高尔夫代码:友好号码的缩写 [英] Code-Golf: Friendly Number Abbreviator

问题描述

基于此问题: 挑战-已更新！ (已从规范中删除了数百个缩写)

按字符计数的最短代码，它将缩写为整数(无小数).

The shortest code by character count that will abbreviate an integer (no decimals).

代码应包含完整程序.

相关范围从0 - 9,223,372,036,854,775,807(带符号的64位整数的上限)开始.

Relevant range is from 0 - 9,223,372,036,854,775,807 (the upper limit for signed 64 bit integer).

缩写的小数位数为正. 您将不需要计算以下内容:920535 abbreviated -1 place(类似于0.920535M).

The number of decimal places for abbreviation will be positive. You will not need to calculate the following: 920535 abbreviated -1 place (which would be something like 0.920535M).

千万不要缩写成几百位数(0-999)(57到1+小数位的缩写是5.7dk-不必要和不友好).

Numbers in the tens and hundreds place (0-999) should never be abbreviated (the abbreviation for the number 57 to 1+ decimal places is 5.7dk - it is unneccessary and not friendly).

请记住从零舍入一半(23.5舍入为24).银行家的舍入是固定的.

Remember to round half away from zero (23.5 gets rounded to 24). Banker's rounding is verboten.

以下是相关的数字缩写:

Here are the relevant number abbreviations:

h = hundred (10 2 )
k = thousand (10 3 )
M = million (10 6 )
G = billion (10 9 )
T = trillion (10 12 )
P = quadrillion (10 15 )
E = quintillion (10 18 )

h = hundred (102)
k = thousand (103)
M = million (106)
G = billion (109)
T = trillion (1012)
P = quadrillion (1015)
E = quintillion (1018)

样品输入/输出(输入 可以作为单独的参数传递):

SAMPLE INPUTS/OUTPUTS (inputs can be passed as separate arguments):

第一个参数是要缩写的整数.第二个是小数位数.

First argument will be the integer to abbreviate. The second is the number of decimal places.

12 1                  => 12 // tens and hundreds places are never rounded
1500 2                => 1.5k
1500 0                => 2k // look, ma! I round UP at .5
0 2                   => 0
1234 0                => 1k
34567 2               => 34.57k
918395 1              => 918.4k
2134124 2             => 2.13M
47475782130 2         => 47.48G
9223372036854775807 3 => 9.223E
// ect...

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相关问题的原始答案(JavaScript，不遵循规范):

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Original answer from related question (JavaScript, does not follow spec):

function abbrNum(number, decPlaces) {
    // 2 decimal places => 100, 3 => 1000, etc
    decPlaces = Math.pow(10,decPlaces);

    // Enumerate number abbreviations
    var abbrev = [ "k", "m", "b", "t" ];

    // Go through the array backwards, so we do the largest first
    for (var i=abbrev.length-1; i>=0; i--) {

        // Convert array index to "1000", "1000000", etc
        var size = Math.pow(10,(i+1)*3);

        // If the number is bigger or equal do the abbreviation
        if(size <= number) {
             // Here, we multiply by decPlaces, round, and then divide by decPlaces.
             // This gives us nice rounding to a particular decimal place.
             number = Math.round(number*decPlaces/size)/decPlaces;

             // Add the letter for the abbreviation
             number += abbrev[i];

             // We are done... stop
             break;
        }
    }

    return number;
}

推荐答案

J，61个 63 65 个字符

J, 61 63 65 characters

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.)

输出:

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.) 1500 0
┌─┬─┐
│2│k│
└─┴─┘

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.) 987654321987654321 4
┌────────┬─┐
│987.6543│P│
└────────┴─┘

(输出之所以被装箱"，是因为J不支持包含不同类型的列表)

(The reason the output is "boxed" like that is because J doesn't support a list consisting of varying types)

说明(从右到左):

(([:<.1000^.{.),{:,{.)

使用,联接([:<.1000^.{.)(第一个参数{.的下层<.以1000为底的对数^.)，我们创建了一个新的3元素列表.我们将其与第二个参数，然后是第一个参数{..

We make a new 3-element list, using , to join ([:<.1000^.{.) (the floored <. base 1000 log ^. of the first param {.. We join it with the second param {: and then the first param {..

因此，在第一行之后，我们将12345 2转换为1 2 12345

So after the first bit, we've transformed say 12345 2 into 1 2 12345

((j.&(1&{)":({.%&1000{:));{&' kMGTPE'@{.)使用;将表达式的两半结合在一个盒子中以产生最终输出.

((j.&(1&{)":({.%&1000{:));{&' kMGTPE'@{.) uses ; to join the two halves of the expression together in a box to produce the final output.

前半部分是((j.&(1&{)":({.%&1000{:))，它将最后一个输入数字({:)除以(%)，即第一次输入的次数.然后，它使用输入列表中的第二个数字(1&{)设置精度":.

The first half is ((j.&(1&{)":({.%&1000{:)) which divides (%) the last input number ({:) by 1000, the first number of times. Then it sets the precision ": using the second number in the input list (1&{).

后半部{&' kMGTPE'@{.-它使用第一个数字从0索引的缩写列表中选择({)适当的字符.

The second half {&' kMGTPE'@{. - this uses the first number to select ({) the appropriate character from the 0-indexed list of abbreviations.

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