Determine sentence type
You are encouraged to solve this task according to the task description, using any language you may know.
Use these sentences: "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it."
- Task
- Search for the last used punctuation in a sentence, and determine its type according to its punctuation.
- Output one of these letters
- "E" (Exclamation!), "Q" (Question?), "S" (Serious.), "N" (Neutral).
- Extra
- Make your code able to determine multiple sentences.
Don't leave any errors!
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contains most consonants
- Find words which contains more than 3 vowels
- Find words which first and last three letters are equals
- Find words which odd letters are consonants and even letters are vowels or vice_versa
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 Bottles of Beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
11l
<lang 11l>F sentenceType(s)
I s.empty R ‘’
[Char] types L(c) s I c == ‘?’ types.append(Char(‘Q’)) E I c == ‘!’ types.append(Char(‘E’)) E I c == ‘.’ types.append(Char(‘S’))
I s.last !C ‘?!.’ types.append(Char(‘N’))
R types.join(‘|’)
V s = ‘hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it’ print(sentenceType(s))</lang>
- Output:
Q|S|E|N
ALGOL 68
Classifies an empty string as "". <lang algol68>BEGIN # determuine the type of a sentence by looking at the final punctuation #
CHAR exclamation = "E"; # classification codes... # CHAR question = "Q"; CHAR serious = "S"; CHAR neutral = "N"; # returns the type(s) of the sentence(s) in s - exclamation, question, # # serious or neutral; if there are multiple sentences # # the types are separated by | # PROC classify = ( STRING s )STRING: BEGIN STRING result := ""; BOOL pending neutral := FALSE; FOR s pos FROM LWB s TO UPB s DO IF pending neutral := FALSE; CHAR c = s[ s pos ]; c = "?" THEN result +:= question + "|" ELIF c = "!" THEN result +:= exclamation + "|" ELIF c = "." THEN result +:= serious + "|" ELSE pending neutral := TRUE FI OD; IF pending neutral THEN result +:= neutral + "|" FI; # if s was empty, then return an empty string, otherwise remove the final separator # IF result = "" THEN "" ELSE result[ LWB result : UPB result - 1 ] FI END # classify # ; # task test case # print( ( classify( "hi there, how are you today? I'd like to present to you the washing machine 9001. " + "You have been nominated to win one of these! Just make sure you don't break it" ) , newline ) )
END</lang>
- Output:
Q|S|E|N
AutoHotkey
<lang autohotkey>Sentence := "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" Msgbox, % SentenceType(Sentence)
SentenceType(Sentence) { Sentence := Trim(Sentence) Loop, Parse, Sentence, .?! { N := (!E && !Q && !S) , S := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), ".")) , Q := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), "?")) , E := (InStr(SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3), "!")) , type .= (E) ? ("E|") : ((Q) ? ("Q|") : ((S) ? ("S|") : "N|")) , D := SubStr(Sentence, InStr(Sentence, A_LoopField)+StrLen(A_LoopField), 3) } return (D = SubStr(Sentence, 1, 3)) ? RTrim(RTrim(type, "|"), "N|") : RTrim(type, "|") }</lang>
- Output:
Q|S|E|N
AWK
<lang AWK>
- syntax: GAWK -f DETERMINE_SENTENCE_TYPE.AWK
BEGIN {
str = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" main(str) main("Exclamation! Question? Serious. Neutral") exit(0)
} function main(str, c) {
while (length(str) > 0) { c = substr(str,1,1) sentence = sentence c if (c == "!") { prn("E") } else if (c == ".") { prn("S") } else if (c == "?") { prn("Q") } str = substr(str,2) } prn("N") print("")
} function prn(type) {
gsub(/^ +/,"",sentence) printf("%s %s\n",type,sentence) sentence = ""
} </lang>
- Output:
Q hi there, how are you today? S I'd like to present to you the washing machine 9001. E You have been nominated to win one of these! N Just make sure you don't break it E Exclamation! Q Question? S Serious. N Neutral
CLU
<lang clu>% This iterator takes a string and yields one of 'E', 'Q', % 'S' or 'N' for every sentence found. % Because sentences are separated by punctuation, only the % last one can be 'N'.
