Appending Numbers

Question

A fairly simple challenge: You will receive two inputs, a string and a number (the number may be taken as a string, ie "123" instead of 123)

If the string does not end in a number (ie, it does not match the regex \d$), just append the number to the end of the string.

If the string does end in a number (ie, it matches the regex \d+$), you should first delete that and then append the number.

Neither of the inputs will ever be invalid or empty (invalid is defined by the numerical input not containing only digits)

The number will never contain a - or a ..

The string will never contain a newline, or unprintable non-whitespace characters.

Test Cases:

abc123 + 345 -> abc345
123 + 1 -> 1
hello + 33 -> hello33
123abc123 + 0 -> 123abc0
ab3d5 + 55 -> ab3d55
onetwo3 + 3 -> onetwo3
99ninenine + 9999 -> 99ninenine9999

Answer 1

Retina, 5 bytes

\d*¶

Takes two strings as input, separated by a newline.

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Answer 2

Python 2, 30 bytes

lambda a,b:a.rstrip(`56**7`)+b

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Answer 3

PHP, 37 36 35 33 bytes

<?=chop($argv[1],3**39),$argv[2];

Saved 1 byte thanks to Jörg Hülsermann.

Answer 4

Perl 5, 12 bytes

11 bytes code + 1 for -p.

s/\d*$/<>/e

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Answer 5

Java 8, 32 bytes

a->b->a.replaceAll("\\d*$","")+b

Takes input a as a String, and for b it doesn't matter whether it's a String or integer (although I use Integer in the TIO-link below).

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Answer 6

Python 2, 32 bytes

import re
re.compile("\d*$").sub

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Takes the inputs in reverse order, both as string.

Answer 7

05AB1E, 9 bytes

DvTFNÜ}}«

Try it online! Probably a bad solution but it's the best I could come up with

Explanation

DvTFNÜ}}«
Dv     }  # For each character in input1
  TF  }   # 10 times
    NÜ    # Get 0-9 and trim it from input1
        « # Concatenate with input2

Answer 8

Japt, 10 bytes

r"%d*$" +V

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 r"%d*$" +V
Ur"%d*$" +V # Implicit input (U and V)
 r          # Remove
  "%d*$"    #   any trailing digits
U           #   from the first input
         +V # And append the second input
            # Implicit output

Answer 9

Python 2, 44 41 39 bytes

EDIT: -4 bytes thanks to @Dom Hastings. I don't use regular expressions much.

EDIT 2: -2 bytes thanks to @totallyhuman pointing out that the number could be taken as a string

Had to be expected...

lambda x,y:re.sub("\d*$",y,x)
import re

Just removes the digits at the end of the string and appends the number

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Answer 10

Pip, 9 7 bytes

a<|XD.b

@DLosc saved me 2 bytes!

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Explanation

a<|         Strip all matches off the end of 'a' (the first cmd line arg)
   XD         of the pattern \d (ordinarily, a regex would be entered as '\d', but digits 
              have a pre-existing constant XD)
     .b     Then concatenate 'b' (the second cmd line arg)
            PIP implicitly prints the results of the last operation.

Answer 11

Jelly, 5 bytes

œrØD;

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How it works

œrØD;  Main link. Left argument: s. Right argument: t

  ØD   Digits; yield "0123456789".
œr     Trim these characters from the right of s.
    ;  Append t.

Answer 12

JavaScript (ES6), 28 26 25 bytes

x=>y=>x.replace(/\d*$/,y)

• 1 byte saved thanks to Neil reminding me why I shouldn't golf early in the morning!

Answer 13

Ruby, 21 bytes

->a,b{a=~/\d*$/;$`+b}

Takes two strings in input.

