Задача состоит в том, чтобы отобразить таблицу ascii для данного массива.

вход

Вход представляет собой 2D матрицу. Длина строки матрицы совпадает с длиной массива. В качестве альтернативы вы можете взять ввод двумерной матрицы с первой строкой заголовка. Внешнее измерение - это строка.

Пример ввода:

[["License","2008-05-08","2009-03-11","2011-11-22","2013-08-12","2015-11-19"],["GPLv2","58.69%","52.2%","42.5%","33%","23%"],["GPLv3","1.64%","4.15%","6.5%","12%","9%"],["LGPL 2.1","11.39%","9.84%","?","6%","5%"],["LGPL 3.0","? (<0.64%)","0.37%","?","3%","2%"],["GPL family together","71.72% (+ <0.64%)","66.56%","?","54%","39%"]]

Выход

Вывод таблицы выглядит следующим образом.

+---------------------+-------------------+------------+------------+------------+------------+
| License             | 2008-05-08        | 2009-03-11 | 2011-11-22 | 2013-08-12 | 2015-11-19 |
+---------------------+-------------------+------------+------------+------------+------------+
| GPLv2               | 58.69%            | 52.2%      | 42.5%      | 33%        | 23%        |
| GPLv3               | 1.64%             | 4.15%      | 6.5%       | 12%        | 9%         |
| LGPL 2.1            | 11.39%            | 9.84%      | ?          | 6%         | 5%         |
| LGPL 3.0            | ? (<0.64%)        | 0.37%      | ?          | 3%         | 2%         |
| GPL family together | 71.72% (+ <0.64%) | 66.56%     | ?          | 54%        | 39%        |
+---------------------+-------------------+------------+------------+------------+------------+

Каждая ячейка имеет ровно один пробел слева и как минимум один пробел справа, дополняемый до выравнивания полос. Однако, по крайней мере, одна ячейка имеет только один пробел справа.

Контрольные примеры

Input:
[["Hello", "World", "!!!"],["aa", "bbbbbbbbb", "CcC"], ["Pyth",  "CJam", "GolfScript"]]

Output:
+-------+-----------+------------+
| Hello | World     | !!!        |
+-------+-----------+------------+
| aa    | bbbbbbbbb | CcC        |
| Pyth  | CJam      | GolfScript |
+-------+-----------+------------+

Пример представления

function ascii_table(array, header) {
    var lengths = array[0].map(function(_, i) {
        var col = array.map(function(row) {
            if (row[i] != undefined) {
                return row[i].length;
            } else {
                return 0;
            }
        });
        return Math.max.apply(Math, col);
    });
    array = array.map(function(row) {
        return '| ' + row.map(function(item, i) {
            var size = item.length;
            if (size < lengths[i]) {
                item += new Array(lengths[i]-size+1).join(' ');
            }
            return item;
        }).join(' | ') + ' |';
    });
    var sep = '+' + lengths.map(function(length) {
        return new Array(length+3).join('-');
    }).join('+') + '+';
    if (header) {
        return sep + '\n' + array[0] + '\n' + sep + '\n' +
            array.slice(1).join('\n') + '\n' + sep;
    } else {
        return sep + '\n' + array.join('\n') + '\n' + sep;
    }
}

Это код-гольф , поэтому выигрывает представление с наименьшим количеством байтов!

CJam, 58 байт

Кто-нибудь знает хитрость игры в гольф в CJam?

q~z_{:,:e>)}%_{)'-*'++}%'+\+N+@@.{f{Se]"| "\+}}z'|N+f+1$f+

vim, 139 138 134

$x0xqq%ls<cr><esc>@qq@q:se nosol|%s/,/\t/g|%s/]/\t./|%!column -t -s'<C-v><Tab>'<cr>qwf";;h<C-v>GI|<esc>@wq@wll<C-v>Gls|<esc>0<C-v>Gs| <esc>:%s/"//g<cr>Yp:s/[^|]/-/g|s/|/+/g<cr>YggpkP

Принимает ввод в форме, показанной в тестовых примерах. Может или не может быть действительным, так как это зависит от входной строки никогда не содержащей какие - либо ", ,, ]или |символов.

Если вход должен содержать ]s, :%s/]/\t./<cr>его можно заменить qe$s<Tab>.<esc>j@eq@eна 2 дополнительных символа. Нет простого способа разрешить ",|ввод данных.

Должен работать в среде Unix, так как он опирается на columnинструмент командной строки.

