Sort spelled-out serial numbers

Question

Given a list of two or more spelled-out serial numbers of equal length greater than two, e.g.

[[ "three" , "one"  , "four"  ],
 [ "one"   , "five" , "nine"  ],
 [ "two"   , "six"  , "five"  ],
 [ "three" , "five" , "eight" ]]

sort the list by the numbers that the words represent:

[[ "one"   , "five" , "nine"  ],
 [ "two"   , "six"  , "five"  ],
 [ "three" , "one"  , "four"  ],
 [ "three" , "five" , "eight" ]]

You may require the numbers to be spelled in lower or upper, but not mixed, case.

Test cases

[["three","one","four"],["one","five","nine"],["two","six","five"],["three","five","eight"]]
gives
[["one","five","nine"],["two","six","five"],["three","one","four"],["three","five","eight"]]

[["two","seven"],["one","eight"],["two","eight"],["one","eight"],["two","eight"],["four","five"]]
gives
[["one","eight"],["one","eight"],["two","seven"],["two","eight"],["two","eight"],["four","five"]]

[["one","four","one","four","two"],["one","three","five","six","two"],["three","seven","three","zero","nine"]]
gives
[["one","three","five","six","two"],["one","four","one","four","two"],["three","seven","three","zero","nine"]]

[["zero","six","one"],["eight","zero","three"],["three","nine","eight"],["eight","seven","four"],["nine","eight","nine"],["four","eight","four"]]
gives
[["zero","six","one"],["three","nine","eight"],["four","eight","four"],["eight","zero","three"],["eight","seven","four"],["nine","eight","nine"]]

Answer 1

Husk, 9 8 bytes

Ö†€¨tfṡn

Try it online!

Algorithm "inspired" by recursive's Stax answer (I've just changed the lookup string a bit), go upvote him!

The trick is mapping each letter to its position in the string tfsen (compressed at the end of this program). Husks lists are 1-based, and missing items return 0, so we get this mapping:

"one"        [0,5,4]
"two"        [1,0,0]
"three"      [1,0,0,4,4]
"four"       [2,0,0,0]
"five"       [2,0,0,4]
"six"        [3,0,0]
"seven"      [3,4,0,4,5]
"eight"      [4,0,0,0,1]
"nine"       [5,0,5,4]

As you can see, the lists are perfectly ordered.

---

In order to be clear, here's how list comparison works in Husk (and in many other languages):

1. If one of the two lists is empty, that's the smaller one.
2. If the first elements of the two lists are different, the one with the smaller first element is the smaller list.
3. Otherwise, discard the first element from both lists and go back to point 1.

Answer 2

Stax, 24 22 17 16 14 bytes

▄Ωφ▐╧Kìg▄↕ñ▼!█

Run and debug it

This program takes arrays of lowercase spelled digits for input. The output is newline-separated like so.

one five nine
two six five
three one four
three five eight

This program sorts the inputs using the ordering obtained under a specific transformation. Each character in each word is replaced by its index in the string "wo thif sen". The original arrays are sorted by this ordering. Then the results are printed after joining with a space.

The spaces serve no purpose, but actually allow greater compression in the string literal.

Answer 3

Jelly, 12 bytes

OḌ%⁽Т%147µÞ

A monadic link.

Try it online! ...or see the test-suite

How?

Converting the digits to ordinals and then from base 10 then taking modulos by 4752 then 147 gives an ascending order:

 zero            , one         , two         , three               , four
[122,101,114,111],[111,110,101],[116,119,111],[116,104,114,101,101],[102,111,117,114]
 133351          , 12301       , 12901       , 1276511             , 114384
 295             , 2797        , 3397        , 2975                , 336
 1               , 4           , 16          , 35                  , 42

 five            , six         , seven               , eight               , nine
[102,105,118,101],[115,105,120],[115,101,118,101,110],[101,105,103,104,116],[110,105,110,101]
 113781          , 12670       , 1263920             , 1126456             , 121701
 4485            , 3166        , 4640                , 232                 , 2901
 75              , 79          , 83                  , 85                  , 108

This can then be used as a key function by which to sort:

OḌ%⁽Т%147µÞ - Link: list of lists of lists of characters
          µÞ - sort (Þ) by the mondadic chain to the left (µ):
O            -   ordinals of the characters
 Ḍ           -   convert from base 10
   ⁽Т       -   literal 4752
  %          -   modulo
      %147   -   modulo by 147

Answer 4

Python, 62 bytes

lambda m:sorted(m,key=lambda r:[int(s,36)%6779%531for s in r])

Try it online! ...or see the test-suite

Note:

lambda m:sorted(m,key=lambda r:[map("trfsen".find,s)for s in r])

which works in Python 2 (but not 3) is longer by two bytes.

Answer 5

APL (Dyalog Classic), 12 bytes

'nesft'∘⍒⌷¨⊂

Try it online!

This is how I found a suitable left argument for dyadic ⍒ (I tried ⍋ and length 6 first):

A←⊖a←↑'zero' 'one' 'two' 'three' 'four' 'five' 'six' 'seven' 'eight' 'nine'
{(a≡a[⍵⍒a;])∧A≡a[⍵⍒A;]:⎕←⍵}¨,⊃∘.,/5⍴⊂∪∊a

Answer 6

Perl 6, 37 bytes

*.sort:{('digit 'X~$_)».parse-names}

Try it

Expanded:

*\     # WhateverCode lambda (this is the parameter)
.sort:
{  # block with implicit parameter ｢$_｣
  (
    'digit ' X~ $_  # concatenate 'digit ' to each named digit
  )».parse-names    # call .parse-names on each
}

The code block will take a value of the form ("three","one","four") and translate it to ("3","1","4") which is a value that .sort can easily use.

