Ten-row bar chart

Question

This is Hole-1 from The Autumn Tournament of APL CodeGolf. I am the original author of the problem there, and thus allowed to re-post it here.

---

Given a list of numbers, produce a horizontal bar chart of # characters for how many numbers fit into each of ten equal-sized groups. For example, if the data ranges from 0-100, the ranges will be 0–9.9, 10–19.9, …, 90–100. (Formally, [0,10), [10,20), …, [90,100].). You may assume that there will be at least two numbers and that not all numbers will be the same.

Examples:

[1,0,0,0,0,0,0,0,0,0] gives:

#########








#        

[0,1,2,3,4,5,6,7,8,9] gives:

#
#
#
#
#
#
#
#
#
#

[0,1,2,3,4,5,6,7,8,9,10] gives:

#
#
#
#
#
#
#
#
#
##

[0,1,2,3,4,5,6,7,8,9,10,11] gives:

##
#
#
#
#
#
#
#
#
##

[0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,-4,-4.5,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,-4,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,1.5,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,2,1.5,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,2.5,2,1.5,1,0.5,0,-0.5,-1,-1.5,-2,3,2.5,2,1.5,1,0.5,0,-0.5,-1,-1.5,3.5,3,2.5,2,1.5,1,0.5,0,-0.5,-1,4,3.5,3,2.5,2,1.5,1,0.5,0,-0.5,4.5,4,3.5,3,2.5,2,1.5,1,0.5,0] gives:

###                
#######            
###########        
###############    
#########          
###################
###############    
###########        
#######            
###                

[9014,9082,9077,9068,8866,8710,9049,8364,8867,9015,9064,9023,9024,8804,8805,8800,8744,8743,8714,9076,8593,8595,9075,9675,8968,8970,8711,8728,8834,8835,8745,8746,8869,8868,9073,9074,9042,9035,9033,9021,8854,9055,9017,9045,9038,9067,9066,8801,8802,9496,9488,9484,9492,9532,9472,9500,9508,9524,9516,9474,8739,9079,8900,8592,8594,9053,9109,9054,9059] gives:

#                        
####                     
#########                
############             
######                   
#########################


###########              
#                        

[0,8,10,13,32,12,6,7,27,9,37,39,95,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,1,2,175,46,48,49,50,51,52,53,54,55,56,57,3,165,36,163,162,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,4,5,253,183,127,193,194,195,199,200,202,203,204,205,206,207,208,210,211,212,213,217,218,219,221,254,227,236,240,242,245,123,125,168,192,196,197,198,201,209,214,216,220,223,224,225,226,228,229,230,231,232,233,234,235,237,238,239,241,91,47,92,60,61,62,45,43,247,215,63,126,42,40,124,59,44,33,243,244,246,248,34,35,30,38,180,64,249,250,251,94,252,96,182,58,191,161,41,93,31,160,167] gives:

#############             
######################    
##########################
######################### 
######################### 
#                         
########                  
################          
########################  
##########################

Answer 1

Python 2, 107 95 bytes

l=input();m=min(l);s=(max(l)-m)/10.;c=0
exec'print"#"*sum(m<=v-c*s<m+s+c/9for v in l);c+=1;'*10

Try it online!

Answer 2

R, 77 81 bytes

+4 bytes to fix some test cases

for(i in hist(x<-scan(),seq(min(x),max(x),,11),r=F)$c)cat(rep('#',i),'\n',sep='')

Try it online!

Link is to a version of the code that takes comma-separated input; this version takes space-separated.

Reads from stdin, prints to stdout.

R is a statistical programming language that does its best to give high-quality results, which is sometimes frustrating:

hist bins the inputs into a histogram with breaks as its second argument. Normally, one would expect that you could specify that the number of breaks to be 10. Indeed, this is the case:

breaks

one of:

• a vector giving the breakpoints between histogram cells,
• a function to compute the vector of breakpoints,
• a single number giving the number of cells for the histogram,
• a character string naming an algorithm to compute the number of cells (see ‘Details’),
• a function to compute the number of cells.

(emphasis added).

The next sentence, however, says:

In the last three cases the number is a suggestion only; as the breakpoints will be set to pretty values, the number is limited to 1e6 (with a warning if it was larger).

So I looked at the documentation of pretty and it just doesn't work for our situation, because it picks break points thusly:

Compute a sequence of about n+1 equally spaced ‘round’ values which cover the range of the values in x. The values are chosen so that they are 1, 2 or 5 times a power of 10.

