Numbers in 2050

Question

It's 2050, and people have decided to write numbers in a new way. They want less to memorize, and number to be able to be written quicker.
For every place value(ones, tens, hundreds, etc.) the number is written with the number in that place, a hyphen, and the place value name. "zero" and it's place value does not need to be written.
The number 0 and negative numbers do not need to be handled, so don't worry about those.

Input:

The input will be a positive integer up to 3 digits.

Output:

The output should be a string that looks like something below.

Test cases:

56 => five-ten six
11 => ten one
72 => seven-ten two
478 => four-hundred seven-ten eight
754 => seven-hundred five-ten four
750 => seven-hundred five-ten
507 => five-hundred seven

This question would not be a dupe because that one has a bit more requirements.
"One" does not need to be printed unless it is in the ones place value.
This is code-golf, so normal golfing rules apply.
Shortest answer wins!
Good luck!

Imagine if numbers are actually written like this in the future...

Answer 1

Perl 5 -p, 100 bytes

$a="{,{one,two,three,four,five,six,seven,eight,nine}";$_=(<"$a-hundred }$a-ten }$a}">)[$_];s/one-//g

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Readable version:

#!perl -pl
$a="{,{one,two,three,four,five,six,seven,eight,nine}";
$_=(<"$a-hundred }$a-ten }$a}">)[$_];
s/one-//g

The idea here is simple: instead of trying to generate the number on the fly, generate a list of all 2050 numbers using a glob expression (which is much easier) and then index into that list. If $n is the list of numbers, we construct the glob {,{$n}-hundred }{,{$n}-ten }{,{$n}}. The only problem is that this glob expression will generate one-ten and one-hundred, which we can simply strip at the end with s/one-//g.

Answer 2

Vyxal S, 25 bytes

ẏṘ↵Z'h;ƛ∆ċ1∆ċ-';\-jȧ1∆ċ$∨

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or try a test suite.

Thank goodness for built-in number to word support

Explained

ẏṘ↵Z'h;ƛ∆ċ1∆ċ-';\-jȧ1∆ċ$∨ # example input → 701
ẏṘ                        # range(len(input) - 1, -1, -1) → [2, 1, 0]
  ↵                       # 10 to the power of each item in that range → [100, 10, 1]
   Z                      # zip that with the digits of the input → [[7, 100], [0, 10], [1, 1]]
    'h;                   # keep only items where the digit of the input isn't 0 → [[7, 100], [1, 1]]
       ƛ                  # to each item in the remaining [digit, power10] list:
        ∆ċ                #   convert both digit and power10 to words → [["seven", "one hundred"], ["one", "one"]]
          1∆ċ-            #   remove any occurances of the word "one" from each → [["seven", " hundred"], ["", ""]]
              ';          #   and remove any empty strings - this makes it so that you have singleton lists for hundreds and tens and an empty list for ones if one is present → [["seven", " hundred"], []]
                \-j       #   join the result of that on "-" → ["seven- hundred", ""]
                   ȧ      #   remove whitespace → ["seven-hundred", ""]
                    1∆ċ$∨ #   get the first truthy item from that and "one" - this works because an empty string means that we want the word "one" → ["seven-hundred", "one"]
                          # The -S joins the result on spaces → "seven-hundred one"

Answer 3

Python + num2words, 162 138 137 bytes

lambda N:' '.join([((num2words(n)+'-'*(w>''),'')[n=='1'<w]+w)for n,w in zip('%3s'%N,['hundred','ten',''])if'0'<n])
from num2words import*

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Requires the package num2words to be installed, so won't run in the ATO link.

-24 bytes thans to Jonathan Allan

-1 byte, replaced '%03s'%N with '%3s'%N

Answer 4

Charcoal, 74 bytes

⪫Φ⮌Ｅ⮌Ｓ⪫Φ⟦§⪪”↶⌈↶？#λ<∕CＳＧ∨⮌№¤"~^hφ﹪▶U⊕hU⦄w↷” ∧∨⊖ι¬κＩι§⪪”↶↧WaＫπ⁵1✂” ∧Ｉικ⟧λ-ι 

Try it online! Link is to verbose version of code. Explanation:

