28L, 29L, 30L, 31L, 32L), X = c(26.892, 23.904, 23.904, 
23.904, 23.904, 23.904, 23.904, 23.904, 23.904, 20.916, 20.916, 
20.916, 20.916, 20.916, 20.916, 20.916, 20.916, 20.916, 20.916, 
20.916, 29.88, 20.916, 14.94, 8.964, 8.964, 5.976, 5.976, 5.976, 
5.976, 5.976, 5.976, 5.976, 5.976, 5.976, 857.56, 860.54, 857.56, 
857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 
857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 
857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 857.56, 
857.56, 857.56, 857.56, 857.56, 857.56), Y = c(167.33, 167.33, 
164.34, 164.34, 164.34, 164.34, 164.34, 164.34, 164.34, 143.42, 
143.42, 143.42, 143.42, 143.42, 143.42, 143.42, 143.42, 143.42, 
143.42, 143.42, 176.29, 182.27, 185.26, 188.24, 188.24, 188.24, 
188.24, 188.24, 188.24, 188.24, 188.24, 188.24, 188.24, 188.24, 
256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 
256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 
256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 
256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97, 256.97
), V = c(2.1128, 1.494, 0, 0, 0, 0, 0, 10.564, 10.564, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 17.034, 19.422, 8.7114, 6.6814, 3.3407, 
1.494, 1.494, 0, 0, 0, 0, 0, 0, 0, 0, 20.1, 0, 1.494, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 1.494, 1.494), P = c(-135, -90, 0, 0, 0, 0, 0, -98.13, 
-98.13, 0, 0, 0, 0, 0, 0, 0, 0, 0, 74.745, 90, 149.04, 153.43, 
153.43, 180, 180, 0, 0, 0, 0, 0, 0, 0, 0, 41.987, 0, 180, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 90, 90)), .Names = c("K", "T", "X", "Y", "V", 
"P"), row.names = c(NA, 66L), class = "data.frame")

由于我有X，Y位置，要制作两个向量，我应该考虑三个X，Y位置能够计算一个角度。我知道：

theta <- acos( sum(a*b) / ( sqrt(sum(a * a)) * sqrt(sum(b * b)) ) )

来自另一个stackoverflow答案（Angle between two vectors in R）。

我还需要将每个T的角度计算为每个K中的一个因子。我知道我可以使用split。

但是如何定义用于角度计算的矢量和角度本身的函数？感谢。

在此图中，我有一个运动的X，Y位置，我需要计算运动的角度。我希望有所帮助。还应注意，第一个和最后一个X，Y位置没有可能的角度。感谢

Answer 1

所以你的数据框有6行。前3组（X，Y）定义直角（th = 90）。接下来的三组（X，Y），第4-6行与第3行相同。因此这些点位于彼此的顶部并且没有角度。此外，K只有一个值，所以很难用K来证明聚合。

然而，这似乎有效：

df <- rbind(df,df,df)     # replicate the original data 3 times
df$K <- rep(1:3,each=6)   # K = 1, 2, 3
# theta in degrees
theta <- function(a,b)(180/pi)*(acos( sum(a*b) / ( sqrt(sum(a * a)) * sqrt(sum(b * b)))))
# this returns a vector of the angles between successive line segmeents
get.angles <- function(df.split){
  dx<- diff(df.split$X)
  dy<- diff(df.split$Y)
  sapply(1:(nrow(df.split)-2),function(i){
    a <- c(dx[i],dy[i])
    b <- c(dx[i+1],dy[i+1])
    theta(a,b)
  }) 
}
# this calls get.angles(...) for each subset of df, based on K
sapply(split(df,df$K),get.angles)
#        1   2   3
# [1,]  90  90  90
# [2,] NaN NaN NaN
# [3,] NaN NaN NaN
# [4,] NaN NaN NaN

编辑（对OP的其他数据和评论的回应）

因此，随着问题的相当大的改变，这种重新设计的解决方案似乎有效。使用df的新定义，

theta <- function(a,b)(180/pi)*(acos(sum(a*b)/(sqrt(sum(a*a))*sqrt(sum(b*b)))))
get.angles <- function(df.split){
  dx<- diff(df.split$X)
  dy<- diff(df.split$Y)
  stops <- which(dx^2+dy^2==0)
  dx<- dx[-stops]
  dy<- dy[-stops]
  df<- df.split[-(stops+1),]
  sapply(1:(length(dx)-1),function(i){
    a <- c(dx[i],dy[i])
    b <- c(dx[i+1],dy[i+1])
    return(cbind(df[i+1,],angle=180-theta(a,b)))
  })
}
result <- t(do.call(cbind,lapply(split(df,df$K),get.angles)))
result
#      K T  X      Y      V      P      angle   
# [1,] 1 2  23.904 167.33 1.494  -90    90      
# [2,] 1 3  23.904 164.34 0      0      171.8714
# [3,] 1 10 20.916 143.42 0      0      7.125665
# [4,] 1 21 29.88  176.29 8.7114 149.04 108.4535
# [5,] 1 22 20.916 182.27 6.6814 153.43 172.8726
# [6,] 1 23 14.94  185.26 3.3407 153.43 179.9233
# [7,] 1 24 8.964  188.24 1.494  180    153.4963
# [8,] 2 2  860.54 256.97 1.494  180    0       

此版本删除了dx ² + dy ² = 0（长度为0的行）的点，换句话说，在同一位置重复的点，并计算其余点的角度。请注意，我正在使用“内部”角度（＆lt; 180）。最后，我们绘制数据以显示这些确实是正确的角度：

library(ggplot2)
ggplot(df[df$K==1,],aes(x=X,y=Y))+
  geom_path()+geom_point(colour="red")+coord_fixed()

请注意coord_fixed()的使用。这迫使纵横比为1：1。否则角度会扭曲。

Answer 2

您可以使用＆＃39; lapply`包裹split来执行此操作：

getAngle <- function(X, Y) acos( sum(X*Y) / ( sqrt(sum(X * X)) * sqrt(sum(Y * Y)) ) )
lapply(split(df[, c("X", "Y")], f = list(df$K)), 
       FUN = function(x){ getAngle(x[, 1], x[, 2])})

# $`1`
#[1] 0.04074904