Which simply won't do.

So the seq(min(x),max(x),,11) specifies 11 equally-spaced points as the breaks, hist(x,breaks,r=F)$c gives the counts, r=F ensures that the bins are right-open intervals, and the for loop takes care of the rest.

Answer 3

Python 2, 96 bytes

l=input()
m=min(l)
w=max(l)-m
for i in range(10):print"#"*sum(w*i<=(x-m)*10<w*-~i+i/9for x in l)

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Answer 4

Mathematica, 152 bytes

(Do[Print[""<>Table["#",Length@Select[s=#,Min@s+(t=#2-#&@@MinMax@s/10)(i-1)<=#<Min@s+t*i&]]],{i,9}];Print[""<>Table["#",Length@Select[s,Max@s-t<=#&]]])&

Try it online!

Answer 5

C (gcc), 241 bytes

#define P(x)for(;x--;putchar('#'));puts("");
double a[999],u,l,x;i,j,n[9];main(k){for(;scanf("%lf",&x)>0;u=u>x?u:x,l=l<x?l:x,a[i++]=x);for(;j<i;++j)for(k=0;k<9;)if(a[j]<l+(++k)*(u-l)/10){n[k-1]++;break;}for(k=0;k<9;++k){i-=n[k];P(n[k])}P(i)}

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Answer 6

JavaScript (ES6), 99 bytes

Edit 2 bytes save thx @JustinMariner

A function returning an array of strings

l=>l.map(v=>o[i=(v-n)/(Math.max(...l)-n)*10|0,i>9?9:i]+='#',o=Array(10).fill``,n=Math.min(...l))&&o

Less golfed

list => {
   var max = Math.max(...list),
       min = Math.min(...list),
       output = Array(10).fill(''),
       index;

   list.forEach( value => (
      index = (value - min) / (max - min) * 10 | 0,
      output [index > 9 ? 9 : index] += '#'
   ) )
   return output
}

Test

var F=
l=>l.map(v=>o[i=(v-n)/(Math.max(...l)-n)*10|0,i>9?9:i]+='#',o=Array(10).fill``,n=Math.min(...l))&&o

var test=[
[1,0,0,0,0,0,0,0,0,0],
[0,1,2,3,4,5,6,7,8,9],
[0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,-4,-4.5,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,-4,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,-3.5,1.5,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,-3,2,1.5,1,0.5,0,-0.5,-1,-1.5,-2,-2.5,2.5,2,1.5,1,0.5,0,-0.5,-1,-1.5,-2,3,2.5,2,1.5,1,0.5,0,-0.5,-1,-1.5,3.5,3,2.5,2,1.5,1,0.5,0,-0.5,-1,4,3.5,3,2.5,2,1.5,1,0.5,0,-0.5,4.5,4,3.5,3,2.5,2,1.5,1,0.5,0],
[0,8,10,13,32,12,6,7,27,9,37,39,95,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,1,2,175,46,48,49,50,51,52,53,54,55,56,57,3,165,36,163,162,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,4,5,253,183,127,193,194,195,199,200,202,203,204,205,206,207,208,210,211,212,213,217,218,219,221,254,227,236,240,242,245,123,125,168,192,196,197,198,201,209,214,216,220,223,224,225,226,228,229,230,231,232,233,234,235,237,238,239,241,91,47,92,60,61,62,45,43,247,215,63,126,42,40,124,59,44,33,243,244,246,248,34,35,30,38,180,64,249,250,251,94,252,96,182,58,191,161,41,93,31,160,167],
[9014,9082,9077,9068,8866,8710,9049,8364,8867,9015,9064,9023,9024,8804,8805,8800,8744,8743,8714,9076,8593,8595,9075,9675,8968,8970,8711,8728,8834,8835,8745,8746,8869,8868,9073,9074,9042,9035,9033,9021,8854,9055,9017,9045,9038,9067,9066,8801,8802,9496,9488,9484,9492,9532,9472,9500,9508,9524,9516,9474,8739,9079,8900,8592,8594,9053,9109,9054,9059]];

output=x=>O.textContent+=x+'\n\n'

test.forEach(t=>output(t+'\n'+F(t).join`\n`))

<pre id=O></pre>

Answer 7

Python 2, 126 121 bytes

def f(a):
 m=min(a);r=range(10);A=[sum(x>=m+i*(max(a)-m)/10.for x in a)for i in r]+[0]
 for i in r:print'#'*(A[i]-A[i+1])

Try it online!

