pla*_*er0 20 string-formatting google-sheets number-formatting google-apps-script array-formulas
有没有办法在谷歌表格中自定义大数字的格式(两种方式至少最多 10^100):
thousands > K
millions > M
billions > B
trillions > T
etc...
negative quadrillions > Q
decillions > D
Run Code Online (Sandbox Code Playgroud)
可以通过:
10000.1 10.0K
100 100.0
1000 1.0K
10000 10.0K
-100000 -100.0K
45646454 45.6M
5654894844216 5.7T
4655454544 4.7B
46546465455511 46.5T
-46546465455511 -46.5T
4654646545551184854556546454454400000000000000000000000000010000000 4.7U
-1000.9999 -1.0K
-100.8989 -100.9
-20.354 -20.4
1.03 1.0
22E+32 2.2D
Run Code Online (Sandbox Code Playgroud)
pla*_*er0 37
遗憾的是,谷歌表格中的内部格式默认只能处理 3 种类型的数字:
可以调整它以显示自定义格式,例如数千K,百万M和常规小数字:
[>999999]0.0,,"M";[>999]0.0,"K";0
Run Code Online (Sandbox Code Playgroud)
或者只有数千K、数百万M、数十亿B
[<999950]0.0,"K";[<999950000]0.0,,"M";0.0,,,"B"
Run Code Online (Sandbox Code Playgroud)
或者只有负千K、负数百万M、负数十亿B
[>-999950]0.0,"K";[>-999950000]0.0,,"M";0.0,,,"B"
Run Code Online (Sandbox Code Playgroud)
或者只有数百万M、数十亿B、数万亿T:
[<999950000]0.0,,"M";[<999950000000]0.0,,,"B";0.0,,,,"T"
Run Code Online (Sandbox Code Playgroud)
或仅从负数百万M到正数百万的数字M:
[>=999950]0.0,,"M";[<=-999950]0.0,,"M";0.0,"K"
Run Code Online (Sandbox Code Playgroud)
但你总是只有 3 个插槽可以使用,这意味着你不能拥有数万亿个插槽作为第四种类型/插槽。仅供参考,第四个插槽存在,但它是为文本保留的。要了解有关 Google 表格中内部格式的更多信息,请参阅:
公式方法更加通用...首先,您需要决定要使用的系统/标准(美国、欧洲、希腊、国际、非官方等...):
之后尝试:
=INDEX(REGEXREPLACE(IFNA(TEXT(A:A/10^(VLOOKUP(LEN(TEXT(INT(ABS(A:A)), "0"))-1,
SEQUENCE(35, 1,, 3), 1, 1)), "#.0")&VLOOKUP(ABS(A:A)*1, {{10^SEQUENCE(34, 1, 3, 3)},
{"K "; "M "; "B "; "T "; "Qa "; "Qi "; "Sx "; "Sp "; "O "; "N "; "D "; "Ud ";
"Dd "; "Td "; "Qad"; "Qid"; "Sxd"; "Spd"; "Od "; "Nd "; "V "; "Uv "; "Dv "; "Tv ";
"Qav"; "Qiv"; "Sxv"; "Spv"; "Ov "; "Nv "; "Tr "; "Ut "; "Dt "; "Tt "}}, 2, 1),
IF(ISBLANK(A:A),, TEXT(A:A, "0.0 "))), "^0\.0 $", "0 "))
Run Code Online (Sandbox Code Playgroud)
如果您对它的工作原理感兴趣,可以加分......
让我们从虚拟数组开始{{},{}}。SEQUENCE(34, 1, 3, 3)将为我们提供从 number 开始、步长为 number 的列中的34数字:133
10这些将在幂上升时用作指数^
所以我们的虚拟数组将是:
接下来,我们将其插入作为VLOOKUP检查ABSA 列的绝对值(将负值转换为正值)乘以的第二个参数*1,以防 A 列的值不是数字。通过VLOOKUP我们返回第二2列,作为第四个参数,我们使用近似模式1
从 -999 到 999 的数字此时会故意出错,因此我们稍后可以使用翻译IFNA来“修复”我们的错误IF(A:A=IF(,,),, TEXT(A:A, "#.0 ")):如果范围 A:A 确实为空,则不输出ISBLANK任何内容,否则使用提供的模式格式化 A 列,#.0例如。如果单元格 A5 = 空,则输出将为空白单元格...如果 -999 < A5=50 < 999,则输出将为 50.0
最后一部分:
TEXT(A:A/10^(VLOOKUP(LEN(TEXT(INT(ABS(A:A)), "0"))-1,
SEQUENCE(35, 1,, 3), 1, 1)), "#.0")
Run Code Online (Sandbox Code Playgroud)
ABS(A:A)将负数转换为正数。INT删除小数(如果有)。TEXT(, "0")将科学计数法转换3E+8为常规数字300000000。LEN来数数字。-1纠正以 10 为基数的表示法。VLOOKUP上面构造的数字为SEQUENCE列35中的数字1,这次从数字 0 开始,,,步长为3数字。以vlookup 的近似模式返回VLOOKUP第一1列(例如序列) 。当上升功率1时插入这个数字作为指数。并取 A 列中的值并将其除以上述构造的特定指数幂的数字。最后,将其格式化为10^10^TEXT#.0
要将丑陋变为0.0美丽,0我们只需使用REGEXREPLACE. 使用andINDEX代替较长的ARRAYFORMULA.
