Block-sorting rows and columns in a 2D array

Given a 2D array of integers, let's sort its rows and columns in blocks. This means that you only have to sort a given row or column, but applying the transformations needed for sorting it to every other row or column in the 2D array.

4x3

-2

3

Positive indices for rows and negative indices for columns

[5 8 7 6 [1 3 2 4
1 3 2 4 order by -3 (3rd column) --> 9 6 3 0
9 6 3 0] 5 8 7 6]

[5 8 7 6 [9 6 3 0
1 3 2 4 order by -4 (4th column) --> 1 3 2 4
9 6 3 0] 5 8 7 6]

[5 8 7 6 [5 7 8 6
1 3 2 4 order by 2 (2nd row) --> 1 2 3 4
9 6 3 0] 9 3 6 0]

[5 8 7 6 [6 7 8 5
1 3 2 4 order by 3 (3rd row) --> 4 2 3 1
9 6 3 0] 0 3 6 9]

[1 2 [1 2 [3 2
3 2] order by -2 (2nd column) --> 3 2] or 1 2] (both are valid)

[7 5 9 7 [5 7 7 9 [5 7 7 9
1 3 2 4 order by 1 (1st row) --> 3 1 4 2 or 3 4 1 2
9 6 3 0] 6 9 0 3] 6 0 9 3]


This is code-golf, so may the shortest code for each language win!

20 Answers
20

R, 55 bytes

function(x,n)`if`(n>0,x[,+x[n,]],x[+x[,-n],])
`+`=order


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Reassigns the + operator (actually a function in R) to the order function, which returns the indices of a vector from smallest to largest. Then it's just array manipulation.

+

order

R, 55 bytes

f=function(m,i)"if"(i<0,t(f(t(m),-i)),m[,order(m[i,])])


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Alternative to ngm's answer; a recursive function which was inspired by DimChtz's answer

@(m,i)sortrows(m,-i) sortrows(m',i)'(i>0)+1


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-11 bytes thanks to @Giuseppe.

-15 bytes thanks to @LuisMendo.

Japt, 18 17 bytes

negative for rows and positive for columns

>0?VñgUÉ:ßUa Vy)y


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U

ß

U

05AB1E, 25 24 14 bytes

diø}Σ¹Ä<è}¹diø


Whopping -10 bytes thanks to @Emigna.

Uses a positive integer-input to sort the rows, negative for columns.

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Explanation:

di } # If the (implicit) integer input is positive:
ø # Swap the rows and columns of the (implicit) matrix input
# i.e. 3 and [[5,8,7,6],[1,3,2,4],[9,6,3,0]]
# → [[5,1,9],[8,3,6],[7,2,3],[6,4,0]]
Σ } # Sort the rows of this matrix by:
¹Ä # Take the absolute value of the input
# i.e. -3 → 3
< # Decreased by 1 to make it 0-indexed
# i.e. 3 → 2
è # And index it into the current row
# i.e. [5,8,7,6] and 2 → 7
# i.e. [5,1,9] and 2 → 9
# i.e. [[5,1,9],[8,3,6],[7,2,3],[6,4,0]] sorted by [9,6,3,0]
# → [[6,4,0],[7,2,3],[8,3,6],[5,1,9]]
# i.e. [[5,8,7,6],[1,3,2,4],[9,6,3,0]] sorted by [7,2,3]
# → [[1,3,2,4],[9,6,3,0],[5,8,7,6]]
¹di # And if the integer input was positive:
ø # Swap the rows and columns back again now that we've sorted them
# i.e. 3 and [[6,4,0],[7,2,3],[8,3,6],[5,1,9]]
# → [[6,7,8,5],[4,2,3,1],[0,3,6,9]]
# (And implicitly output the now sorted matrix)


diø}Σ¹Ä<è]¹diø

JavaScript (ES6), 90 bytes

t=m=>m[0].map((_,x)=>m.map(r=>r[x]))
f=(m,k)=>k<0?m.sort((a,b)=>a[~k]-b[~k]):t(f(t(m),-k))


