Targeted duplicate removal

Question

The task

In this challenge, you are given a number and a list. Your task is to remove from the list all occurrences of the given number except the first (leftmost) one, and output the resulting list. The other elements of the list should be left intact.

The number will be a positive integer below 1000, and the list will only contain positive integers below 1000.
The list is not guaranteed to contain any occurrences of the given number. It may even be empty. In these cases you should output the list as-is.
Input and output formats are flexible within reason. You can output by modifying the list in place.
The lowest byte count wins.

Test cases

5 [] -> []
5 [5] -> [5]
5 [5,5] -> [5]
10 [5,5] -> [5,5]
10 [5,5,10,10,5,5,10,10] -> [5,5,10,5,5]
2 [1,2,3,1,2,3,1,2,3] -> [1,2,3,1,3,1,3]
7 [9,8,7,6,5] -> [9,8,7,6,5]
7 [7,7,7,7,7,7,7,3,7,7,7,7,7,7,3,7,1,7,3] -> [7,3,3,1,3]
432 [432,567,100,432,100] -> [432,567,100,100]

Are we allowed to take the length of the input list as one of the inputs? — user96495, Oct 12, 2020 at 9:02
@2x-1 No, unless you can't compute it from the list in your language. — Zgarb, Oct 12, 2020 at 9:15

caird coinheringaahin g · Accepted Answer · 2020-12-15 07:48:03Z

10

Husk, 5 4 bytes

üoEė

Try it online!

Saved 1 byte thanks to Razetime
Saved 2 bytes thanks to Jo King

Thanks to Razetime for suggesting ü and Jo King for letting me know I could leave the superscript arguments out, saving me 2 bytes. It removes duplicates with a custom predicate that makes sure both arguments are equal to the number to be removed.

Explanation:

üoEė
ü     Remove duplicates by binary function (implicit second argument)
 o    Compose 2 functions
   ė  Make a list of 3 elements (first element is implicitly added)
  E   Are they all equal?

edited Dec 15, 2020 at 7:48

caird coinheringaahin g♦

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answered Oct 12, 2020 at 17:02

user

2872 gold badges20 silver badges68 bronze badges

1

\$\begingroup\$ Ë= can just be E \$\endgroup\$
– Razetime
Oct 12, 2020 at 17:16
\$\begingroup\$ @Razetime Nice, thanks! \$\endgroup\$
– user
Oct 12, 2020 at 17:23

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Giuseppe · Accepted Answer · 2020-10-12 14:10:07Z

9

R, 30 bytes

function(l,d)unique(l,l[l!=d])

Try it online!

unique() has the signature unique(x,incomparables = FALSE,...); this sets incomparables to the elements that aren't equal to d, so only d is uniquified.

edited Oct 12, 2020 at 14:10

answered Oct 12, 2020 at 12:27

Giuseppe

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xnor · Accepted Answer · 2020-10-12 10:12:39Z

Python 2, 50 bytes

l,n=input()
for x in l:
 if~n-x:print x;n^=-(x==n)

Try it online!

Prints output one entry per line.

The idea is to store whether we have already encountered the entry-to-remove n in the sign of n rather than a separate Boolean variable. When we see a list entry that equals n, we negate n. To decide whether to print the current entry x, we check if it equals -n, which checks that it equals the original n and that we've already negated n due to an earlier match. Note that since n and list entries are positive, there's no way to get x==-n before n is negated.

Well, actually, instead of negating n, it's shorter to bit-complement it to ~n, which is -n-1. To do the conditional complementing, we note that we can convert [x,~x][b] to x^-b (as in this tip), using that bitwise xor ^ has x^0==x and x^-1==~x. So, we do n^=-(x==n).

Arnauld · Accepted Answer · 2020-10-12 09:43:10Z

JavaScript (ES6), 32 30 bytes

Expects (x)(list).

x=>a=>a.filter(v=>v^x||a[a=0])

Try it online!

How?

All values v that are not equal to x are preserved thanks to v^x. The first value that is equal to x is kept as well because a[0] is guaranteed to be a positive integer (except if a is empty, but then we don't enter the .filter() loop to begin with). For the next values that are equal to x, we have a = 0 and a[0] === undefined, so they are rejected. This test doesn't throw an error because Numbers are Objects, so it's legal to access the (non-existent) property '0' of 0.

Kyuuhachi · Accepted Answer · 2020-10-12 23:55:44Z

6

Haskell, 43 bytes

a%(b:c)|a==b=b:filter(/=a)c|1<2=b:a%c
_%x=x

Try it online!

Ungolfed:

dedupl v (x:xs)
  | x == v = x : filter (/= v) xs
  | otherwise = x : dedupl v xs
dedupl _ [] = []

Haskell, 40 bytes

This version takes a (negative) predicate for input instead.

f%(b:c)|f b=b:f%c|1<2=b:filter f c
_%x=x

Try it online!

answered Oct 12, 2020 at 23:55

Kyuuhachi

3912 silver badges2 bronze badges

\$\begingroup\$ Simple and elegant! How I didn't thought about that? How I messed up doing complicated stuffs when the solution was so obviously simple? Haskell journey is getting interesting day by day \$\endgroup\$
– AZTECCO
Oct 13, 2020 at 21:40

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Noodle9 · Accepted Answer · 2020-10-12 21:30:43Z

C (gcc), 60 bytes

Saved 2 bytes thanks to ceilingcat!!!
Saved 2 bytes thanks to ErikF!!!

t;f(d,l)int*l;{for(t=0;*l;++l)*l==d&&t++||printf("%d ",*l);}

Try it online!

Inputs a number and a pointer to a null terminated array (as there's no way to know the length of an array passed into a function in C) and outputs the filtered array to stdout.

Explanation

f(d,                        // function taking the duplicate number d,  
    l)int*l;{               // a null terminated array of int l  
  for(                      // loop...  
      t=0;                  //   init test flag t to 0, this will mark the  
                            //   1st (if any) occurance of d  
          *l;               // ...over the array elements  
              ++l)          // bumping the array pointer each time  
      *l==d                 // if the array element isn't d...  
           &&t              //   or it's the 1st time seeing d  
               ++           //   unmark t by making it non-zero  
      ||printf("%d ",*l);   // ...then print that element  
}

If you use 0 as a sentinel, you don't need to take a length (61 bytes): Try it online! — ErikF, Oct 12, 2020 at 20:40

ovs · Accepted Answer · 2020-10-16 15:45:34Z

5

APL (Dyalog Unicode), 18 15 10 bytes

Thanks to Adám for -8 bytes!!!

∊⊢⊆⍨≠∨<\⍤=

Try it online!

Example inputs: left argument 3, right argument 1 2 3 4 3 4.

= does element-wise not-equal comparison. => 0 0 1 0 1 0
<\ Scans with less-than. This keeps just the first 1, all other places are 0. => 0 0 1 0 0 0
≠∨ does element-wise OR with the ≠ mask. => 1 1 1 1 0 1.
⊢⊆ partitions the input based on the vector, including positions with positive integers. => (1 2 3 4) (4) ∊ flattens the nested array. => 1 2 3 4 4

edited Oct 16, 2020 at 15:45

answered Oct 12, 2020 at 9:36

ovs

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\$\begingroup\$ -3: ∊⊢⊆⍨≠∨¯1↓1,⌊\⍤≠ Try it online! \$\endgroup\$
– Adám
Oct 12, 2020 at 10:05
\$\begingroup\$ @Adam thanks a lot! \$\endgroup\$
– ovs
Oct 12, 2020 at 10:27
\$\begingroup\$ -5: ¯1↓1,⌊\⍤≠ → <\⍤= Try it online! \$\endgroup\$
– Adám
Oct 14, 2020 at 11:12

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Jonathan Allan · Accepted Answer · 2020-10-17 15:27:05Z

Jelly, 6 5 bytes

-1 thanks to Sisyphus's suggestion to use Ẇ in place of W€

Ẇi¦⁹ḟ

A full program accepting the list and the value which prints the Jelly representation of a list with all but the first occurrence of the value removed (empty lists print nothing, lists with one element print that element).

Try it online! Or see the test-suite.

How?