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Answer 14

C#, 45 bytes

s=>n=>s.TrimEnd("0123456789".ToCharArray())+n

Explanation:

s=>n=>                                        // Take input
      s.TrimEnd(                              // Trim the end of the string with the specified chars
                "0123456789"                  // Create a string of the digits
                            .ToCharArray())   // Turn the string into a char array
                                           +n // Append the integer onto the end

Answer 15

V, 7 4 bytes

óä*î

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This uses the same regex as the Retina answer above:

s:/\d*\n//

Answer 16

Perl 6, 20 bytes

->$_,\n{S/\d*$/{n}/}

Answer 17

Tcl 32 bytes

proc s a\ b {regsub \\d*$ $a $b}

I'm not sure about the expected interface. This is done as a procedure that accepts the two inputs as call arguments. To turn it to a standalone script that reads input from stdin and outputs the result to stdout, one would need the extra line:

puts [s [gets stdin] [gets stdin]]

or would do it all "inline":

puts [regsub \\d*$ [gets stdin] [gets stdin]]

regsub takes an RE, the original string and a string to replace the matching part with.

Answer 18

Mathematica, 48 bytes

There is already a Mathematica solution (84 bytes).

StringDrop[StringTrim["."<>#,_?DigitQ..]<>#2,1]&

Answer 19

Carrot, 16 21 bytes

$^//^.*?(?=\d*$)/S0^#

Try it online! (input is linefeed separated)

Explanation

$^                Set the stack-string to be equal to the first line in the input
/                 Set the stack-array to be equal to the matches of this regex:
 /^.*?(?=\d*$)/   The beginning of the string followed by non-digit characters at the end that are not included in the match.
S0                Convert to a string with 0 as the delimiter
^#                Append the rest of the input to the stack-string

I had to increase the bytecount by 5 because the code did not work for testcases like a5b3 with multiple digits.

Answer 20

Haskell, 75 bytes 95 bytes 91 79 61 bytes

b=(`notElem`['0'..'9'])
r=reverse
d#e=r(snd$break b$r d)++e

I tried doing this without regex so maybe that would be a dramatically improved answer. Also there are a couple ways I could go about this so I am unsure if I can shave a few bytes with a different approach.

UPDATE: I went up in bytes because I realized I was failing the test case where numbers exist in the string that are not the suffix. Now I am sure that regex would provide a much better answer.

UPDATE2: After some great feedback, more bytes were golfed!

Answer 21

AWK, 18 bytes

sub("[0-9 ]*$",$2)

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This just replaces a string composed of any number of digits and spaces that ends the arguments to the command with the second argument to the command. It's effectively stripping the commandline and adding back the second argument. And since there's guaranteed to be a non-empty second argument, the result is truthy and $0 is printed, which is the modified commandline arguments.

Answer 22

Mathematica, 84 bytes

(k=#;T=ToCharacterCode;L=T@k;While[47<Last@L<58,w=k~StringDrop~-1;k=w;L=T@w];w<>#2)&

input 2 strings

["ab3d5", "55"]

output

ab3d55

Answer 23

C (gcc), 99 98 96 95 bytes

Should be able to golf this down a bit...

main(c,v)char**v;{for(c=strlen(v[1]);~c*isdigit(v[1][--c]);v[1][c]=0);printf(v[1]),puts(v[2]);}

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Answer 24

Noether, 11 bytes

I"\d+$"-I+P

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When inputting a string, enclose it in quotation marks

Explanation:

I       - Push input to stack
"\d+$"  - Push string to stack
-       - Remove characters from first string which match the regex
I       - Push input to stack
+       - Append to the first string
P       - Print top item of stack

Answer 25

Arturo, 25 bytes

$=>[++replace&{/\d+$}""&]

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Answer 26

Go, 77 bytes

import."strings"
func a(s,n string)string{return TrimRight(s,"0123456789")+n}

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Answer 27

05AB1E, 8 bytes

DþRvyÜ}«

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Explanation:

DþRvyÜ}«
D        Duplicate input.
 þ       Get the digits of the input.
  R      Reverse the digits of the input.
   v  }  For each of the digits in the input:
    y        Push the current digit
     Ü       Trim that from the right of the input.
       « Concatenate to 2nd input.

Answer 28

Google Sheets, 28 bytes

=REGEXREPLACE(A1,"\d+$",)&B1