     Ｓ                                      Input as a string
    ⮌                                       Reversed
   Ｅ                                        Map over digits
        ⟦                            ⟧      Tuple of
           ”...”                            Compressed digits string
          ⪪                                 Split on spaces
         §                                  Indexed by
                    ι                       Current digit
                   ⊖                        Decremented
                  ∨                         Logical Or
                      κ                     Current power of 10
                     ¬                      Is zero
                 ∧                          Logical And
                        ι                   Current digit
                       Ｉ                    Cast to integer
                           ”...”            Compressed places string
                          ⪪                 Split on spaces
                         §                  Indexed by
                                   ι        Current digit
                                  Ｉ         Cast to integer
                                 ∧          Logical And
                                    κ       Current power of 10
       Φ                                    Filtered where
                                      λ     String is not empty
      ⪫                                     Joined with
                                       -    Literal string `-`
  ⮌                                         Reversed
 Φ                                          Filtered where
                                        ι   String is not empty
⪫                                           Joined with spaces
                                            Implicitly print

Answer 5

C (gcc), 193 171 169 bytes

• -2 thanks to ceilingcat

Recursively divides the input number, looking up each digit (except for 1 in non-ones position and 0) and and appending the suffix (except for ones position.)

g(v,s,i){v&&g(v/10,s?"hundred":"ten"),(i=v%10)&&printf("%3$.5s%2$s%s "+(i<2&&s)*10,s?:"",!s+"-","one\0 two\0 threefour\0five\0six\0 seveneightnine"+~-i*5);}f(v){g(v,0);}

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Answer 6

Factor + math.text.english math.text.utils, 138 136 126 bytes

[ 1 digit-groups [ 10^ [ number>text ] bi@ "-"glue R/ one[- ]|-one$|zero.*/ ""re-replace ] map-index reverse harvest " "join ]

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Explanation

Code                     | Data stack                              | Comment
============================================================================================================
                         ! 478
1 digit-groups           ! { 8 7 4 }
           <<inside the map-index quotation now; first iteration>>
                         ! 8 0                                       (digit, index)
10^                      ! 8 1
[ number>text ] bi@      ! "eight" "one"
"-"glue                  ! "eight-one"
R/ one[- ]|-one$|zero.*/ ! "eight-one" R/ one[- ]|-one$|zero.*/      (push a regexp literal on the data stack)
""re-replace             ! "eight"
          <<outside the map-index quotation now>>
map-index                ! { "eight" "seven-ten" "four-hundred" }
reverse                  ! { "four-hundred" "seven-ten" "eight" }
harvest                  ! { "four-hundred" "seven-ten" "eight" }    (remove empty strings)
" "join                  ! "four-hundred seven-ten eight"

Answer 7

JavaScript (V8), 162 153 151 148 bytes

-3 bytes thanks to Arnauld.
-6 bytes by merging the handling of 10's and 100's.
-2 bytes thanks to Arnauld, again.
-3 bytes by moving the floor operation, thanks to Shaggy.

f=(n,p)=>n>9?f(n/(m=n>99?100:10)|0,m%4?'ten ':'hundred ')+f(n%m):n?--n|!p?'one,two,three,four,five,six,seven,eight,nine'.split`,`[n]+[p&&'-'+p]:p:''

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Ungolfed

f = (n, p) =>
  n > 9 ? // check if we should handle 10's or 100's
    f(
      n / (m = n > 99 ? 100 : 10) | 0, // check what mod to use, then floor divide
      m % 4 ? 'ten ' : 'hundred ' // check what place to use
    ) + f(n % m) :
  n ? // if 9 >= n > 0
    --n | !p ? // if n > 1 or p ('ten' or 'hundred') is not defined we need to find the name of n
      'one,two,three,four,five,six,seven,eight,nine'.split`,`[n] +
      [p && '-' + p] : // add p if we have it
    p : // return p if n = 1
  '' // return '' if n = 0

Answer 8

Python, 198 188 bytes

lambda m:" ".join([r[int(i)]for i,r in zip('%3s'%m,[[c+s for c in",one,two,three,four,five,six,seven,eight,nine".split(",")]for s in['-hundred','-ten','']])if"0"<i>" "]).replace("one-","")