Ẇi¦⁹ḟ - Link: list, A; value V
  ¦   - sparse application...
 i ⁹  - ...to indices: first occurrence of V in A  ([0] if no V found)
W     - ...action: all non-empty sublists (since ¦ zips, the element, z, at any
                                           given index of A will be [z])
    ḟ - filter discard occurrence of V (leaves the [z] as is)
      - implicit print

I thought ḟẹḊ¥¦ would work for 5, but it fails with a divide by zero error with [5,5] and 5.

@Sisyphus Nice, it will, since the first n entries of the results of Ẇ for a length n list are the wrapped elements of that list. — Jonathan Allan, Oct 17, 2020 at 15:20

Shaggy · Accepted Answer · 2020-10-12 10:52:19Z

4

Japt, 10 7 bytes

kÈ¶V©T°

Try it

-3 bytes thanks to caffeine!

kÈ¶V©T°     :Implicit input of array U and integer V
k           :Remove the elements in U that return true
 È          :When passed through the following function
  ¶V        :Is equal to V?
    ©       :Logical AND with
     T°     :Postfix increment T (initially 0)

edited Oct 12, 2020 at 10:52

answered Oct 12, 2020 at 9:13

Shaggy

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Kevin Cruijssen · Accepted Answer · 2020-10-12 12:50:37Z

4

05AB1E, 6 bytes

Ê0X.;Ï

Integer as first input, list as second input.

Try it online or verify all test cases.

Explanation:

Ê      # Check for each value in the second (implicit) input-list whether it's NOT equal
       # to the first (implicit) input-integer (1 if NOT equal; 0 if equal)
 0X.;  # Replace the first 0 with a 1
     Ï # And only keep the values in the (implicit) input-list at the truthy (1) indices
       # (after which the result is output implicitly)

edited Oct 12, 2020 at 12:50

answered Oct 12, 2020 at 12:44

Kevin Cruijssen

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Noodle9 · Accepted Answer · 2020-10-13 15:06:45Z

3

Bash + sed, 49 bytes

sed "s/\b$1\b/_/;s/\b$1\b \?//g;s/_/$1/"<<<${*:2}

Try it online!

Takes the first argument as the duplicate and the rest as the array.

edited Oct 13, 2020 at 15:06

answered Oct 12, 2020 at 18:41

Noodle9

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Adám · Accepted Answer · 2020-10-14 11:21:01Z

APL (Dyalog Unicode), 9 bytes (SBCS)

Translation of Galen Ivanov's J solution.

Anonymous tacit infix function, taking number as left argument and list as right argument (though argument order can be switched by changing the ⊢s into ⊣s).

∊⊢⊆⍨≠∨∘≠⊢

Try it online!

⊢ on the right argument

…∘≠ apply nub-sieve (Boolean list with Trues where unique elements occur first), then:

…∨ element-wise OR that with:

≠ Boolean list with Trues where elements in the list are different from the number

…⊆⍨ corresponding to runs of Trues in that, extract runs in:

⊢ the list

∊ ϵnlist (flatten)

att · Accepted Answer · 2020-12-14 05:25:52Z

3

Wolfram Language (Mathematica), 26 24 bytes

#2/.(a=#)/;a++>#:>Set@$&

Try it online!

The pattern (a=#) to be matched is only evaluated once, at the very start. Then, the condition a++># is only evaluated when the pattern is matched - so a will have been incremented on subsequent matches.

edited Dec 14, 2020 at 5:25

answered Oct 12, 2020 at 17:49

att

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Galen Ivanov · Accepted Answer · 2020-10-12 09:19:44Z

2

Red, 46 bytes

func[n b][try[replace/all find/tail b n n[]]b]

Try it online!

answered Oct 12, 2020 at 9:19

Galen Ivanov

21k3 gold badges23 silver badges61 bronze badges

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Jitse · Accepted Answer · 2020-10-12 09:27:05Z

2

Python 3, 62 bytes

f=lambda n,l:l.count(n)>1and f(l.pop(~l[::-1].index(n)),l)or l

Try it online!

This function will recursively pop the last instance of the given value until no more than one instance is present. Then it returns the list.

or, for the same byte count

lambda n,l:[j for i,j in enumerate(l)if j!=n or i==l.index(n)]

Try it online!

