In der Sprache Nim sind die Regeln zur Unterscheidung von Bezeichnern etwas lockerer als in den meisten anderen Sprachen. Zwei Bezeichner sind äquivalent oder adressieren dieselbe Variable, wenn sie diesen Regeln folgen :

• Das erste Zeichen von beiden ist dasselbe (Groß- / Kleinschreibung beachten)
• beide Zeichenfolgen sind gleich (Groß- und Kleinschreibung beachten), nachdem alle Instanzen der Zeichen -und entfernt wurden_

Herausforderung

Schreiben Sie ein Programm / eine Funktion , die zwei Zeichenfolgen verwendet , die Nim-Bezeichner darstellen, und geben Sie einen Wahrheits- oder Falschwert aus, je nachdem, ob sie den obigen Regeln entsprechen oder nicht .

Spezifikationen

• Standard I / O - Regeln gelten .
• Standardlücken sind verboten .
• Die Zeichenfolgen enthalten nur ASCII-Ausdrucke . Sie müssen nicht überprüfen, ob es sich um eine gültige Kennung handelt.
• Die Zeichenfolgen können als zwei separate Eingaben, Liste der Zeichenfolgen usw. verwendet werden (Sie kennen den Drill).
• Leere Zeichenfolgen müssen nicht behandelt werden.
• Die Ausgabe muss konsistent sein für True- als auch für Falsey-Werte konsistent sein.
• Bei dieser Herausforderung geht es nicht darum, den kürzesten Ansatz in allen Sprachen zu finden, sondern darum, den kürzesten Ansatz in jeder Sprache zu finden .
• Ihr Code wird in Bytes bewertet , normalerweise in der Codierung UTF-8, sofern nicht anders angegeben.
• Integrierte Funktionen, die diese Aufgabe ausführen, sind zulässig, es wird jedoch empfohlen, eine Lösung zu verwenden, die nicht auf einer integrierten basiert.
• Erklärungen, auch für "praktische" Sprachen, sind erwünscht .

Testfälle

Input                                    Output

count, Count                             falsey
lookMaNoSeparator, answer                falsey
_test, test                              falsey
test, tset                               falsey
aVariableName, a_variable_name           truthy
numbers_are_cool123, numbersAreCool123   truthy
symbolsAre_too>_>, symbols_areTOO>>      truthy

Ungolfed-Referenzimplementierung

Dies ist in Nim selbst geschrieben.

import strutils, re

proc sameIdentifier(a, b: string): bool =
  a[0] == b[0] and
    a.replace(re"_|–", "").toLower == b.replace(re"_|–", "").toLower

JavaScript (ES6), 62 61 Bytes

1 Byte gespeichert dank @JohanKarlsson gespeichert

Übernimmt Eingaben in der Currying-Syntax (a)(b). Gibt einen Booleschen Wert zurück.

a=>b=>(r=s=>s[0]+s.replace(/-|_/g,'').toUpperCase())(b)==r(a)

Testfälle

let f =

a=>b=>(r=s=>s[0]+s.replace(/-|_/g,'').toUpperCase())(b)==r(a)

console.log(f("count")("Count"))                           // falsey
console.log(f("lookMaNoSeparator")("answer"))              // falsey
console.log(f("_test")("test"))                            // falsey
console.log(f("aVariableName")("a_variable_name"))         // truthy
console.log(f("numbers_are_cool123")("numbersAreCool123")) // truthy

Python 3 , 76 Bytes

lambda a,b:f(*a)==f(*b)
f=lambda f,*r:[f+k.lower()for k in r if~-(k in'-_')]

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-1 Byte dank notjagan
-3 Byte dank Wheat Wizard

Eigentlich 15 Bytes

⌠p"-_"(-Σùo⌡M═Y

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Unterhaltsame Tatsache: Dies funktioniert mit einer beliebigen Anzahl von Eingaben (bei weniger als 2 Eingaben wird immer die Wahrheit zurückgegeben).

Erläuterung:

⌠p"-_"(-Σùo⌡M═Y
⌠p"-_"(-Σùo⌡M    for each input:
 p                 separate the first character
  "-_"(-           remove all dashes and underscores from the rest of the string
        Σù         concatenate the list from the last operation and lowercase the string
          o        append it to the first character
             ═Y  are none of the elements unique?

Pyth , 13 Bytes

qFm[hd-r0d"-_

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Erläuterung

qFm[hd-r0d"-_
  m              For each value in the input (which is a list of two strings):
   [             Create a list consisting of
    hd               the first character of each value
      -r0d"-_        and the lowercase version of the value without "-" or "_"
qF               Fold over equivalence; checks to see if both lists are the same

05AB1E , 12 Bytes

εćs„-_SKl«}Ë

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-1 danke an Adnan .

Theoretisch εćs„-_-«}Ëhätte das für 10 Bytes funktionieren sollen, aber leider ist dieses Verhalten vorerst veraltet.

