algorithm - 编程之谜：如何将 Excel 列名转换为数字？

Question

我最近在一次工作面试中被要求解决一个我认为分享会很有趣的编程难题。这是关于将 Excel 列字母转换为实际数字，如果你还记得的话，Excel 用从 A 到 Z 的字母命名它的列，然后序列为 AA、AB、AC... AZ、BA、BB 等。

您必须编写一个接受字符串作为参数的函数（如“AABCCE”）并返回实际的列号。

解决方案可以是任何语言。

score 36 · Accepted Answer

对我来说听起来像是一个标准的减少：

Python：

def excel2num(x): 
    return reduce(lambda s,a:s*26+ord(a)-ord('A')+1, x, 0)

C＃：

int ExcelToNumber(string x) {
    return x.Aggregate(0, (s, c) => s * 26 + c - 'A' + 1 );
}

score 16 · Accepted Answer

我很久以前为一些 Python 脚本写了这个：

def index_to_int(index):
    s = 0
    pow = 1
    for letter in index[::-1]:
        d = int(letter,36) - 9
        s += pow * d
        pow *= 26
    # excel starts column numeration from 1
    return s

score 6 · Accepted Answer

从 STDIN 中读取一个列名并打印出其对应的编号：

perl -le '$x = $x * 26 - 64 + ord for <> =~ /./g; print $x'

注意事项：假设为 ASCII。

编辑：替换"为'以便您的外壳不会$x在字符串中插入。

score 5 · Accepted Answer

巧合的是，我使用 javascript 解决了同样的问题

$(function() { //shorthand document.ready function
    var getNumber = function(x) {
        var result = 0;
        var multiplier = 1;
        for ( var i = x.length-1; i >= 0; i--)
        { 
            var value = ((x[i].charCodeAt(0) - "A".charCodeAt(0)) + 1);
            result = result + value * multiplier;
            multiplier = multiplier * 26;
        }
        return result;
    };
    
    $('#form').on('submit', function(e) { //use on if jQuery 1.7+
        e.preventDefault();  //prevent form from submitting
        var data = $("#number").val();
        $('#answer').text(getNumber(data));
    });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<form id="form">
<input type="text" id="number"></input>
    <button>submit</button>
</form>
<p id="answer"></p>

    var getNumber = function(x) {
        var result = 0;
        var multiplier = 1;
        for ( var i = x.length-1; i >= 0; i--)
        { 
            var value = ((x[i].charCodeAt(0) - "A".charCodeAt(0)) + 1);
            result = result + value * multiplier;
            multiplier = multiplier * 26;
        }
        return result;
    };

http://jsfiddle.net/M7Xty/1/

score 4 · Accepted Answer

哈 - 已经在我们的代码库中编写了它 - 大约 3 次不同的时间:(

%% @doc Convert an string to a decimal integer
%% @spec b26_to_i(string()) -> integer()

b26_to_i(List) when is_list(List) ->
    b26_to_i(string:to_lower(lists:reverse(List)),0,0).

%% private functions
b26_to_i([], _Power, Value) -> 
    Value;

b26_to_i([H|T],Power,Value)->
    NewValue = case (H > 96) andalso (H < 123) of
                   true ->
                       round((H - 96) * math:pow(26, Power));
                   _    ->
                       exit([H | T] ++ " is not a valid base 26 number")
               end,
    b26_to_i(T, Power + 1, NewValue + Value).