Answer 7

APL (Dyalog), 38 bytes

{⍵[⍋(531∘⊥⍤1)(531|6779|36⊥9+⎕A⍳⊢)¨↑⍵]}

Try it online!

Based of Jonathan Allan's awesome solution.

Answer 8

Ruby, 48 bytes

->x{x.sort_by{|y|y.map{|s|s.to_i(35)**2%47394}}}

Abuses the fact that "zero".to_i(35) is 0 (since 'z' isn't a valid digit in base 35), so it's that much easier to brute-force a formula for the other nine digits.

Answer 9

K (ngn/k), 14 bytes

{x@<"tfsen"?x}

Try it online!

Answer 10

Python 2, 85 81 80 bytes

Simply uses the first two letters of each word to determine the number, then sorts each list using that indexing function as the key.

lambda _:sorted(_,key=lambda L:['zeontwthfofisiseeini'.find(s[:2])/2for s in L])

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Saved 4 bytes, thanks to Jonathan Allan

Answer 11

JavaScript (Node.js), 70 bytes

a=>a.sort((a,b)=>(g=a=>a.map(c=>99+parseInt(2+c,34)%607%292))(a)>g(b))

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Answer 12

Ruby, 47 bytes

->x{x.sort_by{|y|y.map{|s|s.to_i(32)%774%538}}}

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Utilises the fact that using a base less than the maximum digit yields a result of zero (as pointed out by histocrat in their answer)

Answer 13

Ruby, 49 bytes

->x{x.sort_by{|y|y.map{|s|s.to_i(36)**4%463595}}}

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Answer 14

Haskell, 133 122 109 107 106 bytes

import Data.List
sortOn$abs.read.(>>=show.head.(`elemIndices`words"ze on tw th fo fi si se ei ni").take 2)

Ungolfed:

import Data.List

nums = ["ze","on","tw","th","fo","fi","si","se","ei","ni"]

lookup' :: Eq a => a -> [a] -> Int
lookup' v = head . elemIndices v

wordToInt :: String -> Int
wordToInt (x:y:_) = lookup' [x,y] nums

wordsToInt :: [String] -> Int
wordsToInt = read . concatMap (show . wordToInt)

sortN :: [[String]] -> [[String]]
sortN = sortOn wordsToInt

Answer 15

05AB1E, 27 bytes

Σ“¡×€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“#skJ

Try it online!

---

Σ                           # Sort input by...
 “¡×€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“     # "zero one two three four five size seven eight nine"
                       #    # Split on spaces.
                        sk  # Find index of each input...
                          J # Join up.

Answer 16

Python 2, 59 bytes

lambda m:sorted(m,key=lambda r:[hash(s)%2249518for s in r])

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Riffing on Jonathan Allan's Python 3 solution...

Answer 17

Java (JDK 10), 132 bytes

l->{l.sort((a,b)->s(a)-s(b));}
int s(String[]a){int v=0;for(var s:a)v=v*99+"zeontwthfofisiseini".indexOf(s.substring(0,2));return v;}

Try it online!

Answer 18

Ruby, 64 bytes

->a{a.sort_by{|b|b.map{|n|"zeontwthfofisiseeini".index n[0,2]}}}

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A lambda accepting a 2D array of strings and returning a 2D array of strings.

Piggybacking off of mbomb007's Python 2 answer for -26 bytes off of what I was about to post.

Answer 19

Jelly, 30 28 27 bytes

w@€“¡¦ẇṆb~ṇjṚØ%ĖġṘḥḞṾṇJḌ»µÞ

Try it online!

-1 thanks to Jonathan Allan.

Finds the index of each digit in the string ‘onetwo...nine’ then sorts using this as a key function with Þ. Don't need to include 'zero' at the beginning because the search for the first two characters of 'zero' will fail and 0 will be returned instead of an index, making 'zero' lexicographically "early".

Answer 20

Perl 5, 103 bytes

sub c{pop=~s%(..)\w+\s*%zeontwthfofisiseeini=~/$1/;$`=~y///c/2%ger}say s/\R//gr for sort{c($a)<=>c$b}<>

Try it online!

Answer 21

Retina 0.8.2, 38 bytes

T`z\o\wit\hfsen`d
O`
T`d`z\o\wit\hfsen

Try it online! Link includes test suite. Works by temporarily replacing the letters zowithfsen with their position in that string, which allows the numbers to be sorted lexically.

Answer 22

Python 3, 141 bytes

print(sorted(eval(input()),key=lambda l:int(''.join([str("zero one two three four five six seven eight nine".split().index(i))for i in l]))))

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Answer 23

C (clang), 229 bytes

y,j,k,t[9];char *s="zeontwthfofisiseeini";r(char **x){for(j=y=k=0;k+1<strlen(*x);k+=j,y=y*10+(strstr(s,t)-s)/2)sscanf(*x+k,"%2[^,]%*[^,]%*[,]%n",t,&j);return y;}c(*x,*y){return r(x)-r(y);}f(*x[],l){qsort(&x[0],l,sizeof(x[0]),c);}

Try it online!

There is no straightforward way of sending array of array of strings to C functions, so in spirit of code-golf, I have taken a minor liberty in input format.

f() accepts an array of pointers to strings, where each string is a number, represented by comma separated spelled-out-digits in lower-case. Additionally, it needs number of strings in array in second parameter. I hope this is acceptable.

f() replaces the pointers in place in sorted order using qsort().
r() reads input number from comma-separated number string. It only compares first two characters to identify number.
c() is comparison function