Jelly, 11 bytes

ḟ⁾-_Œl,Ḣµ€E

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-2 bytes thanks to Erik the Outgolfer
-1 byte thanks to Jonathan Allan

Ruby, 86 64 63 61 51 bytes

f=->x{x[0]+x.upcase.delete("-_")}
->x,y{f[x]==f[y]}

Try it online!

This feels really long still feels a bit long. I would appreciate the help of any Ruby gurus out there in making this at least a bit shorter.

C++, 288 248 bytes

-5 bytes thanks to Zacharý

#include<string>
#include<algorithm>
#define E(a,v)a.erase(std::remove(a.begin(),a.end(),v),a.end());
#define F(a)for(auto&c:a)c=toupper(c);
int e(std::string&a,std::string&b){if(a[0]!=b[0])return 0;E(a,45)E(b,45)E(a,95)E(b,95)F(a)F(b)return a==b;}

Thanks you, Preprocessor. Also, this code takes advantage of the fact that in C++ the rule to cast int to bool is int_var!=0

CJam, 20 bytes

{_"-_"f-:el:=\:c:=*}

Takes input in the form of ["string1","string2"].

Try it Online (testing version)

{
_      e# make copy of input
"-_"f- e# remove all "-" and "_" from both words in copy
:el    e# convert words in copy to lowercase
:=     e# 1 if both words in copy are equal, 0 if not
\      e# move original version of input to top of stack
:c     e# convert each word in original input to only 1st character
:=     e# 1 if both characters from original input are equal, 0 if not
*      e# multply the two numbers we obtained. If and only if both are 1 (true) we return 1 (true)
}

Haskell, 85 78 76 71 68 bytes

2 bytes saved thanks to Ørjan Johansen

import Data.Char
s(a:x)=a:[toLower a|a<-x,all(/=a)"-_"]
x!y=s x==s y

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Errors on the empty string.

Python 2, 72 bytes

lambda x,y:r(x)==r(y)
r=lambda x:x[0]+x[1:].lower().translate(None,"-_")

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Won't work with Python 3 because of the new translate syntax.

Excel, 105 bytes

=AND(CODE(A1)=CODE(B1),SUBSTITUTE(SUBSTITUTE(A1,"_",""),"-","")=SUBSTITUTE(SUBSTITUTE(B1,"_",""),"-",""))

CODE() returns numeric code of first character.

String comparison in Excel is case insensitive.

Husk, 13 bytes

¤=§,←(m_ω-"-_

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Builds for each string a pair consisting of the first character of the string and the whole string lowercased and with all occurrences of -/_ removed. Then checks if the two pairs are equal.

A particularity is that - in Husk is set difference (i.e. it removes only the first occurrence found): in order to remove all occurrences, the fixed point of -"-_ is found with ω-"-_.

Japt, 14 25 bytes

g ¥Vg ©Uu k"_-" ¥Vu k"_-"

Checks case-insensitive string equality by removing all characters in word 2 from word 1, and removing the -_ characters; that results in an empty string ("") if the words are equal.
Thanks Ørjan Johansen for pointing out the problem with this.

Checks first-char equality and if the uppercased inputs are equal after removing _-.

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Explanation

Implicit input: U and V are input strings

g ¥Vg

Check if first letter of U (implicit) equals (¥) the first char of V.

©Uu k"_-" ¥Vu k"_-"

And (©) check if U, uppercased (u) and with _- removed (k), equals (¥) the same for V. Implicitly return the boolean result.

Python 2, 79 73 bytes

-6 bytes thanks to @notjagan: check the length of set of all reduced names is 1 or not.

lambda*l:len({x[0]+re.sub('-|_','',x[1:].lower())for x in l})<2
import re

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Perl 5, 67 bytes

s/.//,push@a,$&,y/_-//dr for<>;say($a[0]eq$a[2]&&lc$a[3]eq lc$a[1])

Try it online!

Takes the identifiers as input on separate lines.

Explanation:

s/.//,             # remove the first character
push@a,            # storage space, even positions are first character
                   # odd positions are remainder
$&,                # implicit variable holding last matched pattern (first char)
y/_-//dr           # Remove _ and - from remainder of input
for<>;             # iterate over all input lines
say                # output
($a[0]eq$a[2]&&    # check that first character is identical and
lc$a[3]eq lc$a[1]) # so is the lowercase version of the rest

Retina, 28 bytes

T`L\_-`l_`(?<=.).+
^(.*)¶\1$

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Charcoal, 29 bytes

∧⁼§θ⁰§η⁰⁼↧⪫⪪⪫⪪θ_ω-ω↧⪫⪪⪫⪪η_ω-ω

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This prints a - for truthy and nothing for falsey.

Link to the verbose version. It first compares the first character of both input strings (⁼§θ⁰§η⁰) and then compares the rest of both strings after removing the underscores and the hyphens (⪫⪪⪫⪪θ_ω-ω) and converting to lowercase (↧).

