根据最小值和最大值过滤数据帧

4 回答

• 1

  我们可以使用 Map 来自 base R

  Map(function(x, y) x[x >= min(y) & x <= max(y)], df, Neighb)
  #$A
  #numeric(0)
  
  #$B
  #[1] 2 2 1 1 1 1 2
  
  #$C
  #[1] 16
  
  #$D
  #[1] 5 5 6 5
  
  #$E
  #[1] 1.09
  
  #$F
  #[1] 2 2 1 1 1 1 2
  

  ---

  如果我们需要 filter 基于逻辑索引的数据集，即基于与'Neighb'的比较而具有全部 TRUE 的行

  df[Reduce(`&`, Map(function(x, y) x >= min(y) & x <= max(y), df, Neighb)), ]
  

  如果它是任何TRUE

  df[Reduce(`|`, Map(function(x, y) x >= min(y) & x <= max(y), df, Neighb)),]
  

  数据

  df <- structure(list(A = c(24, 21, 12, 39, 51, 24, 48, 44, 14, 34), 
                       B = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2), 
                       C = c(16, 19, 12, 39, 82, 24, 32, 44, 18, 51),
                       D = c(5, 2, 79, 39, 27, 40, 5, 12, 6, 5), 
                       E = c(1.2, 1.09, 0.86, 1.9, 2.3, 1.6, 1.6, 1.7, 0.88, 2.7), 
                       F = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2)), 
                  .Names = c("A","B", "C", "D", "E", "F"), 
                  class = "data.frame", 
                  row.names = c(NA, -10L))
  
  
  Neighb <- list(c(15.7, 15.9, 16.0, 16.1, 16.2),
                 c(0, 1, 2, 3, 4),
                 c(15.0, 15.3, 16.0, 16.3, 16.5),
                 c(3, 4, 5, 6, 7),
                 c(1.08, 1.09, 1.10, 1.11, 1.12),
                 c(0, 1, 2, 3, 4))
  

  回复于 2024-04-28T11:59:52+08:00
• 4

  你可以将 purrr 与 purrr 一起使用 between 来自 dplyr 来获得你想要的结果 .

  library(purrr)
  library(dplyr)
  
  map2(df, Neighb, function(x, y) x[between(x, min(y), max(y))] )
  $A
  numeric(0)
  
  $B
  [1] 2 2 1 1 1 1 2
  
  $C
  [1] 16
  
  $D
  [1] 5 5 6 5
  
  $E
  [1] 1.09
  
  $F
  [1] 2 2 1 1 1 1 2
  

  数据：

  df <- structure(list(A = c(24, 21, 12, 39, 51, 24, 48, 44, 14, 34), 
                       B = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2), 
                       C = c(16, 19, 12, 39, 82, 24, 32, 44, 18, 51),
                       D = c(5, 2, 79, 39, 27, 40, 5, 12, 6, 5), 
                       E = c(1.2, 1.09, 0.86, 1.9, 2.3, 1.6, 1.6, 1.7, 0.88, 2.7), 
                       F = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2)), 
                  .Names = c("A","B", "C", "D", "E", "F"), 
                  class = "data.frame", 
                  row.names = c(NA, -10L))
  
  
  Neighb <- list(c(15.7, 15.9, 16.0, 16.1, 16.2),
                 c(0, 1, 2, 3, 4),
                 c(15.0, 15.3, 16.0, 16.3, 16.5),
                 c(3, 4, 5, 6, 7),
                 c(1.08, 1.09, 1.10, 1.11, 1.12),
                 c(0, 1, 2, 3, 4))
  

  回复于 2024-04-28T11:59:52+08:00
• 2

  可能的解决方案：

  # needed packages
  library(data.table)
  
  # get the minimum and maximum for each list item
  nr <- lapply(Neighb, range)
  
  # create a matrix with the 'inrange' function from 'data.table'
  m <- mapply(function(x, y) x %inrange% y, df, nr)
  

  这给了：

  m
  A B C D E F.
  [1，] FALSE FALSE TRUE FUE FALSE FALSE
  [2，] FALSE TRUE FALSE FALSE TRUE
  [3，] FALSE TRUE FALSE FALSE TRUE
  [4，]错误，错误，错误，错误
  [5，] FALSE TRUE FALSE FALSE TRUE
  [6，]错误，错误，错误，错误
  [7，] FALSE TRUE FUE FUE FALSE TRUE
  [8，] FALSE TRUE FALSE FALSE TRUE
  [9，] FALSE TRUE FUE FUE FALSE TRUE
  [10，] FALSE TRUE FUE FUE FALSE TRUE

  现在您可以使用 rowSums 函数过滤 df ：

  df[rowSums(m) == ncol(df),]
  

  将此应用于所呈现的示例数据（ df ）将导致空数据帧，但在原始数据集上很可能会导致非空数据帧 .

  ---

  使用数据：

  df <- read.table(text="     A     B     C     D     E     F
                     1   24    6   16    5 1.20     6
                     2   21    2   19    2 1.09     2
                     3   12    2   12   79 0.860    2
                     4   39    7   39   39 1.90     7
                     5   51    1   82   27 2.30     1
                     6   24    9   24   40 1.60     9
                     7   48    1   32    5 1.60     1
                     8   44    1   44   12 1.70     1
                     9   14    1   18    6 0.880    1
                     10   34   2   51    5 2.70     2", header=TRUE, stringsAsFactors=FALSE)
  Neighb <- list(c(15.7,15.9,16.0,16.1,16.2),c(0:4),c(15.0,15.3,16.0,16.3,16.5),c(3:7),seq(1.08,1.12,0.01),c(0:4))
  

  回复于 2024-04-28T11:59:52+08:00
• 1

  另一种方法可能是

  #minimum and maximum value from given list
  filter_criteria <- lapply(lookup_list, function(x) c(min(x), max(x)))
  
  df1 <- as.data.frame(mapply(function(x, y) replace(x, !(x>=y[1] & x<=y[2]), NA), 
                              df, filter_criteria))
  
  df1
  #    A  B  C  D    E  F
  #1  NA NA 16  5   NA NA
  #2  NA  2 NA NA 1.09  2
  #3  NA  2 NA NA   NA  2
  #4  NA NA NA NA   NA NA
  #5  NA  1 NA NA   NA  1
  #6  NA NA NA NA   NA NA
  #7  NA  1 NA  5   NA  1
  #8  NA  1 NA NA   NA  1
  #9  NA  1 NA  6   NA  1
  #10 NA  2 NA  5   NA  2
  
  #final output
  df1 <- na.omit(df1)   #as per given sample data it's empty
  

  Sample data

  df <- structure(list(A = c(24, 21, 12, 39, 51, 24, 48, 44, 14, 34), 
      B = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2), C = c(16, 19, 12, 39, 
      82, 24, 32, 44, 18, 51), D = c(5, 2, 79, 39, 27, 40, 5, 12, 
      6, 5), E = c(1.2, 1.09, 0.86, 1.9, 2.3, 1.6, 1.6, 1.7, 0.88, 
      2.7), F = c(6, 2, 2, 7, 1, 9, 1, 1, 1, 2)), class = "data.frame", row.names = c("1", 
  "2", "3", "4", "5", "6", "7", "8", "9", "10"))
  
  lookup_list <- list(c(15.7, 15.9, 16, 16.1, 16.2), c(0, 1, 2, 3, 4), c(15, 15.3, 
  16, 16.3, 16.5), c(3, 4, 5, 6, 7), c(1.08, 1.09, 1.1, 1.11, 1.12
  ), c(0, 1, 2, 3, 4))
  

  回复于 2024-04-28T11:59:52+08:00