Hi Acidrums, hi Hans, I worked up the code below, put it in core-con.lua, and recompiled ConTeXt with `context --make cont-en`. It gives the following output. Does it look correct? Hans: the `spanishwords` is a bit hacky; probably words should have subtables words.english and words.spanish. Cheers, Sietse 11.111.111 once millón cien once mil cien once 2.221.101 dos millón doscientos veintiuno mil cien uno 1.111 uno mil cien once 1.218 uno mil doscientos dieciocho 1.234 uno mil doscientos treinta y cuatro 12.345 doce mil trescientos cuarenta y cinco 12.345.678.900.000 doce billón trescientos cuarenta y cinco mil millones seiscientos sietenta y ocho millón novecientos mil local spanishwords = { [0] = "zero", [1] = "uno", [2] = "dos", [3] = "tres", [4] = "cuatro", [5] = "cinco", [6] = "seis", [7] = "siete", [8] = "ocho", [9] = "nueve", [10] = "diez", [11] = "once", [12] = "doce", [13] = "trece", [14] = "catorce", [15] = "quince", [16] = "dieciséis", [17] = "diecisiete", [18] = "dieciocho", [19] = "diecinueve", [20] = "veinte", [21] = "veintiuno", [22] = "veintidós", [23] = "veintitrés", [24] = "veinticuatro", [25] = "veinticinquo", [26] = "veintiséis", [27] = "veintisiete", [28] = "veintiocho", [29] = "veintinueve", [30] = "treinta", [40] = "cuarenta", [50] = "cinquenta", [60] = "sesenta", [70] = "sietenta", [80] = "ochenta", [90] = "noventa", [100] = "cien", [200] = "doscientos", [300] = "trescientos", [400] = "cuatrocientos", [500] = "quinientos", [600] = "seiscientos", [700] = "setecientos", [800] = "ochocientos", [900] = "novecientos", [1000] = "mil", [1000^2] = "millón", [1000^3] = "mil millones", [1000^4] = "billón", } function verbose.spanish(n) local w = spanishwords[n] if w then return w end local t = { } local function compose_one(n) local w = spanishwords[n] if w then t[#t+1] = w return end -- a, b = hundreds, remainder local a, b = floor(n/100), n % 100 -- one thousand if a == 10 then t[#t+1] = spanishwords[1] t[#t+1] = spanishwords[1000] -- x hundred (n.b. this will not give thirteen hundred because -- compose_one(n) is only called after -- `n = compose(two(n, 1000^1))`. elseif a > 0 then t[#t+1] = spanishwords[a*100] end -- the remainder if spanishwords[b] then t[#t+1] = spanishwords[b] else -- a, b = tens, remainder a, b = floor(b/10), n % 10 t[#t+1] = spanishwords[a*10] t[#t+1] = "y" t[#t+1] = spanishwords[b] end end -- compose_two handles x billion, ... x thousand. When 1000 or less is -- left, compose_one takes over. local function compose_two(n,m) if n > (m-1) then local a, b = floor(n/m), n % m if a > 0 then compose_one(a) end t[#t+1] = spanishwords[m] n = b end return n end n = compose_two(n,1000^4) n = compose_two(n,1000^3) n = compose_two(n,1000^2) n = compose_two(n,1000^1) if n > 0 then compose_one(n) end return #t > 0 and concat(t," ") or tostring(n) end verbose.es = verbose.spanish ---- to test the code --- \starttext \def\vbes#1% {#1\crlf \ctxcommand{verbose(#1, "spanish")}} \obeylines \vbes{11111111} \vbes{2221101} \vbes{1111} \vbes{1218} \vbes{1234} \vbes{12345} \vbes{12345678900000} \stopcode

they are local tables ... so invisible to users

Fair point. I was concerned more with changeabiltity: this way someone who writes a new function has an obvious name (numberwords.french) for her new wordlist, and an obvious single point of alteration for her copy of verbose.english, namely to change `local words = numberwords.english` to `...numberwords.french`. (Algorithm tweaks will still come after that, of course.) The code needs very little changing. This is a minor point only, of course. --Sietse + local numberwords = { } - words = { + numberwords.english = { [1] = "one", ... } -spanishwords = { + numberwords.spanish = { [1] = "uno", ... } function verbose.english(n) + local words = allwords.english ... `words` already used throughout ... end function verbose.spanish(n) + local words = allwords.spanish -- to use a different table, no need to -- replace `spanishwords` throughout -- the function; just change the line above. ... no more `spanishwords`, just `words` ... end

