Question

我有文本数据（在R中），并希望用数据框中的其他字符替换某些字符。我认为这将是一个简单的任务，使用空间上的strsplit并创建一个矢量，然后我可以使用匹配（％in％）然后可以粘贴在一起。但后来我想到了标点符号。句子的最后一个单词和结尾的标点符号之间没有空格。

我认为可能有一种更简单的方式来实现我想要的东西，而不是变成我的代码的令人费解的混乱。我很欣赏这个问题的方向。

#Character String
x <- "I like 346 ice cream cones.  They're 99 percent good!  I ate 46."

#Replacement Values Dataframe
  symbol text                     
1 "346"  "three hundred forty six"
2 "99"   "ninety nine"            
3 "46"   "forty six" 

#replacement dataframe
numDF <- 
data.frame(symbol = c("346","99", "46"),
           text = c("three hundred forty six", "ninety nine","forty six"),
           stringsAsFactors = FALSE)

期望的结果：

[1] "I like three hundred forty six ice cream cones.  They're ninety nine percent good!  You ate forty six?")

编辑：我原来认为这个有条件的gsub，因为即使没有涉及gsub，它对我来说也是如此。

Answer 1

在Josh O'Brien的回答的启发下，也许这样做：

x <- "I like 346 ice cream cones.  They're 99 percent good!  I ate 46."
numDF <- structure(c("346", "99", "46", "three hundred forty six", "ninety nine", 
"forty six"), .Dim = c(3L, 2L), .Dimnames = list(c("1", "2", 
"3"), c("symbol", "text")))

pat <-  paste(numDF[,"symbol"], collapse="|")
repeat {
    m <- regexpr(pat, x)
    if(m==-1) break
    sym <- regmatches(x,m)
    regmatches(x,m) <- numDF[match(sym, numDF[,"symbol"]), "text"]
}
x

Answer 2

此解决方案在同名包中使用gsubfn：

library(gsubfn)

(pat <-  paste(numDF$symbol, collapse="|"))
# [1] "346|99|46"

gsubfn(pattern = pat,
       replacement = function(x) {
           numDF$text[match(x, numDF$symbol)]
       },
       x)
[1] "I like three hundred forty six ice cream cones.  They're ninety nine percent good!  I ate forty six."

Answer 3

您可以拆分空白或字边界（在单词和标点符号之间匹配）：

> x
[1] "I like 346 ice cream cones.  They're 99 percent good!  I ate 46."
> strsplit(x, split='\\s|\\>|\\<')
[[1]]
 [1] "I"       "like"    "346"     "ice"     "cream"   "cones"   "."      
 [8] ""        "They"    "'re"     "99"      "percent" "good"    "!"      
[15] ""        "I"       "ate"     "46"      "."

然后你可以做替换。

Answer 4

使用Reduce中的base的另一种解决方案。

list_df <- apply(numDF, 1, as.list)
Reduce(function(x, l) gsub(l$symbol, l$text, x), list_df, init = x)

EDIT。以下是直接使用numbers2words函数的完整解决方案..

list_df <- as.numeric(regmatches(x, gregexpr('[0-9]+', x))[[1]])
Reduce(function(x, l) gsub(l, numbers2words(l), x), list_df, init = x)

Answer 5

目前还不清楚你是否真的想将数字转换成它们的alpha等价物。如果是这样，那么这是一个更为一般的策略。在rhelp档案中有（至少）两个数字到文本转换函数：Jim Lemon的digits2text和John Fox的numberstowords。我还切换到gregexpr以获得矢量化方法：

剪切和粘贴Lemon's function from the HTML found here开箱即用：

>     m <- gregexpr("[0-9]+", x)
>     sym <- regmatches(x,m)
>     regmatches(x,m) <- digits2text(as.numeric(sym[[1]]))
illion = 0 
digilen = 3 
digitext = three hundred forty six 
[1] 6 4 3
> 
> x
[1] "I like three hundred forty six ice cream cones.  They're three hundred forty six percent good!  I ate three hundred forty six."

我需要对数字元素进行一些编辑，因为有一些缺少的换行符搞砸了解析（我在此演示下面包含了成功的版本：

>     m <- gregexpr("[0-9]+", x)
>     sym <- regmatches(x,m)
>     regmatches(x,m) <- numbers2words(as.numeric(sym[[1]]))
> 
> x
[1] "I like three hundred forty six ice cream cones.  They're three hundred forty six percent good!  I ate three hundred forty six."

Fox的功能是从http://tolstoy.newcastle.edu.au/R/help/05/04/2715.html

编辑的

numbers2words <- function(x){

    helper <- function(x){

        digits <- rev(strsplit(as.character(x), "")[[1]])
        nDigits <- length(digits)
        if (nDigits == 1) as.vector(ones[digits])
        else if (nDigits == 2)
            if (x <= 19) as.vector(teens[digits[1]])
                else trim(paste(tens[digits[2]], 
                           Recall(as.numeric(digits[1]))))
        else if (nDigits == 3) trim(paste(ones[digits[3]], "hundred", 
            Recall(makeNumber(digits[2:1]))))
        else {
            nSuffix <- ((nDigits + 2) %/% 3) - 1
            if (nSuffix > length(suffixes)) stop(paste(x, "is too large!"))
            trim(paste(Recall(makeNumber(digits[
                nDigits:(3*nSuffix + 1)])),
                suffixes[nSuffix],  
                Recall(makeNumber(digits[(3*nSuffix):1]))))
            }
        }
    trim <- function(text){
        gsub("^\ ", "", gsub("\ *$", "", text))
        }      


    makeNumber <- function(...) as.numeric(paste(..., collapse=""))
     opts <- options(scipen=100)
    on.exit(options(opts))
    ones <- c("", "one", "two", "three", "four", "five", "six", "seven",

        "eight", "nine")
    names(ones) <- 0:9
    teens <- c("ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",

        "sixteen", " seventeen", "eighteen", "nineteen")
     names(teens) <- 0:9
    tens <- c("twenty", "thirty", "forty", "fifty", "sixty",
                 "seventy", "eighty", "ninety")
    names(tens) <- 2:9
    x <- round(x)
    suffixes <- c("thousand", "million", "billion", "trillion")
     if (length(x) > 1) return(sapply(x, helper))
     helper(x)
    }

有条件的gsub替换

5 个答案: