Words that Contain Digits

Question

A digit word is a word where, after possibly removing some letters, you are left with one of the single digits: ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT or NINE (not ZERO).

For example, BOUNCE and ANNOUNCE are digit words, since they contain the digit one.

ENCODE is not a digit word, even though it contains an O, N and E, since they are not in order.

Write a program/function which takes a single (uppercase or lowercase -- you choose) word as input or a parameter and determines if it is a digit word. Make the code as short as possible.

If the word is not a digit word, you should return 'NO', 0 or any 'falsey' value (this may vary based on your language). If the word is a digit word, you should output the digit it contains, as a number.

You can assume that no words contain more than one digit, so you won't have anything like ONFIVE.

Test Cases

BOUNCE
1

ENCODE
NO

EIGHT
8

BLACKJACK
NO

FABULOUS
NO

EXERCISE
NO

DRIFTWOOD
2

SERVICEMAN
7

INSIGNIFICANCE
9

THROWDOWN
2

ZERO
NO

OZNERO
1

This challenge is taken from (and is a very slight modification of) Question 1 from BIO 2009. Most of the test cases are taken from the mark scheme.

Answer 1

Javascript (ES6), 101 99 bytes

f=
s=>-~'ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE'.split` `.findIndex(x=>s.match([...x].join`.*`))

<!-- snippet demo: -->
<input list=l oninput=console.log(f(this.value))>
<datalist id=l><option value=BOUNCE>
<option value=ENCODE>
<option value=EIGHT>
<option value=BLACKJACK>
<option value=FABULOUS>
<option value=EXERCISE>
<option value=DRIFTWOOD>
<option value=SERVICEMAN>
<option value=INSIGNIFICANCE>
<option value=THROWDOWN>
<option value=ZERO>
<option value=OZNERO>

Answer 2

PHP>=7.0, 87 Bytes

<?for(;$i++<9;)levenshtein(IntlChar::charName("$i"),"DIGIT $argn",0,1,1)?:die("$i")?>NO

If only insertions of chars from one digit as word to the input is done exit the program with the digit. Or change the order to levenshtein("DIGIT $argn",IntlChar::charName("$i"),1,1,0) to count not the deletions of chars

levenshtein

IntlChar::charName

PHP>=7.0, 112 Bytes

for(;$i<9;)$r+=++$i*preg_match("#".chunk_split(substr(IntlChar::charName("$i"),6),1,".*")."#",$argn);echo$r?:NO;

IntlChar::charName

PHP, 128 Bytes

foreach([ONE,TWO,THREE,FOUR,FIVE,SIX,SEVEN,EIGHT,NINE]as$v)$r+=++$k*preg_match("#".chunk_split($v,1,".*")."#",$argn);echo$r?:NO;

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143 Bytes for more then 1 digit

foreach([ONE,TWO,THREE,FOUR,FIVE,SIX,SEVEN,EIGHT,NINE]as$v)$r.=++$k*preg_match("#".chunk_split($v,1,".*")."#",$argn);echo+$r?strtr($r,[""]):NO;

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Answer 3

Mathematica, 83 bytes (WindowsANSI encoding)

±w_:=Rest@Position[Subsets@w~Cases~#&/@Characters@*IntegerName~Array~9,Except@{},1]

Defines a unary function ± that takes a list of lowercase characters as input and returns either a digit, in a form like {{7}}, or else an empty list {}. I don't feel like I did a ton of golfy things here, except that Characters@*IntegerName~Array~9 generates the number-name matches to look for without hard-coding them.

Example usage:

±{"i", "n", "s", "i", "g", "n", "i", "f", "i", "c", "a", "n", "c", "e"}

yields {{9}}.

Answer 4

Python 3, 150 bytes

from itertools import*;lambda x:([i for i in range(10)if(*' OTTFFSSENNWHOIIEIIEORUVXVGN  ERE EHE  E   NT '[i::9],)in[*combinations(x+'  ',5)]]+[0])[0]

combinations returns all the combinations of things in order. It would be simpler to have a set number for the second parameter of combinations, so spaces are added onto the end of the original string that is a parameter of my lambda. That's a simple description of how my entry works. Ask if you would like any more clarification.

Answer 5

Jelly, 31 28 bytes

-2 bytes now that lowercase input is acceptable

“¡¦ẇṆb~ṇjṚØ%ĖġṘḥḞṾṇJḌ»Ḳe€ŒPT

A full program that expects lowercase input and prints the result, using 0 for the falsey case.

As a monadic link taking a list of characters it actually returns a list of integers which contains a single 0 in the falsey case, a single integer between 1 and 9 inclusive in the expected use cases and multiple such entries in cases where more than one number exists in the word.

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How?

“¡¦ẇṆb~ṇjṚØ%ĖġṘḥḞṾṇJḌ»Ḳe€ŒPT - Main link: list of characters, s
“¡¦ẇṆb~ṇjṚØ%ĖġṘḥḞṾṇJḌ»       - dictionary lookup of "one two three four five six seven eight nine"
                      Ḳ      - split on spaces
                         ŒP  - partitions of s
                       e€    - exists in? for €ach
                           T - truthy indexes

Answer 6

Ruby + to_words: 49 48+12 = 61 60 bytes

Uses the flags -rto_words -n. Takes lowercase words. Returns nil if no "digit" found.

