使用这些数据:

library(tidyverse)

df <-
  structure(
    list(
      start_depth = c(10, 15, 20, 25, 30),
      end_depth = c(15,
                    20, 25, 30, 35),
      k = c(
        0.136,
        0.135,
        0.133,
        0.139,
        0.132
      )
    ),
    row.names = c(NA,-5L),
    class = c("tbl_df", "tbl", "data.frame"),
    .Names = c("start_depth",
               "end_depth", "k")
  )

df
#> # A tibble: 5 x 3
#>   start_depth end_depth     k
#>         <dbl>     <dbl> <dbl>
#> 1        10.0      15.0 0.136
#> 2        15.0      20.0 0.135
#> 3        20.0      25.0 0.133
#> 4        25.0      30.0 0.139
#> 5        30.0      35.0 0.132

我想使用以下方程为每对 end_depthstart_depth 传播一个值,增量为 1 米。

例如,假设我从 30-35 米级别的 start_val = 0.001 开始:

end_depth = 35
0.001000000 = 0.001000000 * exp(0.132 * (35 - (35)))

end_depth = 34
0.001141108 = 0.001000000 * exp(0.132 * (35 - (34)))

end_depth = 33
0.001302128 = 0.001000000 * exp(0.132 * (35 - (33)))

end_depth = 32
0.001485869 = 0.001000000 * exp(0.132 * (35 - (32)))

end_depth = 31
0.001695538 = 0.001000000 * exp(0.132 * (35 - (31)))

end_depth = 30
0.001934792 = 0.001000000 * exp(0.132 * (35 - (30)))

然后,25-30 米级别,我会再次开始,但使用最后计算的值(即 0.001934792)

end_depth = 30
0.001934792 * exp(0.139 * (30 - (30)))

end_depth = 29
0.001934792 * exp(0.139 * (30 - (29)))

我正在使用 dplyr,但任何其他选项都是有效的(例如:base R.data.table 等)

reprex package (v0.2.0) 于 2018 年 2 月 26 日创建。

最佳答案

这是 for 循环的一种选择

v <- 0.001000000
lst <- vector("list", nrow(df))
 for(i in rev(seq_along(lst))) {
      e1 <- v * exp(df$k[i] *(df$end_depth[i] -
                  seq(df$start_depth[i], df$end_depth[i], by = 1)))
      lst[[i]] <- e1
      v <- e1[1]

 }

-输出
lst
#[[1]]
#[1] 0.02922428 0.02550820 0.02226465 0.01943353 0.01696241 0.01480552

#[[2]]
#[1] 0.014805519 0.012935817 0.011302229 0.009874938 0.008627890 0.007538325

#[[3]]
#[1] 0.007538325 0.006599540 0.005777667 0.005058146 0.004428230 0.003876761

#[[4]]
#[1] 0.003876761 0.003373666 0.002935859 0.002554867 0.002223316 0.001934792

#[[5]]
#[1] 0.001934792 0.001695538 0.001485869 0.001302128 0.001141108 0.001000000

如果我们使用 tidyverse ,那么可以使用 pmapaccumulate_right
library(purrr)
pmap(df, ~ exp(..3 *(..2 - seq(..1, ..2, by = 1))))  %>%
       accumulate_right(~ .x[1] * .y, .init = 0.001000000) %>%
       head(., -1)
#[[1]]
#[1] 0.02922428 0.02550820 0.02226465 0.01943353 0.01696241 0.01480552

#[[2]]
#[1] 0.014805519 0.012935817 0.011302229 0.009874938 0.008627890 0.007538325

#[[3]]
#[1] 0.007538325 0.006599540 0.005777667 0.005058146 0.004428230 0.003876761

#[[4]]
#[1] 0.003876761 0.003373666 0.002935859 0.002554867 0.002223316 0.001934792

#[[5]]
#[1] 0.001934792 0.001695538 0.001485869 0.001302128 0.001141108 0.001000000

关于r - 根据上一组中计算的另一个值传播值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48989773/

10-12 19:22