Немного устаревшее объяснение (одной ревизией, но это было всего лишь небольшим изменением):

:se nosol<cr>    we need this later: G in visual block shouldn't go to BOL
$x0x             delete the surrounding pair of brackets
qq               record a macro
 %ls<cr><esc>    put each element of the big array on its own line
 @qq             recurse
@q               play back the macro until EOF
:%s/,/\t/g<cr>   replace all remaining commas with tabs
:%s/]/\t./<cr>   replace the ] at the end of lines with tabs and a dot
                 we need this for the line at the right edge of the table
:%!column -t     run the whole file through `column' on tabs
 -s'<C-v><Tab>
 '<cr>
qw               record another macro
 f";;            go forward 3 "s--that is, to the next "column"
 h               go back to the middle of the column
 <C-v>GI|<esc>   insert a line behind the cursor from top to bottom
 @wq             recurse
@w               play back until EOF
ll               move right before the line of dots we added earlier
<C-v>Gl          select the dots
s|<esc>          replace with a line (top to bottom)
0<C-v>G          select all the opening brackets
s| <esc>         again, (the leftmost) line
:%s/"//g<cr>     kill all the quotes around the data
Yp               duplicate bottom line
:s/[^|]/-/g<cr>  replace everything that's not a line with a dash
:s/|/+/g<cr>     now replace the lines with plus signs
YggpkP           put the separators before and after the first line

Спасибо smpl за байт!

JavaScript (ES6), 210 212 219

Редактирование 2 байт сохраненного ТНХ @Neil

a=>(J=(m,j)=>j+m.join(j)+j,a.map(r=>r.map((c,i)=>s[i]>(l=c.length)?0:s[i]=l),s=[]),t=J(s.map(n=>'-'.repeat(n+2)),'+'),z=a.map(r=>J(r.map((c,i)=>' '+c+' '.repeat(s[i]+1-c.length)),'|')),z[0]+=`
`+t,t+J(z,`
`)+t)

ТЕСТОВОЕ ЗАДАНИЕ

F=a=>(
  J=(m,j)=>j+m.join(j)+j,
  a.map(r=>r.map((c,i)=>s[i]>(l=c.length)?0:s[i]=l),s=[]),
  t=J(s.map(n=>'-'.repeat(n+2)),'+'),
  z=a.map(r=>J(r.map((c,i)=>' '+c+' '.repeat(s[i]+1-c.length)),'|')),
  z[0]+='\n'+t,
  t+J(z,'\n')+t
)  

Z=[["License","2008-05-08","2009-03-11","2011-11-22","2013-08-12","2015-11-19"],["GPLv2","58.69%","52.2%","42.5%","33%","23%"],["GPLv3","1.64%","4.15%","6.5%","12%","9%"],["LGPL 2.1","11.39%","9.84%","?","6%","5%"],["LGPL 3.0","? (<0.64%)","0.37%","?","3%","2%"],["GPL family together","71.72% (+ <0.64%)","66.56%","?","54%","39%"]]

O.textContent=F(Z)

<pre id=O></pre>

Python 2, 190

Это решение использует списки и выражения генератора. Он принимает список списков и возвращает строку в требуемом формате.

def b(i):
 d=[max(map(len,c))for c in zip(*i)]
 a='+'+''.join('-'*h+'--+'for h in d)
 e=['|'+''.join(' '+f.ljust(h)+' |'for h,f in zip(d,j))for j in i]
 return'\n'.join([a,e[0],a]+e[1:]+[a])

Код перед минификатором:

def mktable(data):
    sizes = [max(map(len, column)) for column in zip(*data)]
    divider = '+' + ''.join('-'*size+'--+' for size in sizes)
    lines = ['|' + ''.join(
                ' ' + value.ljust(size) + ' |' for size, value in zip(sizes, row)
                )
                for row in data]
    return '\n'.join([divider, lines[0], divider] + lines[1:] + [divider])

data = [
    ["License","2008-05-08","2009-03-11","2011-11-22","2013-08-12","2015-11-19"],
    ["GPLv2","58.69%","52.2%","42.5%","33%","23%"],
    ["GPLv3","1.64%","4.15%","6.5%","12%","9%"],
    ["LGPL 2.1","11.39%","9.84%","?","6%","5%"],
    ["LGPL 3.0","? (<0.64%)","0.37%","?","3%","2%"],
    ["GPL family together","71.72% (+ <0.64%)","66.56%","?","54%","39%"]
    ]

table = mktable(data)
print table

какие выводы:

+---------------------+-------------------+------------+------------+------------+------------+
| License             | 2008-05-08        | 2009-03-11 | 2011-11-22 | 2013-08-12 | 2015-11-19 |
+---------------------+-------------------+------------+------------+------------+------------+
| GPLv2               | 58.69%            | 52.2%      | 42.5%      | 33%        | 23%        |
| GPLv3               | 1.64%             | 4.15%      | 6.5%       | 12%        | 9%         |
| LGPL 2.1            | 11.39%            | 9.84%      | ?          | 6%         | 5%         |
| LGPL 3.0            | ? (<0.64%)        | 0.37%      | ?          | 3%         | 2%         |
| GPL family together | 71.72% (+ <0.64%) | 66.56%     | ?          | 54%        | 39%        |
+---------------------+-------------------+------------+------------+------------+------------+

MATLAB, 244 239 229 226

a=eval(regexprep(input(''),{'], *?[','[[',']]','"'},{';','{','}',''''}));s=size(a);c=repmat(' | ',s(1),1);b=c;for i=1:s(2)
x=char(a{:,i});b=[b x c];end
h=b(1,:);r=h*0+'-';r(h=='|')='+';b=[r;h;r;b(2:end,:);r];disp(b(:,2:end-1))

Объяснение, чтобы следовать.

Прецедент:

Входные данные:

'[["License","2008-05-08","2009-03-11","2011-11-22","2013-08-12","2015-11-19"],["GPLv2","58.69%","52.2%","42.5%","33%","23%"],["GPLv3","1.64%","4.15%","6.5%","12%","9%"],["LGPL 2.1","11.39%","9.84%","?","6%","5%"],["LGPL 3.0","? (<0.64%)","0.37%","?","3%","2%"],["GPL family together","71.72% (+ <0.64%)","66.56%","?","54%","39%"]]'

Выход:

+---------------------+-------------------+------------+------------+------------+------------+
| License             | 2008-05-08        | 2009-03-11 | 2011-11-22 | 2013-08-12 | 2015-11-19 |
+---------------------+-------------------+------------+------------+------------+------------+
| GPLv2               | 58.69%            | 52.2%      | 42.5%      | 33%        | 23%        |
| GPLv3               | 1.64%             | 4.15%      | 6.5%       | 12%        | 9%         |
| LGPL 2.1            | 11.39%            | 9.84%      | ?          | 6%         | 5%         |
| LGPL 3.0            | ? (<0.64%)        | 0.37%      | ?          | 3%         | 2%         |
| GPL family together | 71.72% (+ <0.64%) | 66.56%     | ?          | 54%        | 39%        |
+---------------------+-------------------+------------+------------+------------+------------+

Рубин, 129 126 127 126 символов

->a{t=?|
a.transpose.map{|c|t+=" %-#{c.map(&:size).max}s |"}
[d=(t%a[0].map{p}).tr('| ','+-'),a.map{|r|t%r}.insert(1,d),d]*$/}

Образец прогона:

2.1.5 :001 > puts ->a{t=?|;a.transpose.map{|c|t+=" %-#{c.map(&:size).max}s |"};[d=(t%a[0].map{p}).tr('| ','+-'),a.map{|r|t%r}.insert(1,d),d]*$/}[[["License","2008-05-08","2009-03-11","2011-11-22","2013-08-12","2015-11-19"],["GPLv2","58.69%","52.2%","42.5%","33%","23%"],["GPLv3","1.64%","4.15%","6.5%","12%","9%"],["LGPL 2.1","11.39%","9.84%","?","6%","5%"],["LGPL 3.0","? (<0.64%)","0.37%","?","3%","2%"],["GPL family together","71.72% (+ <0.64%)","66.56%","?","54%","39%"]]]
+---------------------+-------------------+------------+------------+------------+------------+
| License             | 2008-05-08        | 2009-03-11 | 2011-11-22 | 2013-08-12 | 2015-11-19 |
+---------------------+-------------------+------------+------------+------------+------------+
| GPLv2               | 58.69%            | 52.2%      | 42.5%      | 33%        | 23%        |
| GPLv3               | 1.64%             | 4.15%      | 6.5%       | 12%        | 9%         |
| LGPL 2.1            | 11.39%            | 9.84%      | ?          | 6%         | 5%         |
| LGPL 3.0            | ? (<0.64%)        | 0.37%      | ?          | 3%         | 2%         |
| GPL family together | 71.72% (+ <0.64%) | 66.56%     | ?          | 54%        | 39%        |
+---------------------+-------------------+------------+------------+------------+------------+