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Inspired by Sisyphus's answer. Generates three lists of the number string for each of the hundreds, tens and ones. Indexes each list by it's respective value in the input, then gets rid of one-

Answer 9

JavaScript (Node.js), 147 bytes

n=>['hundred','ten',''].flatMap(t=>(c=',,two,three,four,five,six,seven,eight,nine'.split`,`[p=(n*=10)/1e3%10|0])?t?c+'-'+t:c:p?t||'one':[]).join` `

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Answer 10

C (gcc), 223 bytes

o,i,z;main(a,c)char**c;{for(o=strlen(c[1]);z=c[1][i]-49,printf("%s%s%s ",z|o-i==1?"zero\0one\0two\0three\0four\0five\0six\0seven\0eight\0nine"+"059=CHMQW]"[z+1]-48:"",z&&o+~i?"-":"",o-i-2?o-i-3?"":"hundred":"ten"),++i-o;);}

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Thanks to Mukul Kumar for basis, fixed what I perceived to be incorrect output, and shaved down by switching to C and embracing the sketchiness of golf C.

Updates based on issues found and golf contributed by Arnauld

Thanks to ceilingcat for some more C golf sketchiness.

Answer 11

Python 3.8, no library, 184 bytes

Late to the party, but here's my answer. jezza_99 answer using library is unbeatable!

n=input()
for i,o in zip([' one two three four five six seven eight nine'.split(' ')[int(d)]for d in n],['-hundred','-ten',''][-len(n):]):
 if i:print((i+o).replace('one-',''),end=' ')

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---

Commented code

# input: get number to be rewritten in words
# n will be a string
n = input('Input number n to be translated into words (0 < n < 1000): ')

# list of literal digit
# specify split(' ') to have empty string corresponding to 0
literal_digit = ' one two three four five six seven eight nine'.split(' ')
# we will get
# ['', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine']

# list of literal digit corresponding to the input number
# expanded list comprehension [literal_digit[int(d)] for d in n]
literal_number = []
for d in n:
    literal_number.append(literal_digit[int(d)])
# for example '438' -> ['four', 'three', 'eight']

# list of literal orders of magnitude
# we have to consider only the elements according to the length of the number
ord_mag = ['-hundred', '-ten', ''][-len(n):]
# for example '43' -> len = 2 -> start at -2
# we will get only ['-ten', '']

# build the number string translated into words
# loop over the literal number and the corresponding order of magnitude
for i, o in zip(literal_number, ord_mag):
    
    # check if digit is 0 -> '': don't print
    if i:
        
        # not 0
        
        # we must print the literal digit and its order of magnitude
        string = i + o
        
        # if digit is one
        # we must print only the order of magnitude,
        # otherwise only 'one' if the magnitude is unit
        string = string.replace('one-', '')
        # this can be done because the '-' is defined in the order of magnitude
        
        # print the string, concatenating the digit on the same line
        print(string, end=' ')

Answer 12

05AB1E, 53 bytes

“0€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“#ISèāR… —¿°¡#sèøʒнa}€á'-ýðý„€µ-K

Try it online or verify all test cases.

Explanation:

“0€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“
               # Push dictionary string "0 one two ... eight nine"
 #             # Split it on spaces
  I            # Push the input-integer
   S           # Convert it to a list of digits
    è          # Index those into the list
ā              # Push a list in the range [1,length] (without popping)
 R             # Reverse it to range [length,1]
  … —¿°¡       # Push dictionary string "  ten hundred"
        #      # Split it on spaces: ["","","ten","hundred"]
         s     # Swap so the [length,1] integer-list is at the top
          è    # 0-based index into the ["","","ten","hundred"]
ø              # Create pairs of the two lists
 ʒ             # Filter, to only keep the pairs where:
  н            #  the first item
   a           #  only contains letters
               #  (this removes ["0","hundred"], ["0","ten"], and ["0",""] pairs)
 }€            # After the filter: map over each remaining pair:
   á           #  Only keep strings consisting solely of letters
               # (this removes the "" from the last pair)
    '-ý       '# Join each inner pair with "-" delimiter
       ðý      # Then join these strings with " " delimiter
         „€µ-  # Push dictionary string "one-"
             K # Remove all those substrings
               # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “0€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“ is "0 one two three four five six seven eight nine"; … —¿°¡ is " ten hundred"; and „€µ- is "one-".