-1 byte now that lowercase input is allowed, allowing for the removal of the i flag on the regex.

p (1..9).find{|i|$_=~/#{i.to_words.chars*".*"}/}

For a more pure Ruby answer without external gems, 91+1 = 92 bytes:

p (1..9).find{|i|$_=~/#{%w"0 ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE"[i].chars*".*"}/}

Answer 7

Python, 148 bytes

from itertools import*
lambda s:[tuple(w)in combinations(s,len(w))for w in("x ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE "+s).split()].index(1)%10

An unnamed function taking an uppercase only word and returning the integer (1 to 9), or 0 for NO.

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How?

For an input string s the function traverses through a list of the strings: "x", "ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX", "SEVEN", "EIGHT" , "NINE", and s itself looking for any matches.*

The comparison used is whether this string, w, is one that may be formed from a combination of letters in order from the input. The function combinations gets these for us (and only the ones of the required length using len(w)), but they are in the form of tuples, so the strings are cast to tuples for the comparison.

Of the eleven results the one for "x" will always be False, while the one for s itself will always be True. The "x" is there to ensure the index of a match with ONE through NINE are the values required (since Python lists are 0-indexed), the s is there to ensure the call to index(1) (synonymous with index(True)) wont fail when no digit word was found, whereupon the resulting 10 is converted to a 0 with a modulo of ten using %10.

* If s contains spaces for some reason, the list of ws will be longer, but the process will still work since the digit word matches will work the same way, and if none match the first space-split substring from s will match, once again giving 10 and returning 0.

If multiple digit words exist the function will return the minimal one.

Answer 8

05AB1E, 26 bytes

The falsy value here is 0.

æ‘€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š‘#å1k>

Explanation:

æ                           # Compute the powerset of the input
 ‘€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š‘       # Push "ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE"
                     #      # Break on spaces
                      å     # Check each for membership
                       1k   # Get the index of 1 in the array (-1 if not found)
                         >  # Increment by one

Uses the 05AB1E encoding. Try it online! or Verify all test cases!

Answer 9

Japt, 52 bytes

1+`e two È(e fŒr five £x  v eight ͍`¸a@v fX¬q".*

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Answer 10

Retina, 160 126 120 bytes

O.*N.*E
1
T.*W.*O
2
T.*H.*R.*E.*E
3
F.*O.*U.*R
4
F.*I.*V.*E
5
S.*I.*X
6
S.*E.*V.*E.*N
7
E.*I.*G.*H.*T
8
N.*I.*N.*E
9
\D

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Returns an empty string if the input does not contain a digit.

-6 bytes thanks to @CalculatorFeline.

Answer 11

Haskell, 113 111 bytes

import Data.List
last.((`elemIndices`([]:words"ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE"))=<<).subsequences

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Returns 0 if no digit is found.

Find all subsequences of the input word in the list of digits. Prepend an empty string [] at index 0 which is part of every subsequence. elemIndices returns a list of indices and =<< flattens them into a single list. Pick the last index.

Answer 12

JavaScript (ES6), 121 bytes

f=
s=>[...s].map(c=>a=a.map(s=>s.slice(s[0]==c)),a=`ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE`.split` `)&&1+a.indexOf(``)

<input oninput=o.textContent=f(this.value)><pre id=o>0

Returns the lowest detected digit or 0 if no digit was detected (+6 if NO is required).

Answer 13

Bash, 163 bytes

a=NO;case $1 in *O*N*E*)a=1;;*T*W*O*)a=2;;*T*H*R*E*E*)a=3;;*F*O*U*R*)a=4;;*F*I*V*E*)a=5;;*S*I*X*)a=6;;*S*E*V*E*N*)a=7;;*E*I*G*H*T*)a=8;;*N*I*N*E*)a=9;;esac;echo $a

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a=NO;
case $1 in
 *O*N*E*)a=1;;
 *T*W*O*)a=2;;
 *T*H*R*E*E*)a=3;;
 *F*O*U*R*)a=4;;
 *F*I*V*E*)a=5;;
 *S*I*X*)a=6;;
 *S*E*V*E*N*)a=7;;
 *E*I*G*H*T*)a=8;;
 *N*I*N*E*)a=9;;
esac;
echo $a

Don't know RETINA but seems a straight port of that answer.

Answer 14

Java (JDK), 150 bytes

s->{for(var i=0;i<9;)if(s.matches("."+"ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE".split(" ")[i++].replaceAll(".","*$0.")+"*"))return i;return"NO";}

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Saves

• 167 -> 153: various optimization thanks to @KevinCruijssen

Answer 15

Brachylog, 57 bytes

∧"ONE
TWO
THREE
FOUR
FIVE
SIX
SEVEN
EIGHT
NINE"ṇi₁Ph⊆?∧Pt

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Explanation

I think this could be shorter if i were reversible, but it doesn't appear to be.

