CJam, 76

{2f*__{(z.4<},,%,4e<"UBT"="EG"\*\$-2=i3e<=}

소스 코드의 길이 는 43 바이트이며 33 개의 목록을 잘못 분류 합니다.

확인

$ count()(sort | uniq -c | sort -nr)
$ cat score.cjam
qN%{',' er[~]
  {2f*__{(z.4<},,%,4e<"UBT"="EG"\*\$-2=i3e<=}
~N}/
$ for list in U T B E G; { echo $list; cjam score.cjam < $list.txt | count; }
U
     92 U
      6 B
      2 T
T
    100 T
B
     93 B
      7 U
E
     92 E
      8 G
G
     90 G
      6 E
      3 T
      1 U

생각

지수와 감마 분포를 나머지 분포와 구별하는 것은 쉽습니다. 왜냐하면 1 보다 큰 값을 갖는 유일한 분포이기 때문 입니다.

감마 , 지수 등 을 결정하기 위해 샘플의 두 번째로 높은 값을 살펴 봅니다.

• 그것이 [1.5, ∞) 안에 있다면 , 우리는 감마 를 추측 합니다.

• 그것이 [1, 1.5) 에 있다면, 우리는 지수 를 추측 합니다 .

• 그것이 [0, 1) 에 있다면, 우리에게는 3 가지 가능성이 있습니다.

  나머지 분포는 평균에 가까운 샘플 값의 백분율 ( 0.5 ) 로 구별 할 수 있습니다 .

  표본 길이를 (0.3, 0.7)에 속하는 값의 수로 나누고 결과 몫을 살펴 봅니다.

  • 그것이 (1, 2] 에 있다면, 삼각형 을 추측 합니다.

  • 그것이 (2, 3] 에 있다면, 우리는 균일 하다고 생각 합니다.

  • 그것이 (3, ∞) 안에 있다면 우리는 베타 를 추측 합니다.

암호

2f*    e# Multiply all sample values by 2.
__     e# Push to copies of the sample.
{      e# Filter; for each (doubled) value in the sample:
  (z   e#   Subtract 1 and apply absolute value.
  .4<  e#   Check if the result is smaller than 0.4.
},     e# If it is, keep the value.
,/     e# Count the kept values (K).
%      e# Select every Kth value form the sample, starting with the first.
,      e# Compute the length of the resulting array.
       e# This performs ceiled division of the sample length by K.
4e<    e# Truncate the quotient at 4.
"UBT"= e# Select 'T' for 2, 'U' for 3 and 'B' for 4.
"EG"\* e# Place the selected character between 'E' and 'G'.
\$     e# Sort the remaining sample.
-2=i   e# Extract the second-highest (doubled) value and cast to integer.
3e<    e# Truncate the result at 3.
=      e# Select 'E' for 3, 'G' for 2 and the character from before for 1.

답변

function r=p(x);
data=sort(x(1:75));
%% cumulative probability distributiosn
fu=@(x)(0<x&x<1).*x+(1<=x).*1;
ft=@(x)(0<x&x< 0.5).* 2.*x.^2+(1-2*(1-x).^2).*(0.5<=x&x<1)+(1<=x);
fb=@(x)(0<x&x<1).*2.*asin(sqrt(x))/pi+(1<=x);
fe=@(x)(0<x).*(1-exp(-2*x));
fg=@(x)(0<x).*(1-exp(-x*6).*(1+x*6+1/2*(6*x).^2));
fdata = @(x)sum(bsxfun(@le,data,x.'),2).'/length(data);
f = {fe,fg,fu,ft,fb};
str='EGUTB';
%calculate distance to the different cdfs at each datapoint
for k=1:numel(f);
dist(k) = max(abs(f{k}(x)-fdata(x)));
end;
[~,i]=min(dist);
r=str(i);
end

function r=p(x);f={@(x)(0<x).*(1-exp(-2*x)),@(x)(0<x).*(1-exp(-x*6).*(1+x*6+18*x.^2)),@(x)(0<x&x<1).*x+(1<=x),@(x)(0<x&x<.5).*2.*x.^2+(1-2*(1-x).^2).*(.5<=x&x<1)+(1<=x),@(x)(0<x&x<1).*2.*asin(sqrt(x))/pi+(1<=x)};s='EGUTB';for k=1:5;d(k)=max(abs(f{k}(x)-sum(bsxfun(@le,x,x.'),2).'/nnz(x)));end;[~,i]=min(d(1:5-3*any(x>1)));r=s(i)

for (@F) {$ b [$ _ * 13] ++; $ o ++ if $ _> 1}
$ h = ($ t = $ b [0]-$ b [6]) + $ b [12];
print $ o? ($ t> -2? "e": "g") : ($ h = 19? "b": "u"));
$ o = @ b = ()

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
    bbbbbbbbbbbbbbbbbbbbbbbbbbubbbbbbbbbbbbbbbbbbbbbbbbb
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
    eeegeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
gggggggegggggggggggggggggggggggggggggggggggggggggggggggg
    gggggggggggggggggggggggggggggggggggggggggggggggggggg
tttttttttttttttttttttttttttttttttttttttttttttttttttttttttt
    tttttttttttttttttttttttttttttttuttttttttttttutttttt
uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuutuuuuuuuuuuuuuuuubuuuuuuu
    uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuutuuuu

from scipy.stats import*
from numpy import*
def f(l):
    r={'U':kstest(l,'uniform')[1],'T':kstest(l,'triang',args=(.5,))[1],'B':kstest(l,'beta',args=(.5,.5))[1],'E':kstest(l,'expon',args=(0,.5,))[1],'G':kstest(l,'gamma',args=(3,0,1/6.0))[1]}
    if sum([x>1 for x in l]): r['U'],r['T'],r['B']=0,0,0
    return max(r,key=r.get)

from scipy.stats import*
from numpy import*
import os
from io import StringIO
dir=os.path.dirname(os.path.abspath(__file__))+"/random-data-master/"

def f(l):
    r={'U':kstest(l,'uniform')[1],'T':kstest(l,'triang',args=(.5,))[1],'B':kstest(l,'beta',args=(.5,.5))[1],'E':kstest(l,'expon',args=(0,.5,))[1],'G':kstest(l,'gamma',args=(3,0,1/6.0))[1]}
    if sum([x>1 for x in l]): r['U'],r['T'],r['B']=0,0,0
    return max(r,key=r.get)

U=[line.rstrip('\n').split(',') for line in open(dir+'U.txt')]
U=[[float(x) for x in r] for r in U]
T=[line.rstrip('\n').split(',') for line in open(dir+'T.txt')]
T=[[float(x) for x in r] for r in T]
B=[line.rstrip('\n').split(',') for line in open(dir+'B.txt')]
B=[[float(x) for x in r] for r in B]
E=[line.rstrip('\n').split(',') for line in open(dir+'E.txt')]
E=[[float(x) for x in r] for r in E]
G=[line.rstrip('\n').split(',') for line in open(dir+'G.txt')]
G=[[float(x) for x in r] for r in G]

i,_u,_t,_b,_e,_g=0,0,0,0,0,0
for u,t,b,e,g in zip(U,T,B,E,G):
    _u+=1 if f(u)=='U' else 0
    _t+=1 if f(t)=='T' else 0
    _b+=1 if f(b)=='B' else 0
    _e+=1 if f(e)=='E' else 0
    _g+=1 if f(g)=='G' else 0
    print f(u),f(t),f(b),f(e),f(g)
print _u,_t,_b,_e,_g,100*5-_u-_t-_b-_e-_g