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JS has no native transposition method, so we need to define one:

t = m => // given a matrix m
m[0].map((_, x) => // for each column at position x in m:
m.map(r => // for each row r in m:
r[x] // map this cell to r[x]
) // end of map() over rows
) // end of map() over columns


Main function:

f = (m, k) => // given a matrix m and an integer k
k < 0 ? // if k is negative:
m.sort((a, b) => // given a pair (a, b) of matrix rows, sort them:
a[~k] - b[~k] // by comparing a[-k - 1] with b[-k - 1]
) // end of sort
: // else:
t(f(t(m), -k)) // transpose m, call f() with -k and transpose the result


Example with $k=2$:

$$M=pmatrix5&8&7&6\1&3&2&4\9&6&3&0rightarrow
t(M)=pmatrix5&1&9\8&3&6\7&2&3\6&4&0rightarrow
f(t(M),-2)=pmatrix5&mathbfcolorred1&9\7&mathbfcolorred2&3\8&mathbfcolorred3&6\6&mathbfcolorred4&0\f(M,2)=t(f(t(M),-2))=pmatrix5&7&8&6\mathbfcolorred1&mathbfcolorred2&mathbfcolorred3&mathbfcolorred4\9&3&6&0$$

MATL, 17 bytes

y0>XH?!]w|2$XSH?!


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Or verify all test cases

y % Implicit inputs: number n, matrix M. Duplicate from below: pushes n, M, n
0> % Greater than 0?
XH % Copy into clipboard H
? % If true
! % Transpose matrix. This way, when we sort the rows it will correspond
% to sorting the columns of the original M
] % End
w % Swap: moves n to top
| % Absolute value
2$XS % Two-input sortrows function: sorts rows by specified column
H % Push contents from clipboard H
? % If true
! % Transpose again, to convert rows back to columns
% Implicit end
% Implicit display


APL (Dyalog Classic), 23 bytes

⍺<0:⍉(-⍺)∇⍉⍵⋄⍵[;⍋⍺⌷⍵]


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Python 2, 71 70 bytes

f=lambda m,n:n<0and sorted(m,key=lambda l:l[~n])or zip(*f(zip(*m),-n))


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If n is negative, the rows are sorted based on column n.

n

n

Otherwise the matrix is transposed, sorted the same way, and transposed back again.

Jelly, 12 bytes

ZṠ}¡
çị@ÞA}ç


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C# (.NET Core), 186 bytes

(x,y)=>Func<int,int>shift=a=> a[0].Select((r,i)=>a.Select(c=>c[i]).ToArray()).ToArray();return y>0?shift(shift(x).OrderBy(e=>e[y-1]).ToArray()):x.OrderBy(e=>e[-y-1]).ToArray();


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Ungolfed:

private static int Blocksort0a(int array, int sortingInstruction)

Func<int, int> shift = a => a[0].Select((r, i) => a.Select(c => c[i]).ToArray()).ToArray();

sortingInstruction++;

array = sortingInstruction < 0 ?
shift(shift(array).OrderBy(e => e[-sortingInstruction]).ToArray())
:
array.OrderBy(e => e[sortingInstruction]).ToArray();

return null;



The shift function we'll use twice, so a function variable will save space. The function iterates through the horizontal dimension of the array on index, and adds every item on that index in of each horizontal array to a new output array (horizontally) - much the same as in Arnoud's JS solution.

Now the ordering is simple, order horizontal array by number-at-index (argument -1), optionally shifting the array before and after sorting.

Seen how the question talks about arrays specifically, we convert to array a few times (very, very wasteful). Feeling a bit silly to use such a verbose language in code golf hehe.