Hi Hans, I think the verbose.spanish code you put in the latest beta broke verbose.english: after verbose.english is defined, a second `local verbose = { }` at the start of the spanish code accidentally overwrites that table. The code below works; as a bonus, it fixes the bug where 900 was printed as 'nine hundred zero'. It can replace the entire swath of verbose-related code, right up to the handlers. (The English part is mostly untouched; only that bugfix, and a tweak to use the `local words = words.english` idiom) Known bugs: * 100 before 'mil' should be 'cien mil' * 100 on its own should be 'cien' * 100 before 'millón' -- should that be 'cien millón', or 'ciento millón'? * what are the rules for e.g. '1001' -- should that be 'mil y una'? And should 1004 also be mil y cuatro? Acidrums: if you put this in core-con.lua, replacing everything between the commands.ordinal and converters.verbose function definitions, then run context --make cont-en, does the output look okay to you? Also: thank you for the day and month tables! Cheers, Sietse -- verbose numbers local verbose = { } -- verbose english local words = { } words.english = { [0] = "zero", [1] = "one", [2] = "two", [3] = "three", [4] = "four", [5] = "five", [6] = "six", [7] = "seven", [8] = "eight", [9] = "nine", [10] = "ten", [11] = "eleven", [12] = "twelve", [13] = "thirteen", [14] = "fourteen", [15] = "fifteen", [16] = "sixteen", [17] = "seventeen", [18] = "eighteen", [19] = "nineteen", [20] = "twenty", [30] = "thirty", [40] = "forty", [50] = "fifty", [60] = "sixty", [70] = "seventy", [80] = "eighty", [90] = "ninety", [100] = "hundred", [1000] = "thousand", [1000^2] = "million", [1000^3] = "billion", [1000^4] = "trillion", } function verbose.english(n) local words = words.english local w = words[n] if w then return w end local t = { } local function compose_one(n) local w = words[n] if w then t[#t+1] = w return end local a, b = floor(n/100), n % 100 if a == 10 then t[#t+1] = words[1] t[#t+1] = words[1000] elseif a > 0 then t[#t+1] = words[a] t[#t+1] = words[100] -- don't say 'nine hundred zero' if b == 0 then return end end if words[b] then t[#t+1] = words[b] else a, b = floor(b/10), n % 10 t[#t+1] = words[a*10] t[#t+1] = words[b] end end local function compose_two(n,m) if n > (m-1) then local a, b = floor(n/m), n % m if a > 0 then compose_one(a) end t[#t+1] = words[m] n = b end return n end n = compose_two(n,1000^4) n = compose_two(n,1000^3) n = compose_two(n,1000^2) n = compose_two(n,1000^1) if n > 0 then compose_one(n) end return #t > 0 and concat(t," ") or tostring(n) end verbose.en = verbose.english -- print(verbose.english(11111111)) -- print(verbose.english(2221101)) -- print(verbose.english(1111)) -- print(verbose.english(1218)) -- print(verbose.english(1234)) -- print(verbose.english(12345)) -- print(verbose.english(12345678900000)) -- verbose spanish (unchecked) words.spanish = { [1] = "uno", [2] = "dos", [3] = "tres", [4] = "cuatro", [5] = "cinco", [6] = "seis", [7] = "siete", [8] = "ocho", [9] = "nueve", [10] = "diez", [11] = "once", [12] = "doce", [13] = "trece", [14] = "catorce", [15] = "quince", [16] = "dieciséis", [17] = "diecisiete", [18] = "dieciocho", [19] = "diecinueve", [20] = "veinte", [21] = "veintiuno", [22] = "veintidós", [23] = "veintitrés", [24] = "veinticuatro", [25] = "veinticinco", [26] = "veintiséis", [27] = "veintisiete", [28] = "veintiocho", [29] = "veintinueve", [30] = "treinta", [40] = "cuarenta", [50] = "cincuenta", [60] = "sesenta", [70] = "setenta", [80] = "ochenta", [90] = "noventa", [100] = "ciento", [200] = "doscientos", [300] = "trescientos", [400] = "cuatrocientos", [500] = "quinientos", [600] = "seiscientos", [700] = "setecientos", [800] = "ochocientos", [900] = "novecientos", [1000] = "mil", [1000^2] = "millón", [1000^3] = "mil millónes", [1000^4] = "billón", } function verbose.spanish(n) local words = words.spanish local w = words[n] if w then return w end local t = { } local function compose_one(n) local w = words[n] if w then t[#t+1] = w return end -- a, b = hundreds, remainder local a, b = floor(n/100), n % 100 -- one thousand if a == 10 then t[#t+1] = words[1] t[#t+1] = words[1000] -- x hundred (n.b. this will not give thirteen hundred because -- compose_one(n) is only called after -- `n = compose(two(n, 1000^1))`. elseif a > 0 then t[#t+1] = words[a*100] end -- the remainder if words[b] then t[#t+1] = words[b] else -- a, b = tens, remainder a, b = floor(b/10), n % 10 t[#t+1] = words[a*10] t[#t+1] = "y" t[#t+1] = words[b] end end -- compose_two handles x billion, ... x thousand. When 1000 or less is -- left, compose_one takes over. local function compose_two(n,m) if n > (m-1) then local a, b = floor(n/m), n % m if a > 0 then compose_one(a) end t[#t+1] = words[m] n = b end return n end n = compose_two(n,1000^4) n = compose_two(n,1000^3) n = compose_two(n,1000^2) n = compose_two(n,1000^1) if n > 0 then compose_one(n) end return #t > 0 and concat(t," ") or tostring(n) end verbose.es = verbose.spanish -- print(verbose.spanish(31)) -- print(verbose.spanish(101)) -- print(verbose.spanish(199)) ---- verbose handlers come next) ----