Answer 8

Jelly, 21 bytes

A monadic link returns a list of strings.

_Ṃµ÷Ṁ×⁵Ḟµ<⁵+ċЀ⁵R¤”#ẋ

Try it online!

Answer 9

Charcoal, 31 bytes

≔Ｉ⪪Ｓ,θＥχ×#ΣＥθ⁼ι⌊⟦⁹⌊×χ∕⁻λ⌊θ⁻⌈θ⌊θ

Try it online! Link is to verbose version of code. Input of variable-length lists seems to a little awkward in Charcoal, so I've had to wrap the list in an array containing a string. Explanation:

   Ｓ                            Input string
  ⪪ ,                           Split on commas
 Ｉ                              Cast elements to integer
≔    θ                          Assign to variable q
      Ｅχ                        Map from 0 to 9
           Ｅθ                   Map over the list
                      ⁻λ⌊θ      Subtract the minimum from the current
                          ⁻⌈θ⌊θ Subtract the minimum from the maximum
                     ∕          Divide
                   ×χ           Multiply by 10
                  ⌊             Floor
               ⌊⟦⁹              Take minimum with 9
             ⁼ι                 Compare to outer map variable
          Σ                     Take the sum
        ×#                      Repeat # that many times
                                Implicitly print on separate lines

Answer 10

Pyth,  32  31 bytes

*R\#_M.++msmgk+JhSQ*dc-eSQJTQTZ

Try it here! or Verify all the test cases. (with pretty-print using j)

How this works

This is a full program that takes input from STDIN. This is for the 32-byte version. I'll update it soon.

*R\#_M.++msmgk+hSQ*dc-eSQhSQTQTZ  ~ Full program.

         m                    T   ~ Map over [0, 10) with var d.
           m                 Q    ~ Map over the input with var k.
            g                     ~ Is higher than or equal to?
             k                    ~ The current element of the input, k.
              +hSQ*dc-eSQhSQT     ~ We'll break this down into pieces:
               hSQ                  ~ The lowest element of the input list.
              +                     ~ Plus:
                  *dc-eSQhSQT       ~ We'll break this down into more pieces:
                  *                   ~ Multiplication.
                   d                  ~ The current element of [0, 10), d.
                    c       T         ~ Float division by 10 of:
                     -eSQhSQ          ~ The difference between the maximum and minium
                                        of the input list.
          s                       ~ Sum. Count the number of truthy results.
        +                      Z  ~ Append a 0.
      .+                          ~ Get the deltas.
    _M                            ~ Gets -delta for each delta in the list above.
  \#                              ~ The literal character "#".
*R                                ~ Vectorized multiplication. Optionally, you can
                                    use j to join by newlines (as the link does).
                                  ~ Output implicitly.

Answer 11

Fortran 2003, 263 bytes

I wrote it on GNU gfortran 5.4.0, and compiled without any additional flags.

It reads from STDIN, one value at a time, and prints to STDOUT.

Here it goes:

program h;real,allocatable::a(:);character f*9;allocate(a(0));do;read(*,*,end=8)r;a=[a,r];enddo;9 format("(",i0,"(""#""))")
8 a=(a-minval(a))+epsilon(1.);a=ceiling(10*a/maxval(a));do i=1,10;j=count(a==i);if(j==0)print*;if(j==0)cycle;write(f,9)j;
print f;enddo;end

Ungolfed explanation: (I don't know if "golfed" can be applied to fortran but either way :P)

program h
real,allocatable::a(:)       ! Create an allocatable array so we can reallocate dynamically
character f*9                ! A character array to format the output
allocate(a(0))               ! Allocate "a" empty at first
do
  read(*,*,end=8)r           ! Read from STDIN. If EOF, goes to 8, otherwise
  a=[a,r]                    ! Appends to "a"
enddo
9 format("(",i0,"(""#""))")  ! A format label
8 a=(a-minval(a))+epsilon(1.)! (8) Normalizes a (adds epsilon to avoid zero-indexing)
a=ceiling(10*a/maxval(a))    ! Normalizing and multiplying by the number of bins
do i=1,10                    ! Looping on all the bins
  j=count(a==i)              ! Counting the number of occurrences
  if(j==0)print*             ! If none, prints empty line
  if(j==0)cycle              ! And skips the rest of the loop
  write(f,9)j                ! Otherwise writes the count(j) to the print label
  print f                    ! And prints to STDOUT
enddo
end

Fun fact: I made a similar code just yesterday to test my implementation of a Weibull random number generator, so it only needed a little adaptation :)

Answer 12

Perl 5, 85 + 1 (-a) = 86 bytes

@F=sort{$a<=>$b}@F;map$r[($_-$F[0])/($F[-1]-$F[0])*10].='#',@F;$r[9].=pop@r;say for@r

Try it online!