sentence_types = iter (s: string) yields (char)
own punct: string := "!?." % relevant character classes own space: string := " \t\n" own types: string := "EQS" % sentence type characters prev_punct: bool := false % whether the previous character was punctuation last_punct: int := 0 % index of last punctuation character encountered sentence: bool := true % whether there are words since the last punctuation for c: char in string$chars(s) do pu: int := string$indexc(c, punct) sp: int := string$indexc(c, space) if pu ~= 0 then prev_punct := true last_punct := pu elseif sp ~= 0 then if prev_punct then % a space after punctuation means a sentence has ended here yield(types[last_punct]) sentence := false end prev_punct := false sentence := false else sentence := true end end % handle the last sentence if prev_punct then yield(types[last_punct]) elseif sentence then yield('N') end
end sentence_types
% Test start_up = proc ()
po: stream := stream$primary_output() test: string := "hi there, how are you today? I'd like to " || "present to you the washing machine 9001. You " || "have been nominated to win one of these! Just " || "make sure you don't break it"
% print the type of each sentence for c: char in sentence_types(test) do stream$putc(po, c) end
end start_up </lang>
- Output:
QSEN
Epoxy
<lang epoxy>const SentenceTypes: { ["?"]:"Q", ["."]:"S", ["!"]:"E" }
fn DetermineSentenceType(Char) return SentenceTypes[Char]||"N" cls
fn GetSentences(Text) var Sentences: [], Index: 0, Length: #Text loop i:0;i<Length;i+:1 do var Char: string.subs(Text,i,1) var Type: DetermineSentenceType(Char) if Type != "N" || i==Length-1 then log(string.sub(Text,Index,i+1)+" ("+Type+")") Index:i+2; cls cls cls
GetSentences("hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it")</lang>
- Output:
hi there, how are you today? (Q) I'd like to present to you the washing machine 9001. (S) You have been nominated to win one of these! (E) Just make sure you don't break it (N)
Factor
This program attempts to prevent common abbreviations from ending sentences early. It also tries to handle parenthesized sentences and implements an additional type for exclamatory questions (EQ).
<lang factor>USING: combinators io kernel regexp sequences sets splitting wrap.strings ;
! courtesy of https://www.infoplease.com/common-abbreviations
CONSTANT: common-abbreviations {
"A.B." "abbr." "Acad." "A.D." "alt." "A.M." "Assn." "at. no." "at. wt." "Aug." "Ave." "b." "B.A." "B.C." "b.p." "B.S." "c." "Capt." "cent." "co." "Col." "Comdr." "Corp." "Cpl." "d." "D.C." "Dec." "dept." "dist." "div." "Dr." "ed." "est." "et al." "Feb." "fl." "gal." "Gen." "Gov." "grad." "Hon." "i.e." "in." "inc." "Inst." "Jan." "Jr." "lat." "Lib." "long." "Lt." "Ltd." "M.D." "Mr." "Mrs." "mt." "mts." "Mus." "no." "Nov." "Oct." "Op." "pl." "pop." "pseud." "pt." "pub." "Rev." "rev." "R.N." "Sept." "Ser." "Sgt." "Sr." "St." "uninc." "Univ." "U.S." "vol." "vs." "wt."
}
- sentence-enders ( str -- newstr )
R/ \)/ "" re-replace " " split harvest unclip-last swap [ common-abbreviations member? ] reject [ last ".!?" member? ] filter swap suffix ;
- serious? ( str -- ? ) last CHAR: . = ;
- neutral? ( str -- ? ) last ".!?" member? not ;
- mixed? ( str -- ? ) "?!" intersect length 2 = ;
- exclamation? ( str -- ? ) last CHAR: ! = ;
- question? ( str -- ? ) last CHAR: ? = ;
- type ( str -- newstr )
{ { [ dup serious? ] [ drop "S" ] } { [ dup neutral? ] [ drop "N" ] } { [ dup mixed? ] [ drop "EQ" ] } { [ dup exclamation? ] [ drop "E" ] } { [ dup question? ] [ drop "Q" ] } [ drop "UNKNOWN" ] } cond ;
- sentences ( str -- newstr )
sentence-enders [ type ] map "|" join ;
- show ( str -- )
dup sentences " -> " glue 60 wrap-string print ;
"Hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" show nl "(There was nary a mouse stirring.) But the cats were going bonkers!" show nl "\"Why is the car so slow?\" she said." show nl "Hello, Mr. Anderson!" show nl "Are you sure?!?! How can you know?" show</lang>
- Output:
Hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it -> Q|S|E|N (There was nary a mouse stirring.) But the cats were going bonkers! -> S|E "Why is the car so slow?" she said. -> S Hello, Mr. Anderson! -> E Are you sure?!?! How can you know? -> EQ|Q
FreeBASIC
<lang freebasic>function sentype( byref s as string ) as string
'determines the sentence type of the first sentence in the string 'returns "E" for an exclamation, "Q" for a question, "S" for serious 'and "N" for neutral. 'modifies the string to remove the first sentence for i as uinteger = 1 to len(s) if mid(s, i, 1) = "!" then s=right(s,len(s)-i) return "E" end if if mid(s, i, 1) = "." then s=right(s,len(s)-i) return "S" end if if mid(s, i, 1) = "?" then s=right(s,len(s)-i) return "Q" end if next i 'if we get to the end without encountering punctuation, this 'must be a neutral sentence, which can only happen as the last one s="" return "N"
end function
dim as string spam = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"
while len(spam)>0
print sentype(spam)
wend</lang>
- Output:
QS E N
Go
<lang go>package main
import (
"fmt" "strings"
)
func sentenceType(s string) string {
if len(s) == 0 { return "" } var types []string for _, c := range s { if c == '?' { types = append(types, "Q") } else if c == '!' { types = append(types, "E") } else if c == '.' { types = append(types, "S") } } if strings.IndexByte("?!.", s[len(s)-1]) == -1 { types = append(types, "N") } return strings.Join(types, "|")
}
func main() {
s := "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" fmt.Println(sentenceType(s))
}</lang>
- Output:
Q|S|E|N
jq
Works with gojq, the Go implementation of jq
The following parses sentences with embedded quotations naively, so that for example the sentence "He asked 'How are you?'." results in: Q S <lang jq>
- Input: a string
- Output: a stream of sentence type indicators
def sentenceTypes:
def trim: sub("^ +";"") | sub(" +$";""); def parse:
capture("(?[^?!.]*)(?
[?!.])(?<remainder>.*)" ) // {p:"", remainder:""}; def encode: if . == "?" then "Q" elif . == "!" then "E" elif . == "." then "S" else "N" end; trim | select(length>0) | parse | (.p | encode), (.remainder | sentenceTypes); def s: "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"; s | sentenceTypes</lang>
- Output:
Q S E N
Julia
<lang julia>const text = """ Hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"""
haspunctotype(s) = '.' in s ? "S" : '!' in s ? "E" : '?' in s ? "Q" : "N"
text = replace(text, "\n" => " ") parsed = strip.(split(text, r"(?:(?:(?<=[\?\!\.])(?:))|(?:(?:)(?=[\?\!\.])))")) isodd(length(parsed)) && push!(parsed, "") # if ends without pnctuation for i in 1:2:length(parsed)-1
println(rpad(parsed[i] * parsed[i + 1], 52), " ==> ", haspunctotype(parsed[i + 1]))
end
</lang>
- Output:
Hi there, how are you today? ==> Q I'd like to present to you the washing machine 9001. ==> S You have been nominated to win one of these! ==> E Just make sure you don't break it ==> N
Lua
<lang lua>text = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" p2t = { [""]="N", ["."]="S", ["!"]="E", ["?"]="Q" } for s, p in text:gmatch("%s*([^%!%?%.]+)([%!%?%.]?)") do
print(s..p..": "..p2t[p])
end</lang>
- Output:
hi there, how are you today?: Q I'd like to present to you the washing machine 9001.: S You have been nominated to win one of these!: E Just make sure you don't break it: N
Perl
<lang perl>use strict; use warnings; use feature 'say'; use Lingua::Sentence;
my $para1 = <<'EOP'; hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it EOP
my $para2 = <<'EOP'; Just because there are punctuation characters like "?", "!" or especially "." present, it doesn't necessarily mean you have reached the end of a sentence, does it Mr. Magoo? The syntax highlighting here for Perl isn't bad at all. EOP