谜底是它实际上不是数字的 Base26 表示（我们在这里的函数名是自欺欺人），因为其中没有 0。

顺序为：A、B、C ... Z、AA、AB、AC

而不是：A、B、C ...Z、BA、BB、BC

（语言是 Erlang，mais oui）。

score 4 · Accepted Answer

您可以像这样在 C 中执行此操作：

unsigned int coltonum(char * string)
{
   unsigned result = 0;
   char ch;

   while(ch = *string++)
      result = result * 26 + ch - 'A' + 1;

  return result;
}

无错误检查，仅适用于大写字符串，字符串必须为空终止。

score 1 · Accepted Answer

假设列 A = 1

int GetColumnNumber(string columnName)
{
  int sum = 0;
  int exponent = 0;
  for(int i = columnName.Length - 1; i>=0; i--)
  {
    sum += (columnName[i] - 'A' + 1) *  (GetPower(26, exponent));
    exponent++;
  }
  return sum;
}

int GetPower(int number, int exponent)
{
  int power = 1;
  for(int i=0; i<exponent; i++)
    power *= number;
  return power;
}

score 1 · Accepted Answer

警告：这两个版本都假定只有大写字母 A 到 Z。其他任何内容都会导致计算错误。添加一些错误检查和/或大写来改进它们并不难。

斯卡拉

def excel2Number(excel : String) : Int = 
  (0 /: excel) ((accum, ch) => accum * 26 + ch - 'A' + 1)

哈斯克尔

excel2Number :: String -> Int
excel2Number = flip foldl 0 $ \accum ch -> accum * 26 + fromEnum ch - fromEnum 'A' + 1

score 1 · Accepted Answer

从名称中获取列号

爪哇：

public int getColNum (String colName) {

    //remove any whitespace
    colName = colName.trim();

    StringBuffer buff = new StringBuffer(colName);

    //string to lower case, reverse then place in char array
    char chars[] = buff.reverse().toString().toLowerCase().toCharArray();

    int retVal=0, multiplier=0;

    for(int i = 0; i < chars.length;i++){
        //retrieve ascii value of character, subtract 96 so number corresponds to place in alphabet. ascii 'a' = 97 
        multiplier = (int)chars[i]-96;
        //mult the number by 26^(position in array)
        retVal += multiplier * Math.pow(26, i);
    }
    return retVal;
}

score 1 · Accepted Answer

从 Java 中的 int 获取列名（在此处阅读更多内容）：

public String getColName (int colNum) {

   String res = "";

   int quot = colNum;
   int rem;        
    /*1. Subtract one from number.
    *2. Save the mod 26 value.
   *3. Divide the number by 26, save result.
   *4. Convert the remainder to a letter.
   *5. Repeat until the number is zero.
   *6. Return that bitch...
   */
    while(quot > 0)
    {
        quot = quot - 1;
        rem = quot % 26;
        quot = quot / 26;

        //cast to a char and add to the beginning of the string
        //add 97 to convert to the correct ascii number
        res = (char)(rem+97) + res;            
    }   
    return res;
}

score 1 · Accepted Answer

另一个德尔福：

function ExcelColumnNumberToLetter(col: Integer): string;
begin
  if (col <= 26) then begin
    Result := Chr(col + 64);
  end
  else begin
    col := col-1;
    Result := ExcelColumnNumberToLetter(col div 26) + ExcelColumnNumberToLetter((col mod 26) + 1);
  end;
end;

score 1 · Accepted Answer

另一个Java：

public static int convertNameToIndex(String columnName) {
    int index = 0;
    char[] name = columnName.toUpperCase().toCharArray();

    for(int i = 0; i < name.length; i++) {
        index *= 26;
        index += name[i] - 'A' + 1;
    }

    return index;
}

score 1 · Accepted Answer

简单的 Java 解决方案 -->

public class ColumnName {

public static int colIndex(String col)
{   int index=0;
    int mul=0;
    for(int i=col.length()-1;i>=0;i--)
    {   
        index  += (col.charAt(i)-64) * Math.pow(26, mul);
        mul++;
    }
    return index;
}

public static void main(String[] args) {

    System.out.println(colIndex("AAA"));

