将十进制转换为二进制幂 bi

Posted 2023-04-14

【中文标题】将十进制转换为二进制幂 bi

【英文标题】：Converting Decimal to Binary Power bi

【发布时间】：2017-10-23 05:19:03

【问题描述】：

我有一个包含十进制数字 1 到 14 的列名 Config。我想创建一个新列 config2 并将 config 列转换为二进制 (base2)。

例如。 config2 =（配置二进制（base2））

 *Config |Config2*
     1   |0001
     2   |0010
     3   |0011

这是我的数据的样子

【参考方案1】：

您可以在 Power Query / M 中使用递归来实现。

Bin = (t as text, n as number) => 
    if n <= 1 
    then Text.From(n) & t
    else @Bin(Text.From(Number.Mod(n, 2)) & t, Number.RoundDown(n/2))

注意Bin 之前的@，它可以使递归发生。

结果：

我的查询供您参考：

let
    Bin = (t as text, n as number) => if n <= 1 then Text.From(n) & t else @Bin(Text.From(Number.Mod(n, 2)) & t, Number.RoundDown(n/2)),

    Source = 0..14,
    #"Converted to Table" = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
    #"Renamed Columns" = Table.RenameColumns(#"Converted to Table","Column1", "Config"),
    #"Changed Type" = Table.TransformColumnTypes(#"Renamed Columns","Config", Int64.Type),
    #"Added New Column" = Table.AddColumn(#"Changed Type", "Config2", each Bin("", [Config]))
in
    #"Added New Column"

---

如果您需要将前导零填充到 Config2，那么您将需要以下 DAX：

Formatted Config2 = FORMAT(VALUE(Query1[Config2]), "0000")

结果：

【参考方案2】：

下面的函数能够将数值转换为字符串，用另一个基数表示数字，反之亦然。您可以使用碱基 2-16、32 和 64。

示例，如果您将函数命名为 NumberBaseConversion：

= NumberBaseConversion(12, 2, 5) 返回“01100”

= NumberBaseConversion("AB", 16) 返回 171

(input as anynonnull, base as number, optional outputlength as number) as any =>
let
    //    input = 10,
    //    base = 2,
    //    outputlength = null,
    Base16 = "0123456789ABCDEF",
    Base32 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ234567",
    Base64 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",
    Lookups = List.Zip(16,32,64,Base16,Base32,Base64),
    Lookup = Text.ToList(List.Last(List.Select(Lookups,each _0 <= List.Max(16, base)))1),
    InputToList = Text.ToList(input),

    // This part will be executed if input is text:
        Reversed = List.Reverse(InputToList),
        BaseValues = List.Transform(Reversed, each List.PositionOf(Lookup,_)),
        Indexed = List.Zip(BaseValues, 0..Text.Length(input)-1),
        Powered = List.Transform(Indexed, each _0*Number.Power(base,_1)),
        Decimal = List.Sum(Powered),
    // So far this part

    // This part will be executed if input is not text:
        Elements = 1+Number.RoundDown(Number.Log(input,base),0),
        Powers = List.Transform(List.Reverse(0..Elements - 1), each Number.Power(base,_)),
        ResultString = List.Accumulate(Powers,
                                      [Remainder = input,String = ""], 
                                      (c,p) => [Remainder = c[Remainder] - p * Number.RoundDown(c[Remainder] / p,0),
                                                String = c[String] & LookupNumber.RoundDown(c[Remainder]/p,0)])[String],    
        PaddedResultString = if outputlength = null then ResultString else Text.PadStart(ResultString,outputlength,Lookup0),
    // So far this part

    Result = if input is text then Decimal else PaddedResultString
in
    Result