This is just a simple filter.

edited Oct 12, 2020 at 9:27

answered Oct 12, 2020 at 8:27

Jitse

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ovs · Accepted Answer · 2020-10-12 10:31:42Z

2

05AB1E, 8 bytes

ʒÊD¾ms_½

Try it online!

Commented:

ʒ          # filter the first input on ...
 Ê         #   not equal to the second input (n)?
  D        #   duplicate this value
   ¾       #   push the counter variable initially 0
    m      #   power (value != n)**(counter)
           #   this is only 0 if value==n and counter is positive
     s     #   swap to (value != n)
      _    #   negate this
       ½   #   increment the counter variable if this is truthy (value == n)

edited Oct 12, 2020 at 10:31

answered Oct 12, 2020 at 10:16

ovs

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Kjetil S · Accepted Answer · 2020-10-12 11:00:56Z

2

Perl 5, 36 bytes

sub{$n=pop;$i=0;grep$n-$_||!$i++,@_}

Try it online!

Pop last input value from @_ into $n. The remaining @_ is the input list. Filter (grep) @_ for the values that either isn't equal to $n ($n-$_ is truthy when $n and current list value $_ is different) or is the first equal to $n since !$i++ is truthy for the first and not for the rest.

edited Oct 12, 2020 at 11:00

answered Oct 12, 2020 at 10:42

Kjetil S

5,4879 silver badges17 bronze badges

1

\$\begingroup\$ There's no need for the $i=0; part, unless you're using the sub multiple times. Which you shouldn't as this is code golf. \$\endgroup\$
– Abigail
Oct 12, 2020 at 21:31

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Galen Ivanov · Accepted Answer · 2020-10-12 13:36:04Z

2

J, 15 10 bytes

-5 bytes thanks to xash!

]#~=<:~:@]

Try it online!

My initial solution:

J, 15 bytes

[#~~:+i.@#@[=i.

Try it online!

edited Oct 12, 2020 at 13:36

answered Oct 12, 2020 at 10:30

Galen Ivanov

21k3 gold badges23 silver badges61 bronze badges

\$\begingroup\$ @xash Really nice, thanks! +. could be >. :) \$\endgroup\$
– Galen Ivanov
Oct 12, 2020 at 13:23
1

\$\begingroup\$ Ah, with that kind of logic it's 10 bytes :-) \$\endgroup\$
– xash
Oct 12, 2020 at 13:31
\$\begingroup\$ @xash That's great, thanks again! \$\endgroup\$
– Galen Ivanov
Oct 12, 2020 at 13:35

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Graham · Accepted Answer · 2020-10-12 13:39:08Z

2

APL+WIN, 19 bytes

Prompts for vector followed by element to be removed:

((v≠n)+<\v=n←⎕)/v←⎕

Try it online! Thanks to Dyalog Classic

edited Oct 12, 2020 at 13:39

answered Oct 12, 2020 at 13:30

Graham

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Razetime · Accepted Answer · 2020-10-12 17:15:56Z

2

Husk, ¹³ 12 bytes

F+ṀΓ·:f≠⁰↕≠⁰

Try it online!

user's answer.(-3 bytes, then -1 byte.)

Husk, 16 bytes

J²fI§e←of≠²→↕≠²⁰

Try it online!

~~Can probably be shortened with Γ.~~

~~There may be an extremely short solution with ü as well.~~user's answer

+2 bytes after supporting numbers not in the list.

edited Oct 12, 2020 at 17:15

answered Oct 12, 2020 at 9:22

Razetime

27k3 gold badges26 silver badges71 bronze badges

2

\$\begingroup\$ The 5-byter doesn't work on test cases 5 and 6. \$\endgroup\$
– Zgarb
Oct 12, 2020 at 15:02
\$\begingroup\$ @user once arguments are used explicitly, they can't be removed. \$\endgroup\$
– Razetime
Oct 12, 2020 at 16:43
1

\$\begingroup\$ @user You don't need to apologize to me for posting your own answer :) \$\endgroup\$
– Razetime
Oct 12, 2020 at 17:14