On Thu, Nov 22, 2012 at 1:03 AM, Acidrums4 wrote:

* 100 before 'millón' -- should that be 'cien millón', or 'ciento millón'? -> It should be 'cien millones'. 'Millones', plural for 'Millón'. Maybe it should be a variable for plurals for millions (the only plural used for numbers in spanish).

Singular and plural don't always suffice. 1M: en milijon 2M: dva milijona 3/4M: [tri|štiri] milijone 5M: pet milijonov (And then of course, if you want more fun, "en milijon", "enega milijona", "enemu milijonu", "en milijon", "enem milijonu", "enim milijonom", "dva milijona", "dveh milijonov", "dvema milijonoma", "dva milijona", "dveh milijonih", "dvema milijonoma", ...) I'm not suggesting to implement the forms (declinations) from the last paragraph, but if you want to introduce separate singular and plural forms, you might want to support more than just two forms from the start: http://www.gnu.org/savannah-checkouts/gnu/gettext/manual/html_node/Plural-fo... Mojca

Dnia 2012-11-22, o godz. 11:55:17 Mojca Miklavec napisał(a):

On Thu, Nov 22, 2012 at 1:03 AM, Acidrums4 wrote:

...

* 100 before 'millón' -- should that be 'cien millón', or 'ciento millón'? -> It should be 'cien millones'. 'Millones', plural for 'Millón'. Maybe it should be a variable for plurals for millions (the only plural used for numbers in spanish).

Singular and plural don't always suffice.

1M: en milijon 2M: dva milijona 3/4M: [tri|štiri] milijone 5M: pet milijonov

Similarly for Polish (which is no surprise), so a more general mechanism might be a good idea. 1M: (jeden) milion 2/3/4M: dwa/trzy/cztery miliony 5/6/7/8/9M: pięć/sześć/siedem/osiem/dziewięć milionów 32M: trzydzieści dwa miliony 57M: pięćdziesiąt siedem milionów And similarly for thousands.

Mojca

Best, -- Marcin Borkowski http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski Adam Mickiewicz University

On Fri, 23 Nov 2012 23:46:44 +0100 Hans Hagen wrote:

first we need working prototypes for several languages, then I can wrap it into some general mechanism. I think all languages are somewhat different so 'general' is 'sort of general'.

For example, in French, 1000 can be une brique, une patate or un bâton, but I guess that this is not quite the same thing as the subject of this thread... :)