}

score 0 · Accepted Answer

0

将字符串视为以 A、B、... Z 表示的以 26 为底的列号的倒数是否有帮助？

于 2009-04-18T16:16:04.630 回答

score 0 · Accepted Answer

这基本上是一个以 26 为基数的数字，不同之处在于该数字不使用 0-9，然后是字母，而只使用字母。

score 0 · Accepted Answer

这是一个CFML：

<cffunction name="ColToNum" returntype="Numeric">
    <cfargument name="Input" type="String" />
    <cfset var Total = 0 />
    <cfset var Pos = 0 />

    <cfloop index="Pos" from="1" to="#Len(Arguments.Input)#">
        <cfset Total += 26^(Pos-1) * ( Asc( UCase( Mid(Arguments.Input,Pos,1) ) ) - 64 ) />
    </cfloop>

    <cfreturn Total />
</cffunction>

<cfoutput>
    #ColToNum('AABCCE')#
</cfoutput>

因为我心情很奇怪，所以这里有一个 CFScript 版本：

function ColToNum ( Input )
{
    var Total = 0;

    for ( var Pos = 1 ; Pos <= Len(Arguments.Input) ; Pos++ )
    {
        Total += 26^(Pos-1) * ( Asc( UCase( Mid(Arguments.Input,Pos,1) ) ) - 64 );
    }

    return Total;
}

WriteOutput( ColToNum('AABCCE') );

score 0 · Accepted Answer

普通 Lisp：

(defun excel->number (string)
  "Converts an Excel column name to a column number."
  (reduce (lambda (a b) (+ (* a 26) b))
          string
          :key (lambda (x) (- (char-int x) 64))))

编辑：逆运算：

(defun number->excel (number &optional acc)
  "Converts a column number to Excel column name."
  (if (zerop number)
      (concatenate 'string acc)
      (multiple-value-bind (rest current) (floor number 26)
        (if (zerop current)
            (number->excel (- rest 1) (cons #\Z acc))
            (number->excel rest (cons (code-char (+ current 64)) acc))))))

score 0 · Accepted Answer

稍微相关，更好的挑战是反过来：给定列号，找到作为字符串的列标签。

我为 KOffice 实现的 Qt 版本：

QString columnLabel( unsigned column )
{
  QString str;
  unsigned digits = 1;
  unsigned offset = 0;

  column--;
  for( unsigned limit = 26; column >= limit+offset; limit *= 26, digits++ )
    offset += limit;

  for( unsigned c = column - offset; digits; --digits, c/=26 )
    str.prepend( QChar( 'A' + (c%26) ) );

  return str;
}

score 0 · Accepted Answer

另一个[更神秘的] erlang 示例：

col2int(String) -> col2int(0,String).
col2int(X,[A|L]) when A >= 65, A =< 90 ->
col2int(26 * X + A - 65 + 1, L);
col2int(X,[]) -> X.

和反函数：

int2col(Y) when Y > 0 -> int2col(Y,[]).
int2col(0,L) -> L;
int2col(Y,L) when Y rem 26 == 0 -> 
   int2col(Y div 26 - 1,[(26+65-1)|L]);
int2col(Y,L) ->
   P = Y rem 26,
   int2col((Y - P) div 26,[P + 65-1|L]).

score 0 · Accepted Answer

德尔福：

// convert EXcel column name to column number 1..256
// case-sensitive; returns 0 for illegal column name
function cmColmAlfaToNumb( const qSRC : string ) : integer;
var II : integer;
begin
   result := 0;
   for II := 1 to length(qSRC) do begin
      if (qSRC[II]<'A')or(qSRC[II]>'Z') then begin
         result := 0;
         exit;
      end;
      result := result*26+ord(qSRC[II])-ord('A')+1;
   end;
   if result>256 then result := 0;
end;