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LiefdeWen · Accepted Answer · 2020-10-13 07:54:27Z

2

C# (Visual C# Interactive Compiler), 42 bytes

a=>b=>a.Where((x,i)=>x!=b|i==a.IndexOf(b))

Try it online!

answered Oct 13, 2020 at 7:54

LiefdeWen

3,86515 silver badges38 bronze badges

\$\begingroup\$ -1 byte with is operator sorcery. \$\endgroup\$
– the default.
Oct 13, 2020 at 13:57
\$\begingroup\$ @thedefault. That is very impressive, I would've never gotten there, you don't want to post that yourself? \$\endgroup\$
– LiefdeWen
Oct 13, 2020 at 14:08

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Tomer Shetah · Accepted Answer · 2020-10-13 12:33:02Z

2

Scala, 62 61 58 bytes

a=>s=>{val(c,d)=s splitAt s.indexOf(a)+1;c++d.filter(a!=)}

Try it online!

Thanks to Galen for -1 character
Thanks to user for -3 characters

edited Oct 13, 2020 at 12:33

answered Oct 13, 2020 at 8:11

Tomer Shetah

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AZTECCO · Accepted Answer · 2020-10-13 22:05:02Z

2

Haskell, 55 bytes

f n=foldl(\a x->if x==n&&x`elem`a then a else a++[x])[]

Try it online!

Previous answer was complicated and not so good so I decided to try a more expressive approach inspired by @Caagr98 answer, mine is still longer but I feel better now =)

# Previous 72 bytes

g b n(h:t)
 |h/=n=h:g b n t
 |b>0=g 1n t
 |1>0=h:g 1n t
g b n _=[]
f=g 0

Try it online!

edited Oct 13, 2020 at 22:05

answered Oct 12, 2020 at 21:48

AZTECCO

10.6k1 gold badge13 silver badges58 bronze badges

1

\$\begingroup\$ Thank you @user ! Anyway this is a mess, compared to codegolf.stackexchange.com/a/212430/84844 mine is not only longer, it's not elegant and and less readable. I think Caagr98 answer is exactly as expressive as Haskell is meant to be \$\endgroup\$
– AZTECCO
Oct 13, 2020 at 21:32

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Julian · Accepted Answer · 2020-12-15 19:57:09Z

2

PowerShell Core, 50 31 bytes

param($a,$b)$b|?{$_-$a-or!$o++}

-19 bytes thanks to mazzy!

Try it online!

Iterates on the array passed as the second parameter and ignores the duplicate occurrence of the first parameter.

edited Dec 15, 2020 at 19:57

answered Dec 15, 2020 at 6:06

Julian

2,6886 silver badges16 bronze badges

1

\$\begingroup\$ because numbers \$\endgroup\$
– mazzy
Dec 15, 2020 at 8:02
1

\$\begingroup\$ foreach-object -> where-object \$\endgroup\$
– mazzy
Dec 15, 2020 at 8:09

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TheCoderPro · Accepted Answer · 2021-07-02 12:33:40Z

2

JavaScript (Node.js), 117 113 93 bytes

function x(i,j){var d,e,o=[];for(x in i){e=i[x]==j;!(e&&d)?o.push(i[x]):0;e?d=1:0;}return o;}

Try it online!

edited Jul 2, 2021 at 12:33

answered Oct 19, 2020 at 3:28

TheCoderPro

716 bronze badges

2

\$\begingroup\$ It's a good practice to hose your programs on tio so people can check: Try it online! \$\endgroup\$
– Razetime
Oct 19, 2020 at 3:44
1

\$\begingroup\$ You can get this to 62 bytes, I think. Probably a lot more with filter and more golfing tricks, but I don't know JS that well \$\endgroup\$
– user
Oct 21, 2020 at 15:00

Add a comment |

Neil · Accepted Answer · 2020-10-12 09:56:50Z

Charcoal, 12 bytes

ＩΦη∨⁻ιθ⁼κ⌕ηι

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

  η             Input list
 Φ              Filtered where
     ι          Current element
    ⁻           Subtract (i.e. does not equal)
      θ         Input integer
   ∨            Logical Or
        κ       Current index
       ⁼        Equals
         ⌕      First index of
           ι    Current element in
          η     Input list
Ｉ               Cast to string
                Implicitly print

The last character could also be θ of course since the two variables are equal at that point.

ovs · Accepted Answer · 2020-10-12 09:59:39Z

1

Python 2, 55 52 bytes

Thanks to xnor for -3 bytes!

