Question

我想编写一个函数将excel列名转换为相应的数字。到目前为止，我想出的只是部分工作。也就是说，低位字母首先出现的输入（＆＃34; AB＆＃34;，AC＆＃34;等）工作正常。但它反过来不起作用（＆＃34; BA＆＃34;，＆＃34; CA＆＃34;等）。我已经将错误跟踪到了行y <- which(base::LETTERS==x)，但我真的不明白这些布尔运算符如何处理向量。有什么建议吗？

#so to pass excel column-names directly, this function should do the trick
LettersToNumbers <- function(input){
    x <- toupper(substring(input, c(1:nchar(input)), c(1:nchar(input)))) #parse input-string
    y <- which(base::LETTERS==x) #letters to numbers
    y <- rev(y) #reverse
    #base26 conversion:
    result <- 0
    for (i in 1:length(y)){
        result <- result + ( y[i]*26^(i-1) )
    }
    return(result)
}

实际上事实证明还有一些例子不起作用。这里有一些，我真的不明白发生了什么。

> which(LETTERS==c("A", "B"))
[1] 1 2
> which(LETTERS==c("A", "C"))
[1] 1
> which(LETTERS==c("A", "D"))
[1] 1 4
> which(LETTERS==c("D", "A"))
integer(0)
>

Answer 1

这很快又脏，但我觉得它可以让你得到你想要的东西。它应该适用于任意字符串长度。

# Input: A string of letters s
# Output: Corresponding column number
LettersToNumbers <- function(s){
  # Uppercase
  s_upper <- toupper(s)
  # Convert string to a vector of single letters
  s_split <- unlist(strsplit(s_upper, split=""))
  # Convert each letter to the corresponding number
  s_number <- sapply(s_split, function(x) {which(LETTERS == x)})
  # Derive the numeric value associated with each letter
  numbers <- 26^((length(s_number)-1):0)
  # Calculate the column number
  column_number <- sum(s_number * numbers)
  column_number
}
# Vectorize in case you want to pass more than one column name in a single call
LettersToNumbers <- Vectorize(LettersToNumbers)

# Quick tests
LettersToNumbers("A")
LettersToNumbers("Z")
LettersToNumbers("AA")
LettersToNumbers("BA")
LettersToNumbers("AAA")
LettersToNumbers(LETTERS)

如上面的评论中所述，您的代码的主要问题是矢量回收，此功能通过使用sapply来避免。

Answer 2

这些情况下的答案通常是使用%in%，而不是==。例如

which(LETTERS %in% c("D", "A"))

生成1 4。它们不是你想要的顺序 - 所以，这将逐个应用这个功能。

sapply(c("D", "A"), function(x){which(LETTERS %in% x)})

生成4 1。

Answer 3

检查字符元素的长度，然后在LETTERS中添加由位置确定的位置：

TwoLet2Num <- function(chars) { if( nchar( substr(chars,2,2)) ){ 
         res <- which(LETTERS==substr(chars,1,1))*26 + which(LETTERS ==substr(chars,2,2)) 
          } else { res= which(LETTERS==substr(chars,1,1) ) } 
         return(res)}

Answer 4

对于那些想要来回转换的人（都在一个函数中）：输入数字（27），输出字母（'AA'）输入字母（'AA'），输出数字（27）

xlcolconv <- function(col){
    # test: 1 = A, 26 = Z, 27 = AA, 703 = AAA
    if (is.character(col)) {
        # codes from https://stackoverflow.com/a/34537691/2292993
        s = col
        # Uppercase
        s_upper <- toupper(s)
        # Convert string to a vector of single letters
        s_split <- unlist(strsplit(s_upper, split=""))
        # Convert each letter to the corresponding number
        s_number <- sapply(s_split, function(x) {which(LETTERS == x)})
        # Derive the numeric value associated with each letter
        numbers <- 26^((length(s_number)-1):0)
        # Calculate the column number
        column_number <- sum(s_number * numbers)
        return(column_number)
    } else {
        n = col
        letters = ''
        while (n > 0) {
            r = (n - 1) %% 26  # remainder
            letters = paste0(intToUtf8(r + utf8ToInt('A')), letters) # ascii
            n = (n - 1) %/% 26 # quotient
        }
        return(letters)
    }
}

Answer 5

    # Setup converter index numbers
    converter <- 1:702
    # Excel column names in order
    names(converter) <- do.call(paste0, expand.grid(LETTERS, c("",LETTERS))[,2:1])
    ExcelColumnNames <- c("A", "Z", "AA", "AZ", "ZZ")
    converter[ExcelColumnNames] # show excel column numbers
#  A   Z  AA  AZ  ZZ 
#  1  26  27  52 702

Answer 6

一种更快和（已经）矢量化的替代解决方案[不需要Vectorize]

letters2numbers <- function(x){
  
  # letters encoding
  encoding <- setNames(seq_along(LETTERS), LETTERS)
  
  # uppercase
  x <- toupper(x)
  
  # convert string to a list of vectors of single letters
  x <- strsplit(x, split = "")
  
  # convert each letter to the corresponding number
  # calculate the column number
  # return a numeric vector
  sapply(x, function(xs) sum(encoding[xs] * 26^((length(xs)-1):0)))
  
}


letters2numbers("Z")
#> [1] 26
letters2numbers(c("A", "BZ", "CBA", "BDWGN"))
#> [1]       1      78    2081 1000000

基准：

microbenchmark::microbenchmark(
  LettersToNumbers("Z"),
  letters2numbers("Z")
)
#> Unit: microseconds
#>                   expr    min      lq     mean  median     uq     max neval
#>  LettersToNumbers("Z") 60.510 61.9065 70.23292 64.0005 67.957 262.051   100
#>   letters2numbers("Z") 20.481 21.4115 26.70360 22.3420 24.204 140.568   100

microbenchmark::microbenchmark(
  LettersToNumbers(c("A", "BZ", "CBA", "BDWGN")),
  letters2numbers(c("A", "BZ", "CBA", "BDWGN"))
)
#> Unit: microseconds
#>                                            expr     min      lq      mean   median       uq     max neval
#>  LettersToNumbers(c("A", "BZ", "CBA", "BDWGN")) 152.669 158.721 206.97909 171.7530 220.8595 581.819   100
#>   letters2numbers(c("A", "BZ", "CBA", "BDWGN"))  30.255  32.582  42.47789  35.1425  43.9865 174.547   100

将excel列名转换为数字

6 个答案: