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Challenges Q&A Sandbox Meta
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Post Challenge

Repeat the characters

+5
−0

Given a string and a non-negative integer $n$, output a new string in which each character is repeated $n$ times.

Test cases:

"abc", 1    -> "abc"
"Hello", 0  -> ""
"double", 2 -> "ddoouubbllee"
"Hi", 4     -> "HHHHiiii"
"", 20      -> ""
code-golf string
posted about 3 years ago
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celtschk‭
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14 answers

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+1
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Python 3, 53 38 bytes

def f(x,y):
	for c in x:print(end=c*y)

Try it online!

I'm sure someone will find a lambda solution to this.

Golfed 15 bytes thanks to @Moshi's advice.

posted about 3 years ago
3y ago
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General Sebast1an‭
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1 comment thread

Just iterate directly (1 comment)
+3
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Haskell, 18 bytes

(.replicate).(>>=)

Try it online!

posted about 3 years ago
CC0
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Hakerh400‭
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+3
−0

Ruby, 34 29 bytes

->s,n{s.chars.map{_1*n}*""}

Try it online!

posted about 3 years ago
3y ago
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snail_‭
681 reputation 4 28 73 18
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+2
−0

APL (Dyalog Unicode), 1 byte

/

Try it online!

/ is called Replicate.

posted about 3 years ago
CC0
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Adám‭
436 reputation 2 11 45 8
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+2
−0

Python 3, 34 bytes

lambda a,b:''.join(b*i for i in a)

Try it online!

posted about 3 years ago
3y ago
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Endercraft2319‭
71 reputation 0 4 7 1
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+2
−0

JavaScript, 36 bytes

s=>n=>s.replace(/./g,`$&`.repeat(n))

Try it online!

posted about 3 years ago
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Shaggy‭
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+2
−0

jq, 32 27 bytes

-5 bytes (thank you Razetime)

.n as$n|.s/""|map(.*$n)|add

Try it online!

posted about 3 years ago
3y ago
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snail_‭
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2 comment threads

18 bytes: [Try it online!](https://tio.run/##yyr8/z9aQ69YX0lJMzpWSy8vtiYxJeX//2ouBQWlYiUrJY/UnJx8hfD8... (1 comment)
you can use `add` in place of join. (1 comment)
+2
−0

Rockstar, 90 bytes

listen to S
split S
listen to N
O's""
while S
roll S into C
let C be*N-0
let O be+C

say O

Try it here - code will need to be pasted in; s goes on the first line of input (leave blank for the empty string) and n goes on the second line.

posted about 3 years ago
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Shaggy‭
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+2
−0

Japt -m, 2 bytes 

pV

Try it

pV     :Implicit map of first input string
p      :Repeat
 V     :Second input times
posted about 3 years ago
3y ago
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Shaggy‭
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+2
−0

C (gcc), 56 bytes

i;f(s,n)char*s,n;{for(;*s;s++)for(i=n;i--;putchar(*s));}

Try it online!

My attempt at recursion ended up at 61 bytes:

o;f(s,n)char*s,n;{for(o=n;*s&&n--;putchar(*s));*s&&f(++s,o);}

I still think there's probably a more elegant way to solve both loops with recursion, but I can't spot it...

posted about 3 years ago
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Lundin‭
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+2
−0

Jelly, 1 byte

x

Try It Online!

x is Jelly's replicate atom. Since Jelly is inspired by languages like J, APL, etc, it is no surprise that is has an atom for a fairly important function.

posted about 3 years ago
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hyper-neutrino‭
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+1
−0

shortC, 40 37 bytes

i;f(C*s,In){O;*s;s++)Oi=n;i--;P*s));}

Try it online!

A shortC conversion of @Lundin's C (gcc) implementation.

posted about 3 years ago
3y ago
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General Sebast1an‭
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+0
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J, 1 byte

#

Try it online!

The copy verb. It works the exact same as APL and Jelly's replicate.

posted almost 3 years ago
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south‭
361 reputation 0 44 36 22
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+0
−0

Ruby, 23 bytes

->{_1.gsub /./,'\0'*_2}

Try this online!

posted almost 3 years ago
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radarek‭
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