compute angle between moving vectors

Question

I have a data frame that looks like this:

    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")

Since I have X, Y positions, to make two vectors, I should consider three X,Y positions to be able to compute for an angle. I know that:

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

from another stackoverflow answer (Angle between two vectors in R).

I need also to compute the angle per T as a factor within each K. I know I could use split in this.

But how do I define the vector for the computation of the angle and the function for the angle itself? Thanks.

On this diagram, I have X,Y positions of a movement and I need to compute the angle of the movement. I hope that helps. It should also be noted that there is no angle possible for the first and last X,Y positions. Thanks

path of movement

Thanks for sharing data in a nice way, but I'm still confused by what you want. You say you have 3 (x,y) positions, so I assume (K,T), (X,Y) and (V,P). (This doesn't actually seem right, but I don't know what else to think.) Which one is the vertex of the angle? And how to you mean "per K as a factor"? In your sample, all `K` values are 1. — Gregor Thomas, Mar 11 '14 at 19:39
that is just the head of my data. I have multiple Ks. I do not have a fixed vertex of a sort, I have a moving vertex. And the K,T,X,Y,V,P are not my X,Y positions, they are my columns. Only the X,Y positions should be used for the computation. — Kaye11, Mar 11 '14 at 19:42
Ok, so, in the head of your data what angles do you want to calculate in terms of row numbers? Some sort of running angle, The angle between the (X,Y)'s in rows 1,2,3, then 2,3,4, then 3,4,5? Every combination of (X,Y)? Do columns T, V, P matter at all to this question? — Gregor Thomas, Mar 11 '14 at 19:50
Yes you are right. T,V,P do not matter. they are just variables in my data frame. — Kaye11, Mar 11 '14 at 19:52
Your question is still a little confusing.. Perhaps, to clarify, you could include a diagram of what your trying to do and an example of your expected outcome. — N8TRO, Mar 11 '14 at 20:11
THe (X,Y) in your data frame are not the same as the (X,Y) in your plot. — jlhoward, Mar 11 '14 at 20:38

jlhoward · Accepted Answer · 2014-03-12T16:33:01.980

So your data frame has 6 rows. The first 3 sets of (X,Y) define a right angle (th=90). The next three sets of (X,Y), rows 4-6, are identical to row 3. So those points sit on top of each other and there is no angle. Also there is only one value of K so it's kind of hard to demonstrate aggregation by K.

Nevertheless, this seems to work:

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

EDIT (Response to OP's additional data, and comments)

So with the rather substantial change to the question, this reworked solution seems to work. Using your new definition of df,

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

This version removes points where dx²+dy²=0 (lines of length=0), in other words repeated points at the same location, and calculates the angles for the remaining points. Note that I'm using "internal" angles (<180). Finally, we plot the data to show that these are indeed the proper angles:

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

Note the use of coord_fixed(). This forces the aspect ration to 1:1. Otherwise the angles are distorted.

Hello, one of the problems is that if I there is a movement, then no movement, then a movement again, it takes NaN as an angle even if there is an angle. I would update my data structure so you could see what I mean.. — Kaye11, Mar 11 '14 at 21:03
I have a sort of stupid question. I checked the values against my whole X, Y positions, I did `r=merge(df,result[, c (1:2, 7)], by=c("K", "T"), all=TRUE)`. First, I noticed that at K=1, T=24, I have a change of X,Y values but no change in angle. Second, `angle=180-theta(a,b)` is for the internal angle right? How could I change it so that angles could be from -180 to 180? Because I also want to have a clear cut line between moving up and down. Thanks! — Kaye11, Mar 12 '14 at 15:48
First, there is a change from T=23 to T=24, but not after, so I don't see the problem. In the plot, you can see there are 7 angles for K=1, and in `results` you can see there are 7 angles for K=1. Regardless, the original setup had `angle` associated with the *first* of the three points that define the angle. I updated this above to associate angle with the middle (vertex) point. Maybe this is clearer. Second, yes, as the answer states, these are internal angles. If you want to know if movement is up or down, you can just look at the value of `dy`. — jlhoward, Mar 12 '14 at 16:39
Sorry I might have missed that. Looking at the value of dy might not be a good idea as I have thousand of Ks. You are right though with the computation of internal angles. Thanks again :) — Kaye11, Mar 13 '14 at 16:24

Christopher Louden · Answer 2 · 2014-03-11T19:56:05.933

0

You can wrap split with 'lapply` to do this:

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

edited Mar 11 '14 at 19:56

answered Mar 11 '14 at 19:41

Christopher Louden

7,540
2
26
29

I get an error: `Error in x[, 1] : incorrect number of dimensions` – Kaye11 Mar 11 '14 at 19:47
`aggregate` turns the data frame into a vector before passing it to `FUN`. I did not realize that. I fixed it so that it uses `lapply` and `split` and actually works (I get 0.04). – Christopher Louden Mar 11 '14 at 19:57
I forgot to add that I want the angle for every `T` also. Sorry, I will fix the question. – Kaye11 Mar 11 '14 at 20:02
`f` in `split` can be a list of factors to get the interaction. – Christopher Louden Mar 11 '14 at 20:07
Hello, I think I did something wrong. If I use `f = list(df$K, df$A)` in the split, I get all 0 values for the angle. – Kaye11 Mar 11 '14 at 20:30
That is because using 'T' as a factor creates one-dimensional vectors. The angle between all one-dimensional vectors, using the formula you gave, is `acos(1)` which is 0. – Christopher Louden Mar 11 '14 at 20:33

compute angle between moving vectors

2 Answers2