用千位后的逗号分离重构复杂因子向量

Posted 2021-04-04

我想重新格式化因子向量，以便它包含的数字有一千个分隔符。向量包含整数和实数，而没有关于值或顺序的任何特定规则。

Data

特别是，我正在使用类似于下面生成的矢量vec：

content <- c("0 - 100", "0 - 100", "0 - 100", "0 - 100",
             "150.22 - 170.33",
             "1000 - 2000","1000 - 2000", "1000 - 2000", "1000 - 2000", 
             "7000 - 10000", "7000 - 10000", "7000 - 10000", "7000 - 10000",
             "7000 - 10000", "1000000 - 22000000", "1000000 - 22000000", 
             "1000000 - 22000000",
             "44000000 - 66000000.8989898989")

vec <- factor(x = content, levels = unique(content))

Desired results

我的目标是重新格式化此向量，以便数字包含类似Excel的1,000分隔符，如下例所示：

100.00 1,000.00 1,000,000.00 1,000,000.56 24,564,000,000.56

---

Tried approach

我正在考虑使用gsubfn和一个可以传递数字的原型对象。然后可能创建另一个3位数的原型对象并替换。如下面的代码所示：

gsubfn(pattern = "[0-9][0-9][0-9]", replacement = ~paste0(x, ','), 
       x = as.character(vec))

这只是部分插入逗号：

"150,.22 - 170,.33"

这显然是错的。我还必须将角色向量转换为因子。因此，我的问题归结为两个要素：

• 我该如何解决逗号问题？
• 如何保持因子的原始结构？ - 我需要以与原始序列相同的方式排序因子向量，但在正确的位置使用逗号。

答案

仅在levels上运行似乎保持您的精度水平，而不是将您的矢量转换为character矢量并且效率更高，因为它将您操作的数据的大小减小到唯一值（而不是整个矢量）

levels(vec) <- sapply(strsplit(levels(vec), " - "), 
                       function(x) paste(prettyNum(x, 
                                            big.mark = ",", 
                                            preserve.width = "none"), 
                                   collapse = " - "))
vec
#  [1] 0 - 100                            0 - 100                            0 - 100                            0 - 100                            150.22 - 170.33                   
#  [6] 1,000 - 2,000                      1,000 - 2,000                      1,000 - 2,000                      1,000 - 2,000                      7,000 - 10,000                    
# [11] 7,000 - 10,000                     7,000 - 10,000                     7,000 - 10,000                     7,000 - 10,000                     1,000,000 - 22,000,000            
# [16] 1,000,000 - 22,000,000             1,000,000 - 22,000,000             44,000,000 - 66,000,000.8989898989
# Levels: 0 - 100 150.22 - 170.33 1,000 - 2,000 7,000 - 10,000 1,000,000 - 22,000,000 44,000,000 - 66,000,000.8989898989 

另一答案

使用正向前瞻性正则表达式...

content <- c("0 - 100", "0 - 100", "0 - 100", "0 - 100",
              "1000 - 2000","1000 - 2000", "1000 - 2000", "1000 - 2000", 
              "7000 - 10000", "7000 - 10000", "7000 - 10000", "7000 - 10000",
              "7000 - 10000", "1000000 - 22000000", "1000000 - 22000000", 
              "1000000 - 22000000")
gsub("(\d)(?=(?:\d{3})+\b)", "\1,", content, perl=T)
# [1] "0 - 100"                "0 - 100"                "0 - 100"               
# [4] "0 - 100"                "1,000 - 2,000"          "1,000 - 2,000"         
# [7] "1,000 - 2,000"          "1,000 - 2,000"          "7,000 - 10,000"        
# [10] "7,000 - 10,000"         "7,000 - 10,000"         "7,000 - 10,000"        
# [13] "7,000 - 10,000"         "1,000,000 - 22,000,000" "1,000,000 - 22,000,000"
# [16] "1,000,000 - 22,000,000"

另一答案

也许你可以使用formatC：

sapply(
  X = lapply(
    X = strsplit(x = content, split = " - "),
    FUN = function(x) {
      formatC(x = as.numeric(x), format = "f", flag = "#", big.mark = ",", 
              decimal.mark = ".", digits = 2, drop0trailing = FALSE)
    }
  ),
  FUN = paste, collapse = " - "
)
# [1] "0.00 - 100.00"                 "0.00 - 100.00"                 "0.00 - 100.00"                
# [4] "0.00 - 100.00"                 "150.22 - 170.33"               "1,000.00 - 2,000.00"          
# [7] "1,000.00 - 2,000.00"           "1,000.00 - 2,000.00"           "1,000.00 - 2,000.00"          
# [10] "7,000.00 - 10,000.00"          "7,000.00 - 10,000.00"          "7,000.00 - 10,000.00"         
# [13] "7,000.00 - 10,000.00"          "7,000.00 - 10,000.00"          "1,000,000.00 - 22,000,000.00" 
# [16] "1,000,000.00 - 22,000,000.00"  "1,000,000.00 - 22,000,000.00"  "44,000,000.00 - 66,000,000.90"