C#, 101 89 bytes

string g(string s)=>string.Concat(s.ToUpper().Split('-','_'));f=>s=>f[0]==s[0]&g(f)==g(s)

Saved 12 bytes thanks to @kusi581.

Java (OpenJDK 8), 95 bytes

a->b->a.charAt(0)==b.charAt(0)&&a.replaceAll("_|-","").equalsIgnoreCase(b.replaceAll("_|-",""))

Try it online! Pretty straight forward.

Pyke, 13 bytes

F'hl1"-_"-)Xq

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F         )   -  for i in input():
 'hl1         -   i[0], i.lower()
     "-_"-    -         ^.remove("-_")
           Xq - equal(^)

C (gcc), 126 114 bytes

#define p(s)do++s;while(*s==45||*s==95);*s>95?*s-=32:0;
f(char*a,char*b){while(*a&&*a==*b){p(a)p(b)}return*a==*b;}

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With whitespace and comments:

#define p(s)                   // Define helper macro p           \
    do ++s;                    // Increment pointer at least once \
    while (*s==45 | *s==95);   // and past any '-' or '_'         \
    *s>95 ? *s -= 32 : 0;      // If lowercase letter, convert to upper

f(char* a, char* b) {          // Define main function f
    while (*a && *a == *b) {   // Loop until end of either string
                               // or a difference found
        p(a)                   // Transform pointer and one char
        p(b)                   // via helper p above
    }
    return *a==*b;             // Test chars equal (on success, both '\0')
}

Dyalog APL, 47 32 28 27 26 22 bytes

-4 bytes thanks to Kritixi Lithos

{(=/⊃¨⍵)∧≡/819⌶⍵~'-_'}   

Takes input as a list of the strings.

Try it online!

How?

{(=/⊃¨⍵)∧≡/819⌶⍵~'-_'}
               ⍵~'-_'   Remove '-' and '_'
           819⌶         Lowercase
         ≡/             Equality between elements
        ∧               And
 (=/⊃¨⍵)                The first element of each element is equal

Common Lisp, 98 bytes

(lambda(x y)(and(eql(elt x 0)(elt y 0))(string-equal(#1=remove #\-(#1##\_ y))(#1##\-(#1##\_ x)))))

Try it online!

Ungolfed (super straightforward!) version:

(defun f(x y)
  (and (eql (elt x 0) (elt y 0))         ; check if initial characters are identical
       (string-equal                     ; string comparison (case insensitive)
         (remove #\- (remove #\_ y))     ; remove from both strings the unwanted chars
         (remove #\- (remove #\_ x)))))

R, 76 bytes

function(l)(g=substr(l,1,1))[1]==g[2]&(h=tolower(gsub('-|_','',l)))[1]==h[2]

Anonymous function that takes input as a list of two strings. Takes advantage of the fact that R's string operations, while quite long in # of characters, are vectorized. Additionally wrapping an assignment in parentheses will bind the variable, so (g=substr(l,1,1)) retains a variable to be reused later in the line and similarly for h.

R returns the last evaluated expression as function output.

Ungolfed:

function(l){
  g <- substr(l,1,1)
  h <- tolower(gsub("_|-","",l))
  (g[1]==g[2])&(h[1]==h[2])
}

Try it online! (all test cases)

Brachylog, 17 bytes

hᵛ&{{¬∈"_-"&ụ}ˢ}ᵛ

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Outputs through predicate success/failure.

h                    The first element
 ᵛ                   is the same for each element of the input,
  &                  and
   {           }ᵛ    for each element of the input the following are the same:
    {      &ụ}ˢ      every element uppercased which satisfies the condition that
     ¬∈              it is not an element of
       "_-"          the string "_-".

Erlang 113 bytes

A=fun(L)->string:to_lower(lists:flatten(string:tokens(L,"-_")))end,fun([C|D],[C|E])->A(D)==A(E);(_,_)->a==b end.

a pair of anonymous functions that compare the two lists. meant to be pasted in the erlang shell.

more readable:

A=fun(L) ->
    string:to_lower( % case insensitive
        lists:flatten( % squash all characters back into one list
            string:tokens(L,"-_") % return a list of list of characters
        )
    )
end.
fun([C|D],[C|E]) -> % are the first characters exactly the same?
    A(D)==A(E); % does the rest compare correctly?
   (_,_) -> % first chars not the same
    a==b % shorter than 'false'
end.

Clip, 25 bytes

&=(x(y=AxAy[Aa--m.L`a'-'_

Explanation:

x, y and z may be referenced in a Clip program to implicitly take up to three inputs. Since this program only references x and y, it takes two inputs which are assigned to x and y.

 =(x(y                    First characters of x and y are equal
&                         And
      =AxAy               A(x) == A(y)
           [Aa            Function A, takes parameter a
                m.L`a     Map all elements of a to lower case
              --     '-'_ Remove all occurrences of '-' and '_'

Takes two strings from standard input, outputs 1 and 0 for true and false respectively.