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我有日期、深度和温度的温度分析仪 (tp) 数据。每个日期的深度并不完全相同，因此我需要将其统一到相同的深度并通过线性近似设置该深度的温度。我能够通过使用“近似”函数的循环来做到这一点（参见随附代码的第一部分）。但我知道我应该在没有循环的情况下做得更好（考虑到我将有大约 600,000 行）。我试图用'by'函数来做，但没有成功地将结果（列表）转换为数据框或矩阵（参见代码的第二部分）。请记住，圆形深度的长度并不总是与示例中的相同。圆形深度在 Depth2 列中，插值温度放在 Temp2 中解决这个问题的“正确”方法是什么？

# create df manually
tp <- data.frame(Date=double(31), Depth=double(31), Temperature=double(31))
tp$Date[1:11] <- '2009-12-17' ; tp$Date[12:22] <- '2009-12-18'; tp$Date[23:31] <- '2009-12-19' 
tp$Depth <- c(24.92,25.50,25.88,26.33,26.92,27.41,27.93,28.37,28.82,29.38,29.92,25.07,25.56,26.06,26.54,27.04,27.53,28.03,28.52,29.02,29.50,30.01,25.05,25.55,26.04,26.53,27.02,27.52,28.01,28.53,29.01)
tp$Temperature <- c(19.08,19.06,19.06,18.87,18.67,17.27,16.53,16.43,16.30,16.26,16.22,17.62,17.43,17.11,16.72,16.38,16.28,16.20,16.15,16.13,16.11,16.08,17.54,17.43,17.32,17.14,16.89,16.53,16.28,16.20,16.13)

# create rounded depth column
tp$Depth2 <- round(tp$Depth)

# loop on date to calculate linear approximation for rounded depth
dtgrp <- tp[!duplicated(tp[,1]),1]
for (i in dtgrp) {
  x1 <- tp[tp$Date == i, "Depth"]  
  y1 <- tp[tp$Date == i, "Temperature"]
  x2 <- tp[tp$Date == i, "Depth2"]
  tpa <- approx(x=x1,y=y1,xout=x2, rule=2)
  tp[tp$Date == i, "Temp2"] <- tpa$y
}
# reduce result to rounded depth
tp1 <- tp[!duplicated(tp[,-c(2:3)]),-c(2:3)]

# not part of the question, but the end need is for a matrix, so this complete it:
library(reshape2)
tpbydt <- acast(tp1, Date~Depth2, value.var="Temp2")

# second part: I tried to use the by function (instead of loop) but got lost when tring to convert it to data frame or matrix
rdpth <- function(x1,y1,x2) {
  tpa <- approx(x=x1,y=y1,xout=x2, rule=2)
  return(tpa)
}
tp2 <- by(tp, tp$Date,function(tp) rdpth(tp$Depth,tp$Temperature,tp$Depth2), simplify = TRUE)
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1 回答 1

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与 call 非常接近，by但请记住它返回一个对象列表。因此，考虑构建一个数据框列表以在最后进行行绑定：

df_list <- by(tp, tp$Date, function(sub) {
  tpa <- approx(x=sub$Depth, y=sub$Temperature, xout=sub$Depth2, rule=2)

  df <- unique(data.frame(Date = sub$Date, 
                          Depth2 = sub$Depth2,
                          Temp2 = tpa$y,
                          stringsAsFactors = FALSE))
  return(df)
})    

tp2 <- do.call(rbind, unname(df_list))

tp2
#          Date Depth2    Temp2
# 1  2009-12-17     25 19.07724
# 2  2009-12-17     26 19.00933
# 5  2009-12-17     27 18.44143
# 7  2009-12-17     28 16.51409
# 9  2009-12-17     29 16.28714
# 11 2009-12-17     30 16.22000
# 12 2009-12-18     25 17.62000
# 21 2009-12-18     26 17.14840
# 4  2009-12-18     27 16.40720
# 6  2009-12-18     28 16.20480
# 8  2009-12-18     29 16.13080
# 10 2009-12-18     30 16.08059
# 13 2009-12-19     25 17.54000
# 22 2009-12-19     26 17.32898
# 41 2009-12-19     27 16.90020
# 61 2009-12-19     28 16.28510
# 81 2009-12-19     29 16.13146

如果你 reset ，这与你的输出row.names完全相同：tp1

identical(data.frame(tp1, row.names = NULL),
          data.frame(tp2, row.names = NULL))
# [1] TRUE
于 2020-02-01T17:19:23.537 回答