Question

我正在查看命令cut()（example(cut)）的示例菜单，特别是此部分：

cut> aaa <- c(1,2,3,4,5,2,3,4,5,6,7)

cut> cut(aaa, 3)
[1] (0.994,3] (0.994,3] (3,5]     (3,5]     (3,5]     (0.994,3]
[7] (3,5]     (3,5]     (3,5]     (5,7.01]  (5,7.01] 
Levels: (0.994,3] (3,5] (5,7.01]

cut> cut(aaa, 3, dig.lab = 4, ordered = TRUE)
[1] (0.994,2.998] (0.994,2.998] (2.998,5.002] (2.998,5.002]
[5] (2.998,5.002] (0.994,2.998] (2.998,5.002] (2.998,5.002]
[9] (2.998,5.002] (5.002,7.006] (5.002,7.006]
Levels: (0.994,2.998] < (2.998,5.002] < (5.002,7.006]

cut> ## one way to extract the breakpoints
cut> labs <- levels(cut(aaa, 3))

cut> cbind(lower = as.numeric( sub("\\((.+),.*", "\\1", labs) ),
cut+       upper = as.numeric( sub("[^,]*,([^]]*)\\]", "\\1", labs) ))
     lower upper
[1,] 0.994  3.00
[2,] 3.000  5.00
[3,] 5.000  7.01

如果区间在右侧关闭（如上所示），那么它会显示一种使用cbind()提取数据断点的方法

现在，我们假设我的数据将被剪切，但表示左侧的关闭。

cut(aaa, 3, dig.lab = 4, ordered = TRUE, right = FALSE)

如何使用相同的命令cbind()现在提取我的断点？（如果有更多方法，欢迎你）

Answer 1

只需对您的模式使用以下内容，然后使用gsub代替："\\[|\\]|\\(|\\)"。

一个例子。

out <- levels(cut(aaa, 3, dig.lab = 4, ordered = TRUE, right = FALSE))
gsub("\\[|\\]|\\(|\\)", "", out)
# [1] "0.994,2.998" "2.998,5.002" "5.002,7.006"

而且，这是一种快速读取数据的方法：

read.csv(text = gsub("\\[|\\]|\\(|\\)", "", out), header = FALSE)
#      V1    V2
# 1 0.994 2.998
# 2 2.998 5.002
# 3 5.002 7.006

仅供参考：无论区间是左侧还是右侧，都可以使用相同的模式。使用您的原始示例：

labs <- levels(cut(aaa, 3))
labs
# [1] "(0.994,3]" "(3,5]"     "(5,7.01]" 
read.csv(text = gsub("\\[|\\]|\\(|\\)", "", labs), header = FALSE)
#      V1   V2
# 1 0.994 3.00
# 2 3.000 5.00
# 3 5.000 7.01

至于替代方案，因为您只需要在使用read.csv之前删除第一个和最后一个字符，您也可以轻松使用substr而不必对正则表达式大惊小怪（如果不是你的东西）：

substr(labs, 2, nchar(labs)-1)
# [1] "0.994,3" "3,5"     "5,7.01"

更新：完全不同的替代

由于很明显R必须计算这些值并将它们存储为函数的一部分才能生成您看到的输出，因此操作函数以使其输出不同的东西并不太难。

查看cut.default的代码，您会发现以下几行：

if (codes.only) 
    code
else factor(code, seq_along(labels), labels, ordered = ordered_result)

很容易更改最后几行以输出list，其中包含cut的输出作为第一项，以及计算的范围（直接来自cut函数，没有从粘贴在一起factor labels中提取。

例如，in the Gist I've posted at this link，我已按以下方式更改了这些行：

if (codes.only) 
  FIN <- code
else FIN <- factor(code, seq_along(labels), labels, ordered = ordered_result)
list(output = FIN, ranges = data.frame(lower = ch.br[-nb], upper = ch.br[-1L]))

现在，比较：

cut(aaa, 3)
#  [1] (0.994,3] (0.994,3] (3,5]     (3,5]     (3,5]     (0.994,3] (3,5]     (3,5]    
#  [9] (3,5]     (5,7.01]  (5,7.01] 
# Levels: (0.994,3] (3,5] (5,7.01]
CUT(aaa, 3)
# $output
# [1] (0.994,3] (0.994,3] (3,5]     (3,5]     (3,5]     (0.994,3] (3,5]     (3,5]    
# [9] (3,5]     (5,7.01]  (5,7.01] 
# Levels: (0.994,3] (3,5] (5,7.01]
# 
# $ranges
#   lower upper
# 1 0.994     3
# 2     3     5
# 3     5  7.01

而且，right = FALSE：

cut(aaa, 3, dig.lab = 4, ordered = TRUE, right = FALSE)
#  [1] [0.994,2.998) [0.994,2.998) [2.998,5.002) [2.998,5.002) [2.998,5.002)
#  [6] [0.994,2.998) [2.998,5.002) [2.998,5.002) [2.998,5.002) [5.002,7.006)
# [11] [5.002,7.006)
# Levels: [0.994,2.998) < [2.998,5.002) < [5.002,7.006)
CUT(aaa, 3, dig.lab = 4, ordered = TRUE, right = FALSE)
# $output
#  [1] [0.994,2.998) [0.994,2.998) [2.998,5.002) [2.998,5.002) [2.998,5.002)
#  [6] [0.994,2.998) [2.998,5.002) [2.998,5.002) [2.998,5.002) [5.002,7.006)
# [11] [5.002,7.006)
# Levels: [0.994,2.998) < [2.998,5.002) < [5.002,7.006)

# $ranges
#   lower upper
# 1 0.994 2.998
# 2 2.998 5.002
# 3 5.002 7.006

在左侧关闭时间间隔提取断点

1 个答案:

更新：完全不同的替代