Question

我有一个包含2列GL和GLDESC的数据框，并希望根据列KIND中的某些数据添加名为GLDESC的第3列。< / p>

数据框如下：

      GL                             GLDESC
1 515100         Payroll-Indir Salary Labor
2 515900 Payroll-Indir Compensated Absences
3 532300                           Bulk Gas
4 539991                     Area Charge In
5 551000        Repairs & Maint-Spare Parts
6 551100                 Supplies-Operating
7 551300                        Consumables

对于数据表的每一行：

如果GLDESC在字符串中的任何位置包含单词Payroll，那么我希望KIND为Payroll
如果GLDESC在字符串中的任何位置包含单词Gas，那么我希望KIND为Materials
在所有其他情况下，我希望KIND为Other

我在stackoverflow上查找了类似的示例，但找不到任何内容，也在R查看了switch，grep，apply和正则表达式上的虚拟对象，以尝试仅匹配GLDESC列的一部分然后使用该帐户填写KIND列，但无法使其正常工作。

Answer 1

由于您只有两个条件，因此可以使用嵌套的ifelse：

#random data; it wasn't easy to copy-paste yours  
DF <- data.frame(GL = sample(10), GLDESC = paste(sample(letters, 10), 
  c("gas", "payroll12", "GaSer", "asdf", "qweaa", "PayROll-12", 
     "asdfg", "GAS--2", "fghfgh", "qweee"), sample(letters, 10), sep = " "))

DF$KIND <- ifelse(grepl("gas", DF$GLDESC, ignore.case = T), "Materials", 
         ifelse(grepl("payroll", DF$GLDESC, ignore.case = T), "Payroll", "Other"))

DF
#   GL         GLDESC      KIND
#1   8        e gas l Materials
#2   1  c payroll12 y   Payroll
#3  10      m GaSer v Materials
#4   6       t asdf n     Other
#5   2      w qweaa t     Other
#6   4 r PayROll-12 q   Payroll
#7   9      n asdfg a     Other
#8   5     d GAS--2 w Materials
#9   7     s fghfgh e     Other
#10  3      g qweee k     Other

编辑 2016年10月3日 （......收到的注意力超出预期）

处理更多模式的可能解决方案可能是迭代所有模式，并且只要匹配，就会逐渐减少比较量：

ff = function(x, patterns, replacements = patterns, fill = NA, ...)
{
    stopifnot(length(patterns) == length(replacements))

    ans = rep_len(as.character(fill), length(x))    
    empty = seq_along(x)

    for(i in seq_along(patterns)) {
        greps = grepl(patterns[[i]], x[empty], ...)
        ans[empty[greps]] = replacements[[i]]  
        empty = empty[!greps]
    }

    return(ans)
}

ff(DF$GLDESC, c("gas", "payroll"), c("Materials", "Payroll"), "Other", ignore.case = TRUE)
# [1] "Materials" "Payroll"   "Materials" "Other"     "Other"     "Payroll"   "Other"     "Materials" "Other"     "Other"

ff(c("pat1a pat2", "pat1a pat1b", "pat3", "pat4"), 
   c("pat1a|pat1b", "pat2", "pat3"), 
   c("1", "2", "3"), fill = "empty")
#[1] "1"     "1"     "3"     "empty"

ff(c("pat1a pat2", "pat1a pat1b", "pat3", "pat4"), 
   c("pat2", "pat1a|pat1b", "pat3"), 
   c("2", "1", "3"), fill = "empty")
#[1] "2"     "1"     "3"     "empty"

基于与其他列匹配的部分字符串在数据框中创建新列

1 个答案: