如何按程序计算大量向量之间的角度？

Question

我有 4 个向量及其在不同时间步长的坐标。

lon <- list(505997.627175236, 505997.627175236, 505997.627175236, 505997.627175236, 505997.064187932, 505997.814896096,
          505997.843587834, 505997.880929633, 505996.906012923, 505998.486599226, 505998.075906002, 505998.079921271)   
lon <- do.call(rbind.data.frame, lon)

lat <- list(7941821.025438220, 7941821.025438220, 7941821.025438220, 7941821.025438220, 7941819.791667340, 7941821.329316000,
            7941821.741379530, 7941821.171989530, 7941819.103811300, 7941821.831421200, 7941822.024924560, 7941822.110412460)
lat <- do.call(rbind.data.frame, lat)

step <- list(1,1,1,1,2,2,2,2,3,3,3,3)
step <- do.call(rbind.data.frame, step)

allbuff <- cbind(lon, lat, step)
colnames(allbuff) <- c("lon", "lat", "STEP")

我已经使用以下脚本计算了 step、step+1、...、step+n 中四个向量之间的角度：

M_PI <- 3.14159265359

output_angle = NULL

for (i in unique(allbuff$STEP)) {

  select = allbuff[allbuff$STEP == i, 1:2]

  result1 = atan2((select[1,2] - select[2,2]), (select[1,1] - select[2,1]))*(180/M_PI) # between 1 & 2
  result2 = atan2((select[1,2] - select[3,2]), (select[1,1] - select[3,1]))*(180/M_PI) # between 1 & 3
  result3 = atan2((select[1,2] - select[4,2]), (select[1,1] - select[4,1]))*(180/M_PI) # between 1 & 4
  result4 = atan2((select[2,2] - select[3,2]), (select[2,1] - select[3,1]))*(180/M_PI) # between 2 & 3
  result5 = atan2((select[2,2] - select[4,2]), (select[2,1] - select[4,1]))*(180/M_PI) # between 2 & 4
  result6 = atan2((select[3,2] - select[4,2]), (select[3,1] - select[4,1]))*(180/M_PI) # between 3 & 4

  result <- rbind(result1,result2,result3,result4,result5,result6)
  STEP <- c(i,i)
  result <- cbind(result, as.data.frame(STEP))

  output_angle = rbind(output_angle,result)

}

output_angle <- as.data.frame(output_angle)

用少量向量编码但用 1000 个向量编码很有效，而且不会太长，这种编码方式可能非常耗时。

因此，无论输入的向量数量是多少，是否有更有效的方法（从程序上讲）来计算第 n 步中所有向量之间的角度？

Answer 1

也许下面的基本 R 代码可以提供一些帮助

dfout <- do.call(rbind,
                 c(make.row.names = F,
                   lapply(split(allbuff,allbuff$STEP), function(v) 
                     data.frame(result = combn(nrow(v),2,function(k) atan2(diff(v[rev(k),2]),diff(v[rev(k),1])) )*(180/M_PI),
                                STEP = unique(v[3]),
                                row.names = NULL))))

你会得到

> dfout
       result STEP
1     0.00000    1
2     0.00000    1
3     0.00000    1
4     0.00000    1
5     0.00000    1
6     0.00000    1
7  -116.02248    2
8  -111.78912    2
9  -120.61296    2
10  -93.98304    2
11  112.76891    2
12   93.75220    2
13 -120.09129    3
14 -111.82589    3
15 -111.32784    3
16  -25.22807    3
17  -34.45110    3
18  -92.68914    3

如何按程序计算大量向量之间的角度？

How to procedurally calculate angles between a great number of vectors?

procedural

r

vector

angle