Question

我需要确定位于特定试金体内的一组微米坐标的一小部分。这一试金字被定义为双方距离协调制度的边缘相当遥远的地区(在这种情况下,协调制度受大约(-50, - 20),(-50, 20),(50,20),(50,20),(50,-20)。此外,我希望能够测试与边缘不同距离的试金星结果。我的方法如下:

  # set initial limits to the coordinate system
  lim.xleft = -50
  lim.xright = 50
  lim.ybottom = -20
  lim.ytop = 20

frac.near.edge <- function(coord.pairs, tolerance){
  # set the coordinates of the rectangle of interest
  exclude.xleft = lim.xleft + tolerance
  exclude.xright = lim.xright - tolerance
  exclude.ybottom = lim.ybottom + tolerance
  exclude.ytop = lim.ytop - tolerance
  out <- vector()
  # loop through the pairs testing whether the point is inside the rectangle or outside
  for(i in 1:nrow(coord.pairs)){
    if(coord.pairs[i, 1] > exclude.xleft & coord.pairs[i, 1] < exclude.xright &  coord.pairs[i, 2] > exclude.ybottom & coord.pairs[i, 2] < exclude.ytop){
      out[i] <- "in"
    } else {
      out[i] <- "out"
    }
  }
  # return how many points were inside the rectangle and how many were outside
  return(table(out))
}

# try it out on something much bigger!
foo <- data.fram(x = runif(100), y = runif(100))
system.time(frac.near.edge(foo,  tolerance = 5))

大型数据集(按10°5×y pairs的顺序排列的地雷)的浓度为very。我如何能够加快这一进程? 环形周围的方法?

Answer 1

exclude.xleft = lim.xleft + tolerance
exclude.xright = lim.xright - tolerance
exclude.ybottom = lim.ybottom + tolerance
exclude.ytop = lim.ytop - tolerance
out <- c("out", "in")[1+( findInterval(coord.pairs[ , 1], c(exclude.xleft, exclude.xright))==1 &
                       findInterval(coord.pairs[ , 2], c(exclude.ybottom,   exclude.ytop))==1)]

在50K试验案例中,你的方法为0.01和19。

coord.pairs<- cbind(rnorm(50000, 0,50), rnorm(1000,0,20)); tolerance=10

Answer 2

这可能更适合《东南欧法律评论》(。我不知道这部编码是否有用,但你对如何进行方案规划提出了真正的疑问,因为这会带来更多的改进。我也产生了更好的数据集,因为你们都作出了反应。

foo <- data.frame(x = sample(-100:100, 100, replace=TRUE), 
    y = sample(-100:100, 100, replace=TRUE))

xleft = -50
xright = 50
ybottom = -2
ytop = 20
foo$x >= xleft & foo$x <= xright  & foo$y >= ybottom & foo$y <= ytop

友情链接