-阿尔。

score 0 · Accepted Answer

def ExcelColumnToNumber(ColumnName):
    ColNum = 0
    for i in range(0, len(ColumnName)):
        # Easier once formula determined: 'PositionValue * Base^Position'
        # i.e. AA=(1*26^1)+(1*26^0)   or  792=(7*10^2)+(9*10^1)+(2*10^0)
        ColNum += (int(ColumnName[i],36) -9) * (pow(26, len(ColumnName)-i-1))
    return ColNum

ps 我的第一个 Python 脚本！

score 0 · Accepted Answer

此版本纯粹是功能性的，并允许替代“代码”序列，例如，如果您只想使用字母“A”到“C”。在 Scala 中，来自 dcsobral 的建议。

def columnNumber(name: String) = {
    val code = 'A' to 'Z'

    name.foldLeft(0) { (sum, letter) =>
        (sum * code.length) + (code.indexOf(letter) + 1)
    }
}

score 0 · Accepted Answer

0

在数学中：

FromDigits[ToCharacterCode@# - 64, 26] &

于 2011-05-17T07:46:07.837 回答

score 0 · Accepted Answer

使用 Mr. Wizard 令人敬畏的 Mathematica 代码，但摆脱了神秘的纯函数！

columnNumber[name_String] := FromDigits[ToCharacterCode[name] - 64, 26]

score 0 · Accepted Answer

维基百科有很好的解释和算法

http://en.wikipedia.org/wiki/Hexavigesimal

public static String toBase26(int value){
    // Note: This is a slightly modified version of the Alphabet-only conversion algorithm

    value = Math.abs(value);
    String converted = "";

    boolean iteration = false;

    // Repeatedly divide the number by 26 and convert the
    // remainder into the appropriate letter.
    do {
        int remainder = value % 26;

        // Compensate for the last letter of the series being corrected on 2 or more iterations.
        if (iteration && value < 25) {
            remainder--;
        }

        converted = (char)(remainder + 'A') + converted;
        value = (value - remainder) / 26;

        iteration = true;
    } while (value > 0);

    return converted;    
}

score 0 · Accepted Answer

…只需要PHP的解决方案。这就是我想出的：

/**
 * Calculates the column number for a given column name.
 *
 * @param string $columnName the column name: "A", "B", …, "Y", "Z", "AA", "AB" … "AZ", "BA", … "ZZ", "AAA", …
 *
 * @return int the column number for the given column name: 1 for "A", 2 for "B", …, 25 for "Y", 26 for "Z", 27 for "AA", … 52 for "AZ", 53 for "BA", … 703 for "AAA", …
 */
function getColumnNumber($columnName){
    //  the function's result
    $columnNumber = 0;

    //  at first we need to lower-case the string because we calculate with the ASCII value of (lower-case) "a"
    $columnName = strtolower($columnName);
    //  ASCII value of letter "a"
    $aAsciiValue = ord('a') - 1;

    //  iterate all characters by splitting the column name
    foreach (str_split($columnName) as $character) {
        //  determine ASCII value of current character and substract with that one from letter "a"
        $characterNumberValue = ord($character) - $aAsciiValue;

        //  through iteration and multiplying we finally get the previous letters' values on base 26
        //  then we just add the current character's number value
        $columnNumber = $columnNumber * 26 + $characterNumberValue;
    }

    //  return the result
    return $columnNumber;
}

当然，只需在 foreach 循环中将一些东西组合成一行代码，就可以稍微缩短脚本：

//  …
$columnNumber = $columnNumber * 26 + ord($character) - ord('a') + 1;
//  …

score 0 · Accepted Answer

在 Python 中，没有减少：

def transform(column_string):
    return sum((ascii_uppercase.index(letter)+1) * 26**position for position, letter in enumerate(column_string[::-1]))

score 0 · Accepted Answer

这是此代码在 Python 中的另一个版本：

keycode=1
for i in range (1,len(word)):
    numtest[i]=word[i-1]
    keycode = keycode*26*int(wordtest[numtest[i]])
last=word[-1:]
keycode=keycode+int(wordtest[last])
print(keycode)
print(bin(keycode))
#Numtest and wordtest are dictionaries.

algorithm - 编程之谜：如何将 Excel 列名转换为数字？

28 回答 28

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