Answer 13

Java (OpenJDK 8), 246 221 209 207 206 163 162 161 157 bytes

l->{String x="";double b=l[0],m=b,q,i;for(double d:l){b=d<b?d:b;m=d>m?d:m;}for(i=(m-b)/10,q=b;q<m;q+=i,x+="\n")for(double d:l)x+=d>=q&d<q+i?"#":"";return x;}

Try it online!

Answer 14

Perl 5, 84 + 19 (-MList::Util=min,max) bytes

$s=-($m=min@_=@ARGV)+max@_;$a[($_-$m)*10/$s]++for@_;$a[9]+=pop@a;print"#"x$_,$/for@a

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Answer 15

Perl 5, 102 bytes

$l=(@n=sort{$a<=>$b}<>)[-1]-($f=$n[0]);$m=$f+$l*$_/10,say'#'x(@n-(@n=grep$_>=$m,@n))for 1..9;say'#'x@n

Try it online.

Ungolfed:

my @n = sort { $a <=> $b } <>;
my $f = $n[0];
my $l = $n[-1] - $n[0];
for (1 .. 9) {
    my $m = $f + $l * ($_ / 10);
    my $c = scalar @n;
    @n = grep { $_ >= $m } @n;
    say('#' x ($c - scalar @n));
}
say('#' x scalar @n);

Answer 16

q/kdb+, 52 bytes

Solution:

{sum[t=/:bin[m+.1*(t:(!)10)*max[x]-m:min x;x]]#'"#"}

Try it online! (Note the TIO link is a 44 byte K (oK) port of this solution as there is no TIO for q/kdb+).

Examples:

q){sum[t=/:bin[m+.1*(t:(!)10)*max[x]-m:min x;x]]#'"#"}1 0 0 0 0 0 0 0 0 0f
"#########"
""
""
""
""
""
""
""
""
,"#

q){sum[t=/:bin[m+.1*(t:(!)10)*max[x]-m:min x;x]]#'"#"}9014 9082 9077 9068 8866 8710 9049 8364 8867 9015 9064 9023 9024 8804 8805 8800 8744 8743 8714 9076 8593 8595 9075 9675 8968 8970 8711 8728 8834 8835 8745 8746 8869 8868 9073 9074 9042 9035 9033 9021 8854 9055 9017 9045 9038 9067 9066 8801 8802 9496 9488 9484 9492 9532 9472 9500 9508 9524 9516 9474 8739 9079 8900 8592 8594 9053 9109 9054 9059f
,"#"
"####"
"#########"
"############"
"######"
"#########################"
""
""
"###########"
,"#"

Explanation:

Most of the code is used creating the buckets that bin buckets the input into.

{sum[t=/:bin[m+.1*(t:til 10)*max[x]-m:min x;x]]#'"#"} / ungolfed solution
{                                                   } / lambda function with implicit x as parameter
                                               #'"#"  / take (#) each-both "#", 1 2 3#'"#" => "#","##","###"
 sum[                                         ]       / sum up everything inside the brackets
         bin[                              ;x]        / binary search each x in list (first parameter)
                                    m:min x           / store minimum of list x in variable m
                             max[x]-                  / subtract from the maximum of list x
                  (t:til 10)*                         / range 0..9 vectorised multiplication against max delta of list
               .1*                                    / multiply by 0.1 (aka divide by 10)
             m+                                       / minimum of list vectorised addition against list
     t=/:                                             / match each-right against range 0..9 (buckets)

Answer 17

Jelly, 19 bytes

_Ṃµ÷Ṁ×⁵Ḟ«9ċЀ⁵Ḷ¤”#ẋ

Try it online!

This is based upon my APL answer for the original problem, which I will post after that competition is over.

How? (I'm not good at explaining things)

_Ṃµ÷Ṁ×⁵Ḟ«9ċЀ⁵Ḷ¤”#ẋ
_Ṃ                  = subtract the minimum
  µ                 = Sort of like a reverse-order compose
   ÷Ṁ               = divide by the max
     ×⁵             = Multiply by 10
       Ḟ            = Take the floor
        «9          = x => min(x,9)
          ċЀ⁵Ḷ¤    = count occurrences of [0,...,9]
                ”#ẋ = create the list