如何选择r中最小值和最大值之间尽可能均匀间隔的n个值？

Question

考虑

n

结果 (gp) 的组 (

ngp

)，如何选择/子集给定数量的结果 (

nesgp

)，这些结果在最小值和最大值（都必须包括在内）之间尽可能均匀地间隔在新专栏中

selec

？
编辑：理想情况下，未选择的结果应在新的

NA

列中显示为

selec

，而不是重复。

> print(dat, n=56)
# A tibble: 56 x 4
   gp    result   ngp nesgp
   <chr>  <dbl> <dbl> <dbl>
 1 CA      1.64    24    15
 2 CA      1.69    24    15
 3 CA      1.71    24    15
 4 CA      1.74    24    15
 5 CA      1.78    24    15
 6 CA      1.82    24    15
 7 CA      1.86    24    15
 8 CA      1.9     24    15
 9 CA      1.94    24    15
10 CA      1.98    24    15
11 CA      2.6     24    15
12 CA      2.65    24    15
13 CA      2.71    24    15
14 CA      2.76    24    15
15 CA      2.83    24    15
16 CA      2.89    24    15
17 CA      2.94    24    15
18 CA      3       24    15
19 CA      3.22    24    15
20 CA      3.42    24    15
21 CA      3.47    24    15
22 CA      3.68    24    15
23 CA      3.85    24    15
24 CA      4.38    24    15
25 ASAT    9       20    12
26 ASAT   11       20    12
27 ASAT   51       20    12
28 ASAT   61       20    12
29 ASAT   69       20    12
30 ASAT   78       20    12
31 ASAT   89       20    12
32 ASAT  102       20    12
33 ASAT  111       20    12
34 ASAT  120       20    12
35 ASAT  146       20    12
36 ASAT  163       20    12
37 ASAT  189       20    12
38 ASAT  208       20    12
39 ASAT  218       20    12
40 ASAT  304       20    12
41 ASAT  332       20    12
42 ASAT  345       20    12
43 ASAT  362       20    12
44 ASAT  402       20    12
45 ORO     0.56    12     8
46 ORO     0.7     12     8
47 ORO     0.77    12     8
48 ORO     0.78    12     8
49 ORO     0.82    12     8
50 ORO     0.82    12     8
51 ORO     0.92    12     8
52 ORO     0.94    12     8
53 ORO     1.16    12     8
54 ORO     1.46    12     8
55 ORO     1.54    12     8
56 ORO     1.77    12     8 

数据

dat <-
structure(list(gp = c("CA", "CA", "CA", "CA", "CA", "CA", "CA", 
"CA", "CA", "CA", "CA", "CA", "CA", "CA", "CA", "CA", "CA", "CA", 
"CA", "CA", "CA", "CA", "CA", "CA", "ASAT", "ASAT", "ASAT", "ASAT", 
"ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", 
"ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", "ASAT", 
"ORO", "ORO", "ORO", "ORO", "ORO", "ORO", "ORO", "ORO", "ORO", 
"ORO", "ORO", "ORO"), result = c(1.64, 1.69, 1.71, 1.74, 1.78, 
1.82, 1.86, 1.9, 1.94, 1.98, 2.6, 2.65, 2.71, 2.76, 2.83, 2.89, 
2.94, 3, 3.22, 3.42, 3.47, 3.68, 3.85, 4.38, 9, 11, 51, 61, 69, 
78, 89, 102, 111, 120, 146, 163, 189, 208, 218, 304, 332, 345, 
362, 402, 0.56, 0.7, 0.77, 0.78, 0.82, 0.82, 0.92, 0.94, 1.16, 
1.46, 1.54, 1.77), ngp = c(24, 24, 24, 24, 24, 24, 24, 24, 24, 
24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 20, 
20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 
20, 20, 20, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12), 
    nesgp = c(15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 
    15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 12, 12, 12, 
    12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 
    12, 12, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -56L)) 

感谢您的帮助。

Answer 1

以下解决方案使用了辅助功能

fun

，也许还有更简单的方法可以做到这一点。
该函数创建一系列

n

值，并使用

findInterval

来查看这些值在

x

中的位置。然后，在

for

循环中，检查到每个区间极值的距离，并将最小值分配给返回值

y

。

suppressPackageStartupMessages(
  library(tidyverse)
)

fun <- function(x, n, na.rm = FALSE) {
  xmin <- min(x, na.rm = na.rm)
  xmax <- max(x, na.rm = na.rm)
  ref <- seq(xmin, xmax, length.out = n)
  x <- sort(x)
  j <- findInterval(ref, x)
  y <- numeric(n)
  y[1L] <- xmin
  y[n] <- xmax
  for(i in seq_len(n)[-c(1L, n)]) {
    if(abs(ref[i] - x[ j[i] ]) < abs(ref[i] - x[ j[i + 1L] ])) {
      y[i] <- x[ j[i] ]
    } else y[i] <- x[ j[i + 1L] ]
  }
  y
}
dat %>%
  reframe(selec = fun(result, first(ngp)), .by = gp)
#> # A tibble: 56 × 2
#>    gp    selec
#>    <chr> <dbl>
#>  1 CA     1.64
#>  2 CA     1.74
#>  3 CA     1.86
#>  4 CA     1.98
#>  5 CA     1.98
#>  6 CA     1.98
#>  7 CA     1.98
#>  8 CA     1.98
#>  9 CA     2.71
#> 10 CA     2.71
#> # ℹ 46 more rows

^{创建于 2024-02-04，使用 reprex v2.0.2}

Answer 2

我不确定你所说的“尽可能均匀地间隔”是什么意思，但我写了一个例子，使用点的采样来最小化它们之间的分布，这对你来说可能是一个很好的起点：

par(mfrow = c(length(unique(dat$gp)), 1))
dat$selec <- NA
# for each group, 
groups <- unique(dat$gp)
for(gp in groups){
  x <- dat$result[dat$gp == gp]
  minmax_x <- range(x)
  possible_xs <- x[!(x %in% minmax_x)]
  # run a lot of samples of different possible lengths to test
  r <- replicate(20000, sort(c(minmax_x,
                               sample(possible_xs, 
                                      size = sample(3:length(possible_xs),1)
                                      )
                               )
                             )
                 )

  spreads <- sapply(r, function(obj) var(diff(obj)))
  minimized_variance_index <- which.min(spreads)
  dat$selec[which(dat$result %in% r[[minimized_variance_index]])] <- 1
  # visualize
  plot(x, rep(1, length(x)), yaxt = "n", ylab = "", xlab = "result",
       main = paste(gp,", spread =", round(var(diff(r[[minimized_variance_index]])),5)))
  abline(v= r[[minimized_variance_index]])
}

在这种情况下，选择的点并不像您想要的那么多。