Find The Least Common Characters

Question

Given a string, find the 10 least common alpha-numeric characters within that string.

Output should be formatted with numbers left-padded with spaces so they look like this:

e:      1
t:     16
c:     42
b:    400
z:  50000
a: 100000

Use this to test your program: here

Output for test case:

j:  2
x:  3
q:  9
k: 32
v: 67
c: 73
w: 79
f: 93
b: 94
p: 96

Shortest code wins. You can submit a program or a function.

Answer 1

K, 60

{(-:|/#:'f)$f:10#$(!x)[i]!r i:<r:. x:#:'=_x@&x in,/.Q`a`A`n}

Answer 2

Mathematica, 102 93:

v2:

  Grid[SortBy[
             Tally@StringCases[ToLowerCase@#,x_?LetterQ:>x<>":"],
              Last@# &],
      Alignment->Right]&@s

v1:

TableForm[
         SortBy[Tally@StringCases[ToLowerCase@#, x_?LetterQ :> x <> ":"],
                #[[2]] &], 
         TableAlignments -> Right] &

Answer 3

Bash, 72

Assuming no single-letter-named files in the current directory; input from stdin:

grep -io [a-z]|sort -f|uniq -ci|sort|head|sed -E 's/(.*)([a-z])/\2:\1/'

Works on Linux, but for some reason, my mac's grep seems to ignore the -i when used with -o (bug or undocumented gotcha?), but the following (78) works:

tr A-Z a-z|grep -o [a-z]|sort|uniq -c|sort|head|sed -E 's/(.*)([a-z])/\2:\1/'

If the output were allowed to be reversed, the sed part could be dropped to save 30 characters in each of the previous solutions.

Answer 4

Python, 206 199 195 characters

import sys
t=sys.stdin.read().lower()
f={}
for c in t:
 if c.isalnum():f[c]=f.get(c,0)+1
s=lambda x:x[1]
h=sorted(f.items(),key=s)[:10]
for n in h:print n[0]+': '+`n[1]`.rjust(len(`max(h,s)[1]`))

Undoubtedly there is a lot of room for improvement, but here's my first attempt. Produces identical output for the test case provided.

206 -> 199: Removed unnecessary newline and keyword argument

199 -> 195: iteritems()

Answer 5

Javascript, 271 characters

I'm not sure whether this breaks the rules of not (it doesn't justify the numbers in the output, but here's my solution anyway:

c=[];t=128;while(t--)c[t]=[0,t];i=prompt().toLowerCase();for(n=0;n<i.length;x=i.charCodeAt(n++),c[x][0]+=x>96&&x<123);for(n=i=0,c=c.sort(function(a,b){return(a[0]-b[0])});n<10;i++)if(c[i][0]>0){console.log(String.fromCharCode(c[i][1])+": "+("  "+c[i][0]).slice(-2));n++}

I'm pretty sure it can be simplified (I won't win with this character count :P), but now I can't find anything.

Okay, so I completely forgot that the padding was actually a requirement :O Corrected that and fixed a bug. (250 -> 271)

Answer 6

Haskell (229)

import Char
import Data.List
z=filter
t=show
l=length
k x=[(l$z(==m)x,m)|m<-x]
s z=[b:':':take(1+(maximum$map(l.t.fst)z)-l(t a))(repeat ' ')++t a++"\n"|(a,b)<-z]
main=interact$concat.s.take 10.sort.nub.k.map toLower.z isAlphaNum

Answer 7

Perl, 86 chars

given string is in $_

s/[a-z]/$h{lc($&)}++/eig;printf"$_:%9d$/",$h{$_}for(sort{$h{$a}<=>$h{$b}}keys%h)[0..9]

Answer 8

Python, 144

import os
s=os.read(0,10000).lower()
n=s.count
a=sorted(filter(str.isalnum,set(s)),key=n)[:10]
for c in a:print c+': %%%ii'%len(`n(a[-1])`)%n(c)