my $splitter = Lingua::Sentence->new("en"); for my $text ($para1, $para2) {
for my $s (split /\n/, $splitter->split( $text =~ s/\n//gr ) { print "$s| "; if ($s =~ /!$/) { say 'E' } elsif ($s =~ /\?$/) { say 'Q' } elsif ($s =~ /\.$/) { say 'S' } else { say 'N' } }
}</lang>
- Output:
hi there, how are you today?| Q I'd like to present to you the washing machine 9001.| S You have been nominated to win one of these!| E Just make sure you don't break it.| N Just because there are punctuation characters like "?", "!" or especially "." present, it doesn't necessarily mean you have reached the end of a sentence, does it Mr. Magoo?| Q The syntax highlighting here for Perl isn't bad at all.| S
Phix
with javascript_semantics constant s = `hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it` sequence t = split_any(trim(s),"?!."), u = substitute_all(s,t,repeat("|",length(t))), v = substitute_all(u,{"|?","|!","|.","|"},"QESN"), w = join(v,'|') ?w
- Output:
"Q|S|E|N"
Python
<lang python>import re
txt = """ Hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it"""
def haspunctotype(s):
return 'S' if '.' in s else 'E' if '!' in s else 'Q' if '?' in s else 'N'
txt = re.sub('\n', , txt) pars = [s.strip() for s in re.split("(?:(?:(?<=[\?\!\.])(?:))|(?:(?:)(?=[\?\!\.])))", txt)] if len(pars) % 2:
pars.append() # if ends without punctuation
for i in range(0, len(pars)-1, 2):
print((pars[i] + pars[i + 1]).ljust(54), "==>", haspunctotype(pars[i + 1]))
</lang>
- Output:
Hi there, how are you today? ==> Q I'd like to present to you the washing machine 9001. ==> S You have been nominated to win one of these! ==> E Just make sure you don't break it ==> N
Or for more generality, and an alternative to hand-crafted regular expressions:
<lang python>Grouping and tagging by final character of string
from functools import reduce from itertools import groupby
- tagGroups :: Dict -> [String] -> [(String, [String])]
def tagGroups(tagDict):
A list of (Tag, SentenceList) tuples, derived from an input text and a supplied dictionary of tags for each of a set of final punctuation marks. def go(sentences): return [ (tagDict.get(k, 'Not punctuated'), list(v)) for (k, v) in groupby( sorted(sentences, key=last), key=last ) ] return go
- sentenceSegments :: Chars -> String -> [String]
def sentenceSegments(punctuationChars):
A list of sentences delimited by the supplied punctuation characters, where these are followed by spaces. def go(s): return [ .join(cs).strip() for cs in splitBy( sentenceBreak(punctuationChars) )(s) ] return go
- sentenceBreak :: Chars -> (Char, Char) -> Bool
def sentenceBreak(finalPunctuation):
True if the first of two characters is a final punctuation mark and the second is a space. def go(a, b): return a in finalPunctuation and " " == b return go
- ------------------------- TEST -------------------------
- main :: IO ()
def main():
Join, segmentation, tags
tags = {'!': 'E', '?': 'Q', '.': 'S'}
# Joined by spaces, sample = ' '.join([ "Hi there, how are you today?", "I'd like to present to you the washing machine 9001.", "You have been nominated to win one of these!", "Might it be possible to add some challenge to this task?", "Feels as light as polystyrene filler.", "But perhaps substance isn't the goal!", "Just make sure you don't break off before the" ])
# segmented by punctuation, sentences = sentenceSegments( tags.keys() )(sample)
# and grouped under tags. for kv in tagGroups(tags)(sentences): print(kv)
- ----------------------- GENERIC ------------------------
- last :: [a] -> a
def last(xs):
The last element of a non-empty list. return xs[-1]
- splitBy :: (a -> a -> Bool) -> [a] -> a
def splitBy(p):
A list split wherever two consecutive items match the binary predicate p. # step :: (a, [a], a) -> a -> (a, [a], a) def step(acp, x): acc, active, prev = acp
return (acc + [active], [x], x) if p(prev, x) else ( (acc, active + [x], x) )
# go :: [a] -> a def go(xs): if 2 > len(xs): return xs else: h = xs[0] ys = reduce(step, xs[1:], ([], [h], h)) # The accumulated sublists, and the final group. return ys[0] + [ys[1]]
return go
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
('E', ['You have been nominated to win one of these!', "But perhaps substance isn't the goal!"]) ('S', ["I'd like to present to you the washing machine 9001.", 'Feels as light as polystyrene filler.']) ('Q', ['Hi there, how are you today?', 'Might it be possible to add some challenge to this task?']) ('Not punctuated', ["Just make sure you don't break off before the"])
Raku
<lang perl6>use Lingua::EN::Sentence;
my $paragraph = q:to/PARAGRAPH/; hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it
Just because there are punctuation characters like "?", "!" or especially "."
