R/splitLines.R

#' splitLines
#' @name splitLines
#' @description This function splits each line of an "sp" object of class SpatialLines into segments of a given length.
#' @param spatial_line An "sp" object of class SpatialLines or SpatialLinesDataFrame.
#' @param split_length The length of the segments to split the lines into, in units of the SpatialLines. Default 10.
#' @param return.dataframe If TRUE, the functions returns a dataframe with IDs and coordinates, otherwise returns a SpatialLines object. Default FALSE.
#' @param plot.results Visual representation of the results. If TRUE, the functions plots the original object in black and the vertex of the new segments in red. Default FALSE.
#' @references http://math.stackexchange.com/questions/175896/finding-a-point-along-a-line-a-certain-distance-away-from-another-point?newreg=468f66d7274f449b8ecf3fa4e63f41fe
#' @references http://tutorial.math.lamar.edu/Classes/CalcII/Vectors_Basics.aspx
#' @examples library("sp", "sptools")
#' c1 = cbind(c(0,2,5,7), c(0,2,3,3))
#' l1 = Line(c1)
#' sl = list(Lines(list(l1), ID = 1))
#' SP = SpatialLines(sl)
#' plot(SP)
#' SPs = splitLines(SP, 0.4, plot.results = TRUE)

splitLines = function(spatial_line,
                      split_length = 10,
                      return.dataframe = FALSE,
                      plot.results = FALSE) {
  #### Split the lines ####
  # Convert the input SpatialLine object into a dataframe and create an empty output dataframe
  linedf = SpatialLines2df(spatial_line)
  df = data.frame(
    id = character(),
    fx = numeric(),
    fy = numeric(),
    tx = numeric(),
    ty = numeric(),
    stringsAsFactors = FALSE
  )


  for (i in 1:nrow(linedf)) {
    # For each line of the dataframe, corresponding to a single line of the spatial object
    # skips if length is less then split_length
    v_seg = linedf[i, ]
    seg_length = sqrt((v_seg$fx - v_seg$tx) ^ 2 + (v_seg$fy - v_seg$ty) ^
                        2) # segment length
    if (seg_length <= split_length) {
      df[nrow(df) + 1,] = c(paste0(v_seg$id, '_', '0000'),
                            v_seg$fx,
                            v_seg$fy,
                            v_seg$tx,
                            v_seg$ty)
      next()
    }
    # Create a vector of direction as the line and unit length
    # vector v corresponding to the line
    v = c(v_seg$tx  -  v_seg$fx,
          v_seg$ty  -  v_seg$fy)
    # vector of direction v and unit length
    u = c(v[1]  /  sqrt(v[1]  ^  2 + v[2]  ^  2), v[2]  /  sqrt(v[1]  ^  2 + v[2]  ^ 2))
    # Calculates how many segment the line is split into and the leftover
    num_seg = floor(seg_length  /  split_length)
    seg_left = seg_length - (num_seg  *  split_length)

    # Add to the output dataframe each segment plus the leftover
    for (i in 0:(num_seg  -  1)) {
      # Add num_seg segments
      df[nrow(df)  +  1,] = c(
        paste0(v_seg$id, '_', formatC(i, width = 4, flag = '0')),
        v_seg$fx + u[1]  *  split_length  *  i,
        v_seg$fy + u[2]  *  split_length  *  i,
        v_seg$fx + u[1]  *  split_length  *  (i  +  1),
        v_seg$fy + u[2]  *  split_length  *  (i  +  1)
      )
    }
    df[nrow(df) + 1,] = c(
      paste0(v_seg$id, '_', formatC(
        num_seg, width = 4, flag = '0'
      )),
      # Add leftover segment
      v_seg$fx + u[1] * split_length * num_seg,
      v_seg$fy + u[2] * split_length * num_seg,
      v_seg$tx,
      v_seg$ty
    )

  }

  #### Visualise the results to check ####
  if (plot.results) {
    plot(spatial_line)
    coords = cbind(as.numeric(df$fx), as.numeric(df$fy))
    coords = rbind(coords, as.numeric(df$tx[nrow(df)]), as.numeric(df$ty)[nrow(df)])
    sp_points = SpatialPoints(coords)
    plot(sp_points, col = 'red', add = T)
  }

  #### Output ####
  df$fx = as.numeric(df$fx)
  df$fy = as.numeric(df$fy)
  df$tx = as.numeric(df$tx)
  df$ty = as.numeric(df$ty)
  if (return.dataframe) {
    return(df)
  } # Return a dataframe
  sl = linedf2SpatialLines(df)
  return(sl) # Return a SpatialLine object
}
duccioa/sptools documentation built on May 3, 2019, 1:48 p.m.