Output is newline-separated.

n,l=input()
x=1
for d in l:
 if x|d-n:print d;x*=d-n

Try it online!

edited Oct 12, 2020 at 9:59

answered Oct 12, 2020 at 8:41

ovs

58.1k3 gold badges42 silver badges154 bronze badges

1

\$\begingroup\$ It looks like it's shorter to just write d-n in place of w. \$\endgroup\$
– xnor
Oct 12, 2020 at 9:51
\$\begingroup\$ @xnor It is indeed. I initially had d!=n and it was the same length, but I forgot to check again. Thanks a lot! \$\endgroup\$
– ovs
Oct 12, 2020 at 10:00

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Galen Ivanov · Accepted Answer · 2020-10-12 13:09:04Z

1

K (Kona), 21 bytes

{y@&(~x=y)+(!#y)=y?x}

Try it online!

answered Oct 12, 2020 at 13:09

Galen Ivanov

21k3 gold badges23 silver badges61 bronze badges

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Conor O'Brien · Accepted Answer · 2020-10-13 16:38:11Z

Stacked, 28 bytes

[@y:0@b[b\y=:b+@b*¬]"!keep]

Try it online!

Explanation

[@y:0@b[b\y=:b+@b*¬]"!keep]
[                         ]   anonymous function (expects 2 args)
 @y                           save top as y
    0@b                       initialize b = 0
   :   [           ]"!        for each element E in the input array:
        b\                      save the current value of b for later computation
          y= b+@b               b = max(b, y == E)
        b y=:    *¬             not both (old b) and (y == E) are true
                                for y != E, and for the first y == E, this is 1, else 0
                              this generates a mask of 1s and 0s
                      keep    keep only the elements in the input which correspond to a 1

PHP 63 Bytes

Number provided in $n, list provided in $a,

$p=explode($n,$a,2);echo$p[0].$n.str_replace("$n,", '', $p[1]);

Ungolfed:

$p = explode($n,$a,2);
echo $p[0].$n.str_replace("$n,", '', $p[1]);

e.g.

$n=432;
$a="[432,567,100,432,100]";
$p = explode($n,$a,2);
echo $p[0].$n.str_replace("$n,", '', $p[1]);

(I'm unsure if it's ok not to count the input into the bytes, or the opening '<?php' for that matter...)

answered Oct 17, 2020 at 23:05

Zsolt Szilagy

7734 silver badges10 bronze badges

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Targeted duplicate removal

The task

Test cases

34 Answers 34

Husk, 5 4 bytes

R, 30 bytes

Python 2, 50 bytes

JavaScript (ES6), 32 30 bytes

How?

Haskell, 43 bytes

Haskell, 40 bytes

C (gcc), 60 bytes

Explanation

APL (Dyalog Unicode), 18 15 10 bytes

Jelly, 6 5 bytes

How?

Japt, 10 7 bytes

05AB1E, 6 bytes

Bash + sed, 49 bytes

APL (Dyalog Unicode), 9 bytes (SBCS)

Wolfram Language (Mathematica), 26 24 bytes

Red, 46 bytes

Python 3, 62 bytes

05AB1E, 8 bytes

Perl 5, 36 bytes

J, 15 10 bytes

My initial solution:

J, 15 bytes

APL+WIN, 19 bytes

Husk, 13 12 bytes

Husk, 16 bytes

C# (Visual C# Interactive Compiler), 42 bytes

Scala, 62 61 58 bytes

Haskell, 55 bytes

PowerShell Core, 50 31 bytes

JavaScript (Node.js), 117 113 93 bytes

Charcoal, 12 bytes

Python 2, 55 52 bytes

K (Kona), 21 bytes

Stacked, 28 bytes

Explanation

Other Solutions

PHP 63 Bytes

Ungolfed:

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