旁注:要删除尾随空格(可以添加漂亮的对齐方式,哈哈),可以从公式中删除它们,也可以TRIM在INDEX.
感谢@TheMaster对此进行报道
这是它的一个模组:
/**
* formats various numbers according to the provided short format
* @customfunction
* @param {A1:C100} range a 2D array
* @param {[X1:Y10]} database [optional] a real/virtual 2D array
* where the odd column holds exponent of base 10
* and the even column contains format suffixes
* @param {[5]} value [optional] fix suffix to fixed length
* by padding spaces (only if the second parameter exists)
*/
// examples:
// =CSF(A1:A)
// =CSF(2:2; X5:Y10)
// =CSF(A1:3; G10:J30)
// =CSF(C:C; X:Y; 2) to use custom alignment
// =CSF(C:C; X:Y; 0) to remove alignment
// =INDEX(TRIM(CSF(A:A))) to remove alignment
// =CSF(B10:D30; {3\ "K"; 4\ "TK"}) for non-english sheets
// =CSF(E5, {2, "deci"; 3, "kilo"}) for english sheets
// =INDEX(IF(ISERR(A:A*1); A:A; CSF(A:A))) to return non-numbers
// =INDEX(IF((ISERR(A:A*1))+(ISBLANK(A:A)), A:A, CSF(A:A*1))) enforce mode
function CSF(
range,
database = [
[3, 'K' ], //Thousand
[6, 'M' ], //Million
[9, 'B' ], //Billion
[12, 'T' ], //Trillion
[15, 'Qa' ], //Quadrillion
[18, 'Qi' ], //Quintillion
[21, 'Sx' ], //Sextillion
[24, 'Sp' ], //Septillion
[27, 'O' ], //Octillion
[30, 'N' ], //Nonillion
[33, 'D' ], //Decillion
[36, 'Ud' ], //Undecillion
[39, 'Dd' ], //Duodecillion
[42, 'Td' ], //Tredecillion
[45, 'Qad'], //Quattuordecillion
[48, 'Qid'], //Quindecillion
[51, 'Sxd'], //Sexdecillion
[54, 'Spd'], //Septendecillion
[57, 'Od' ], //Octodecillion
[60, 'Nd' ], //Novemdecillion
[63, 'V' ], //Vigintillion
[66, 'Uv' ], //Unvigintillion
[69, 'Dv' ], //Duovigintillion
[72, 'Tv' ], //Trevigintillion
[75, 'Qav'], //Quattuorvigintillion
[78, 'Qiv'], //Quinvigintillion
[81, 'Sxv'], //Sexvigintillion
[84, 'Spv'], //Septenvigintillion
[87, 'Ov' ], //Octovigintillion
[90, 'Nv' ], //Novemvigintillion
[93, 'Tr' ], //Trigintillion
[96, 'Ut' ], //Untrigintillion
[99, 'Dt' ], //Duotrigintillion
[100, 'G' ], //Googol
[102, 'Tt' ], //Tretrigintillion or One Hundred Googol
],
value = 3
) {
if (
database[database.length - 1] &&
database[database.length - 1][0] !== 0
) {
database = database.reverse();
database.push([0, '']);
}
const addSuffix = num => {
const pad3 = (str = '') => str.padEnd(value, ' ');
const decim = 1 // round to decimal places
const separ = 0 // separate number and suffix
const anum = Math.abs(num);
if (num === 0)
return '0' + ' ' + ' '.repeat(separ) + ' '.repeat(decim) + pad3();
if (anum > 0 && anum < 1)
return String(num.toFixed(decim)) + ' '.repeat(separ) + pad3();
for (const [exp, suffix] of database) {
if (anum >= Math.pow(10, exp))
return `${(num / Math.pow(10, exp)).toFixed(decim)
}${' '.repeat(separ) + pad3(suffix)}`;
}
};
return customFunctionRecurse_(
range, CSF, addSuffix, database, value, true
);
}
function customFunctionRecurse_(
array, mainFunc, subFunc, ...extraArgToMainFunc
) {
if (Array.isArray(array))
return array.map(e => mainFunc(e, ...extraArgToMainFunc));
else return subFunc(array);
}
Run Code Online (Sandbox Code Playgroud)
旁注 1:此脚本不需要在使用前获得授权
旁注 2:单元格格式需要设置为“自动”或“数字”,否则使用强制模式