present, it doesn't necessarily mean you have reached the end of a sentence,
does it Mr. Magoo? The syntax highlighting here for Raku isn't the best.
PARAGRAPH
say join "\n\n", $paragraph.&get_sentences.map: {
/(<:punct>)$/; $_ ~ ' | ' ~ do given $0 { when '!' { 'E' }; when '?' { 'Q' }; when '.' { 'S' }; default { 'N' }; }
}</lang>
- Output:
hi there, how are you today? | Q I'd like to present to you the washing machine 9001. | S You have been nominated to win one of these! | E Just make sure you don't break it | N Just because there are punctuation characters like "?", "!" or especially "." present, it doesn't necessarily mean you have reached the end of a sentence, does it Mr. Magoo? | Q The syntax highlighting here for Raku isn't the best. | S
Wren
<lang ecmascript>var sentenceType = Fn.new { |s|
if (s.count == 0) return "" var types = [] for (c in s) { if (c == "?") { types.add("Q") } else if (c == "!") { types.add("E") } else if (c == ".") { types.add("S") } } if (!"?!.".contains(s[-1])) types.add("N") return types.join("|")
}
var s = "hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it" System.print(sentenceType.call(s))</lang>
- Output:
Q|S|E|N
The following alternative version takes the simplistic view that (unless they end the final sentence of the paragraph) ?, ! or . will only end a sentence if they're immediately followed by a space. This of course is nonsense, given the way English is written nowadays, but it's probably an improvement on the first version without the need to search through an inevitably incomplete list of abbreviations. <lang ecmascript>import "./pattern" for Pattern import "./trait" for Indexed
var map = { "?": "Q", "!": "E", ".": "S", "": "N" } var p = Pattern.new("[? |! |. ]") var paras = [
"hi there, how are you today? I'd like to present to you the washing machine 9001. You have been nominated to win one of these! Just make sure you don't break it", "hi there, how are you on St.David's day (isn't it a holiday yet?), Mr.Smith? I'd like to present to you (well someone has to win one!) the washing machine 900.1. You have been nominated by Capt.Johnson('?') to win one of these! Just make sure you (or Mrs.Smith) don't break it. By the way, what the heck is an exclamatory question!?"
]
for (para in paras) {
para = para.trim() var sentences = p.splitAll(para) var endings = p.findAll(para).map { |m| m.text[0] }.toList var lastChar = sentences[-1][-1] if ("?!.".contains(lastChar)) { endings.add(lastChar) sentences[-1] = sentences[-1][0...-1] } else { endings.add("") } for (se in Indexed.new(sentences)) { var ix = se.index var sentence = se.value System.print("%(map[endings[ix]]) <- %(sentence + endings[ix])") } System.print()
}</lang>
- Output:
Q <- hi there, how are you today? S <- I'd like to present to you the washing machine 9001. E <- You have been nominated to win one of these! N <- Just make sure you don't break it Q <- hi there, how are you on St.David's day (isn't it a holiday yet?), Mr.Smith? S <- I'd like to present to you (well someone has to win one!) the washing machine 900.1. E <- You have been nominated by Capt.Johnson('?') to win one of these! S <- Just make sure you (or Mrs.Smith) don't break it. Q <- By the way, what the heck is an exclamatory question!?