使用R中的dplyr在2D中插入来自数据帧的数据

时间:2015-05-25 12:56:10

标签: r interpolation dplyr

我有两个数据框:参考插值。这是参考的一瞥():

$ Value    (dbl) 62049.67, 62040.96, 62053.02, 62039.31, 62020.82, 62001.73,...
$ X       (dbl) -10.14236, -10.14236, -10.14236, -10.14236, -10.14236, -10....
$ Y       (dbl) -12.68236, -12.64708, -12.61181, -12.57653, -12.54125, -12....

这是 Interpolated

$ X       (dbl) -10.1346, -10.0838, -10.0330, -9.9822, -9.9314, -9.8806, -9...
$ Y       (dbl) -12.6746, -12.6746, -12.6746, -12.6746, -12.6746, -12.6746,...

我想使用 Reference 中的2D插值在 Interpolated 中获取变量 Value

我正在考虑使用 akima 包中的 bicubic()函数,类似于bicubic(Reference$X, Reference$Y, Reference$Value, Interpolated$X, Interpolated$Y)。但是 bicubic()需要 Reference $ Value 中的矩阵。

有没有简单的方法在2D中插入数据帧中的数据,最好使用 dplyr

1 个答案:

答案 0 :(得分:0)

不知道你是否收到过这个答案。我一直在寻找相同的东西,不得不创建自己的功能来做到这一点。请参阅以下内容:

interpolate <- function(x, x1, x2, y1, y2) {
  # Interpolates between two points.
  #
  # Args:
  #   x: Corresponding x value of y value to return.
  #   x1: Low x-value.
  #   x2: High x-value.
  #   y1: Low y-value.
  #   y2: High y-value.
  #
  # Returns:
  #   Interpolated value corresponding to x between the two points.
  y <- y1 + (y2-y1)*(x-x1)/(x2-x1)
  return(y)
}

doubleinterpolate <- function(x, y, z, xout, yout) {
  # Returns a double interpolated value among three vectors with two
  # values in two of the vectors.
  #
  # Args:
  #   x: Vector containing a known value.
  #   y: Vector containing a known value.
  #   z: Vector containing an unknown value.
  #   xout: Known value in x-vector.
  #   yout: Known value in y-vector.
  #
  # Returns:
  #   Double interpolated value in z of the points xout and yout.

  # Determine adjacent values in the table
  x_low <- max(x[x < xout])
  x_high <- min(x[x > xout])
  y_low <- max(y[y < yout])
  y_high <- min(y[y > yout])

  # Create df and subset
  df <- data_frame(x = x, y = y, z = z)
  df_low <- df[x == x_low, ]
  df_high <- df[x == x_high, ]

  # Interpolate low x-values
  yint1 <- as.numeric(spline(df_low$y, df_low$z, xout = yout)[2])
  yint2 <- as.numeric(spline(df_high$y, df_high$z, xout = yout)[2])

  #Interpolate to get last value
  zout <- interpolate(xout, x_low, x_high, yint1, yint2)

  return(zout)
}