∧"..."ṇi₁Ph⊆?∧Pt
∧                  Break unification with input
 "..."             String containing words ONE through NINE, newline-separated
      ṇ            Split on newlines
       i₁          [item, index] pair for some item in that list (1-based index)
         P         Call that pair P
          h        Its first element (the word)
           ⊆       is a non-contiguous substring of
            ?      the input
             ∧     And
               Pt  P's last element (the index) is the output

Answer 16

Python, 125 123 bytes

f=lambda b,x=9,i=0:x and f(b,x-1)+x*all(i:=b.find(c,i)+1for c in"NINE EIGHT SEVEN SIX FIVE FOUR THREE TWO ONE".split()[-x])

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-2 bytes from @xnor, also fixing a misalignment with the spec

Answer 17

Factor + math.combinatorics math.text.english, 53 bytes

[ all-subsets 9 [1,b] [ number>text ] map disjoint? ]

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Returns f for 'is a digit word' and t for 'is not a digit word.'

• all-subsets Get all the subsets of the input.
• 9 [1,b] [ number>text ] map A shorter way to write { "one" "two" "three" "four" "five" "six" "seven" "eight" "nine" }.
• disjoint? Do the sets not overlap? (This is 2 bytes shorter than saying intersects?.)

Answer 18

PHP, 134 132 128 bytes

<?=preg_match(strtr(chunk_split("#(ONE0TWO0THREE0FOUR0FIVE0SIX0SEVEN0EIGHT0NINE",1,".*"),[")|("]).")#",$argn,$m)?count($m)-1:NO;

run as pipe with -nF or try it online.

Creates a regex with the words in parentheses; i.e. each word N is in the Nth sub-expression.
If a word is found, the matched string will be in $m[0] and in the the Nth element, with the elements between them empty and no empty string behind; i.e. $m has N+1 elements.

Answer 19

Javascript, 121 bytes

(s,a=0)=>'ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE'.split``.map((p,i)=>RegExp(p.split``.join`.*`).exec(s)?a=i+1:0)&&a

or 116

(s,a=0)=>' ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE'.split` `.map((p,i)=>a|=~s.search(p.split``.join`.*`)&&i)&&a

But just recycling material at this point.

Answer 20

Pyth, -44- 41 bytes

+1xm}dyQc."EX%~)Û#lº,îQÓCe¯4Aô4"\P1

Takes a quoted string, outputs 0 for NO.

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explanation

+1xm}dyQc."EX%~)Û#lº,îQÓCe¯4Aô4"\P1
         ."EX%~)Û#lº,îQÓCe¯4Aô4"      # Compressed string: "ONEPTWOPTHREEPFOURPFIVEPSIXPSEVENPEIGHTPNINE" (P is close to the other letters, makes the compression better)
        c                       \P    # split on P: ['ONE', 'TWO', 'THREE', 'FOUR', 'FIVE', 'SIX', 'SEVEN', 'EIGHT', 'NINE']
   m                                  # map over this list (variable: d)
    }dyQ                              # is d a (ordered) subset of the input (Q)? (= element of the powerset)
  x                                1  # get the index of the first true
+1                                    # add one, because the list was indexed at 0 and conveniently -1 (not found) becomes 0 

Answer 21

Java, 254 bytes

int g(String a,String b){int i=0,p=0;for(char c:a.toCharArray()){p=i;i=b.indexOf(c);if(i<=p)return 0;}return 1;}
String f(String a){String[]T={"ONE","TWO","THREE","FOUR","FIVE","SIX","SEVEN","EIGHT","NINE"};for(String t:T)if(g(t,a)>0)return t;return"NO";}

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boolean g(String a,String b)
{
    int i = 0, p = 0;
    for(char c:a.toCharArray())
    {
        p = i;
        i = b.indexOf(c);
        if(i <= p) return false;
    }
    return true;
}

String f(String a)
{
    String[]T={"ONE","TWO","THREE","FOUR","FIVE","SIX","SEVEN","EIGHT","NINE"};
    for(String t:T)if(g(t,a))return t;
    return"NO";
}

Answer 22

C, 198 bytes

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char*T[]={"ONE","TWO","THREE","FOUR","FIVE","SIX","SEVEN","EIGHT","NINE"};
g(char*a,char*b){while(*a&*b)a+=(*b++==*a);return!*a;}
i;char*f(char*b){for(i=0;i<9;i++)if(g(T[i],b))return T[i];return"NO";}

Answer 23

Python 2, 155 bytes

import re;lambda k:next((i for i,j in enumerate([re.search('.*'.join(list(s)),k)for s in'ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE'.split()])if j),-1)+1

An anonymous function searching for regex group. Not the best solution here in Python but an alternative way.

Answer 24

Vyxal, 9 8 bytes

9'∆ċ?ṗṠc

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-1 thanks to EmanresuA

9 bytes but faster:

9'∆ċ:?Þ∩=

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The joys of having number to word built-ins!

Explained

9'∆ċ:?Þ∩=
9'        # From the range [1, 9], keep only items n where:
  ∆ċ      #   n as a word
     ?Þ∩  #   under multiset intersection with the input
    :   = #   is the same.