Question

我有一个像这样的数据框：

df
      A     B     C     D     E     F
   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 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.
# ... with 4,688 more rows

我想根据列表过滤此数据框，这样对于每列df，最小值和最大值将根据列表的最小值和最大值Neighb：

[[1]]
[1] 15.7 15.9 16.0 16.1 16.2

[[2]]
[1] 0 1 2 3 4

[[3]]
[1] 15.0 15.3 16.0 16.3 16.5

[[4]]
[1] 3 4 5 6 7

[[5]]
[1] 1.08 1.09 1.10 1.11 1.12

[[6]]
[1] 0 1 2 3 4

有没有办法用dplyr / base R有效地做到这一点？到目前为止，我一直使用循环并过滤每列df

Answer 1

我们可以使用Map

中的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＆＃39; Neighb＆＃39;

的比较而拥有全部TRUE的行

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

如果是任何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))

Answer 2

您可以使用map2中的purrr和between中的dplyr来获得您想要的结果。

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))

Answer 3

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)

这给出了：

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

现在，您可以使用df - 功能：

过滤rowSums

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

在呈现的示例数据（df）上应用此选项将导致空数据帧，但在原始数据集上很可能会导致非空数据帧。

使用过的数据：

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))

Answer 4

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

示例数据

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)), 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))

根据最小值和最大值过滤数据帧

4 个答案:

数据