Answer 13

Retina 0.8.2, 102 bytes

\B
-ten¶
ten(¶..)
hundred$1
A`0
¶
 
1-

1
one
2
two
3
three
4
four
5
five
6
six
7
seven
8
eight
9
nine

Try it online! Link includes test cases. Explanation:

\B
-ten¶

If there is more than one digit then assume the leading digit is a ten. Also split the digits onto their own line.

ten(¶..)
hundred$1

If there's a second digit then correct it to a hundred.

A`0

Delete any lines containing a 0.

¶
 

Join the lines together again.

1-

10 and 100 should just be ten and hundred.

1
one
2
two
3
three
4
four
5
five
6
six
7
seven
8
eight
9
nine

Turn the remaining digits into words.

Bonus: Here's a version that works up to 9999 for 116 bytes: Try it online! Link includes test cases. (Sadly the three digit version is 103 bytes.)

Answer 14

Python 2, 209 bytes

i=input();t,n="0 one two three four five six seven eight nine".split(),map(int,"00"+`i`)[-3:]
print ("",t[n[0]]+"-hundred ")[n[0]and i>99]+("",(t[n[1]]+"-","")[n[1]<2]+"ten ")[n[1]and i>9]+("",t[n[2]])[n[2]>0]

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Python 2 because:

1. Accepts integers directly as input

2. `` converts integer to string

3. Can directly index the result of map without converting to a list

4. No () required by the print

Answer 15

PHP, 195 bytes

<?php foreach(str_split(substr("00".fgets(STDIN),-3))as$i=>$c)echo trim($c>0?[0,$i>1?"one":"","two","three","four","five","six","seven","eight","nine"][$c].["-hundred","-ten",""][$i]." ":"","-");

Input via STDIN, output via echo

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---

De-golf:

<?php
    $inp = substr("00".fgets(STDIN),-3);
        // here I take input and append leading zeros, so as the total lengh is 3
        //   (for example, 13 becomes 013)
    $arr = str_split($inp);
        // string to array of chars
    
    foreach($arr as $index => $digit){
        $num_array = [0, $index > 1 ? "one":"","two","three","four","five","six","seven","eight","nine"]; // here one is hidden unless it's on the last position
        $name_array = ["-hundred","-ten",""];
        $chunk = $num_array[$digit].$name_array[$index]." ";
        $chunk = $digit > 0 ? $chunk : ""; // empty string if it's zero
        $trimmed = trim($chunk, "-"); // to dispose of "-" at the beginning of the string (this could happen if we drop "one" in "one-ten", so just "-ten" left)
        echo $trimmed;
    }
?>

---

Fun note - I didn't expect PHP doesn't complain here:

Answer 16

C++, 277 bytes

#include <iostream>
main(int a,char*c[]){char s[9];int q[]={0,4,7,10,15,19,23,26,31,36,40},n,m,o=strlen(c[1]),i=0;do{m=c[1][i]-48;n=q[m];strcpy_s(s,"zeroonetwothreefourfivesixseveneightnine"+n,q[m+1]-n);std::cout<<((o-i==2)?"ten-":(o-i==3)?"hundred-":"")<<s<<' ';}while(++i<o);}

Ungolfed

#include <iostream>

int main(int a, char* c[]) {
    char s[9];
    int q[] = { 0,4,7,10,15,19,23,26,31,36,40 }, n, m, o = strlen(c[1]), i = 0;
    do {
        m = c[1][i] - 48;
        n = q[m];
        strncpy_s(s, 9, "zeroonetwothreefourfivesixseveneightnine" + n, q[m + 1] - n);
        std::cout << ((o - i == 2) ? "ten-" : (o - i == 3) ? "hundred-" : "") << s << ' ';
    } while (++i < o);
}  

Input comes from command-line arguments

Reason for using C++ over C

• apparantly some compilers include string.h header in iostream

Answer 17

Julia, ¹⁸⁵ 158 bytes

n+m+s=!(n÷m)*"-$s "*!(n%m)
!n=replace(n<10 ? split(" one two three four five six seven eight nine"," ")[-~n] :
n<100 ? n+10+:ten : n+100+:hundred,"one-"=>"")

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