将编号转换成R中的编号

有人知道将一个数字的文本表示形式转换为实际数字的功能,例如“二万三百五十”到20305年.我在数据帧行中编写了数字,并将其转换为数字.

在包qdap中,您可以用字代替数字表示的数字(例如,1001变为一千个),但不能相反：

library(qdap)
replace_number("I like 346457 ice cream cones.")
[1] "I like three hundred forty six thousand four hundred fifty seven ice cream cones."

解决方法

这是一个开始,应该让你成千上万.

word2num <- function(word){
    wsplit <- strsplit(tolower(word)," ")[[1]]
    one_digits <- list(zero=0,one=1,two=2,three=3,four=4,five=5,six=6,seven=7,eight=8,nine=9)
    teens <- list(eleven=11,twelve=12,thirteen=13,fourteen=14,fifteen=15,sixteen=16,seventeen=17,eighteen=18,nineteen=19)
    ten_digits <- list(ten=10,twenty=20,thirty=30,forty=40,fifty=50,sixty=60,seventy=70,eighty=80,ninety=90)
    doubles <- c(teens,ten_digits)
    out <- 0
    i <- 1
    while(i <= length(wsplit)){
        j <- 1
        if(i==1 && wsplit[i]=="hundred")
            temp <- 100
        else if(i==1 && wsplit[i]=="thousand")
            temp <- 1000
        else if(wsplit[i] %in% names(one_digits))
            temp <- as.numeric(one_digits[wsplit[i]])
        else if(wsplit[i] %in% names(teens))
            temp <- as.numeric(teens[wsplit[i]])
        else if(wsplit[i] %in% names(ten_digits))
            temp <- (as.numeric(ten_digits[wsplit[i]]))
        if(i < length(wsplit) && wsplit[i+1]=="hundred"){
            if(i>1 && wsplit[i-1] %in% c("hundred","thousand"))
                out <- out + 100*temp
            else
                out <- 100*(out + temp)
            j <- 2
        }
        else if(i < length(wsplit) && wsplit[i+1]=="thousand"){
            if(i>1 && wsplit[i-1] %in% c("hundred","thousand"))
                out <- out + 1000*temp
            else
                out <- 1000*(out + temp)
            j <- 2
        }
        else if(i < length(wsplit) && wsplit[i+1] %in% names(doubles)){
            temp <- temP*100
            out <- out + temp
        }
        else{
            out <- out + temp
        }
        i <- i + j
    }
    return(list(word,out))
}

结果：

> word2num("fifty seven")
[[1]]
[1] "fifty seven"

[[2]]
[1] 57

> word2num("four fifty seven")
[[1]]
[1] "four fifty seven"

[[2]]
[1] 457

> word2num("six thousand four fifty seven")
[[1]]
[1] "six thousand four fifty seven"

[[2]]
[1] 6457

> word2num("forty six thousand four fifty seven")
[[1]]
[1] "forty six thousand four fifty seven"

[[2]]
[1] 46457

> word2num("forty six thousand four hundred fifty seven")
[[1]]
[1] "forty six thousand four hundred fifty seven"

[[2]]
[1] 46457

> word2num("three forty six thousand four hundred fifty seven")
[[1]]
[1] "three forty six thousand four hundred fifty seven"

[[2]]
[1] 346457

我可以告诉你,这对于word2num(“四十万五十”)是不行的,因为它不知道如何处理连续的“百”和“千”字,但算法可以大概修改.任何人都应该随意修改,如果他们有改进或建立在他们自己的答案.我只是认为这是一个有趣的问题,玩一会儿.

编辑：显然,Bill Venables有一个名为english的包,可以实现这个比上面的代码更好.

原文地址：https://www.jb51.cc/html/229621.html