将书面数字转换为 R 中的数字

问题描述

有谁知道将数字的文本表示转换为实际数字的函数，例如'20305' 变成 20305.我在数据帧行中写了数字，想将它们转换为数字.

Does anybody know a function to convert a text representation of a number into an actual number, e.g. 'twenty thousand three hundred and five' into 20305. I have written numbers in dataframe rows and want to convert them to numbers.

在 qdap 包中，您可以用单词替换数字表示的数字(例如，1001 变成一千)，但反过来不行:

In package qdap, you can replace numeric represented numbers with words (e.g., 1001 becomes one thousand one), but not the other way around:

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."

推荐答案

这是一个可以让您达到数十万的开始.

Here's a start that should get you to hundreds of thousands.

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

我已经可以告诉你，这不适用于 word2num("40555")，因为它不知道如何处理连续的一百"和千"项，但算法可能会被修改.如果他们有改进或在他们自己的答案中建立这些改进，任何人都应该随意编辑它.我只是觉得这是一个有趣的问题(一小会儿).

I can tell you already that this won't work for word2num("four hundred thousand fifty"), because it doesn't know how to handle consecutive "hundred" and "thousand" terms, but the algorithm can be modified probably. Anyone should feel free to edit this if they have improvements or build on them in their own answer. I just thought this was a fun problem to play with (for a little while).

显然 Bill Venables 有一个名为 english 的包这可能比上面的代码更好地实现这一点.

Apparently Bill Venables has a package called english that may achieve this even better than the above code.

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