C# (.NET Core), 142/139 138/135 bytes (and yet another -1 by Kevin)

(a,s)=>s<0?a.OrderBy(e=>e[~s]).ToArray():a.Select(f=>a[s-1].Select((v,j)=>newv,j).OrderBy(e=>e.v).Select(e=>f[e.j]).ToArray()).ToArray()


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Ungolfed:

private static int Blocksort0b(int array, int sortingInstruction)

if (sortingInstruction < 0) return array.OrderBy(e => e[-sortingInstruction - 1]).ToArray();
var rowIndices = array[sortingInstruction - 1].Select((value, index) => (value, index)).OrderBy(e => e.value);
var newRow = new int[array[0].Length];
for (var i = 0; i < array.Length; i++)

int horizontalIndexer = 0;
foreach (var e in rowIndices)

newRow[horizontalIndexer++] = array[i][e.index];

array[i] = newRow.ToArray();

return array;



New all-inline approach; negative answer still orders arrays by element-at-index. Otherwise, a collection of value-index-pair is created of the array-at-index and sorted by value. This effectively creates a collection of indices in order of having-to-be-added. Then for each array, the elements in the predetermined positions are selected. Quite some trimming of code and ugly, ugly, ugly **silently sobs** reuse of input parameters is involved, and there you go ... 142 bytes.

Again, the arrays argument is strictly enforced, adding quite some overhead for .ToArray() calls.

135 bytes claim, eh?! C# 7.2 inferred value-tuples would trim an additional three bytes, but tio.run doesn't allow. Therefor, this is the answer i decided to post for easy verification.

(a,s)=>

a=>s=>

(s<0)?

-s-1

~s

Java (OpenJDK 8), 326 bytes

(a,b)->int l=a.length,w=a[0].length,k,m,t,i;if(b>0)for(i=0;i<w;i++)for(k=1;k<(w-i);k++)if(a[b-1][k-1]>a[b-1][k])for(m=0;m<l;m++)t=a[m][k];a[m][k]=a[m][k-1];a[m][k-1]=t;elseb*=-1;for(i=0;i<l;i++)for(k=1;k<(l-i);k++)if(a[k-1][b-1]>a[k][b-1])for(m=0;m<w;m++)t=a[k][m];a[k][m]=a[k-1][m];a[k-1][m]=t;return a;


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Well guys, this question was very frustrating for me, and I posted my answer KNOWING I was forgetting something, luckily we have legends like Kevin Cruijssen out here to help us out :)

Java (OpenJDK 8), 281 bytes

a->b->int l=a.length,w=a[0].length,k,m,t,i;if(b>0)for(i=0;i<w;i++)for(k=0;++k<w-i;)for(m=0;a[b-1][k-1]>a[b-1][k]&m<l;a[m][k]=a[m][k-1],a[m++][k-1]=t)t=a[m][k];else for(b*=-1,i=0;i<l;i++)for(k=0;++k<l-i;)for(m=0;a[k-1][b-1]>a[k][b-1]&m<w;a[k][m]=a[k-1][m],a[k-1][m++]=t)t=a[k][m];


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a->b->

(a,b)->

return

Clean, 95 bytes

import StdEnv,Data.List,Data.Func
$n#f=if(n>0)transpose id
=f o sortBy(on(<)u=u!!(abs n-1))o f


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Kotlin, 192 bytes

m:Array<Array<Int>>,s:Int->if(s<0)m.sortByit[-s-1]elseval a=Array(m[0].size)c->Array(m.size)m[it][c]
a.sortByit[s-1]
(0..m.size-1).mapr->(0..m[0].size-1).mapm[r][it]=a[it][r]


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Ruby, 69 bytes

->a,nf=->x[0,x.transpose,x][n<=>0];f[f[a].sort_byx[n.abs-1]]


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VBA (Excel), 205 bytes

Yay! 2nd longest byte count! I didn't completely lose :D

Golfed:

Sub d(a)
With ActiveSheet.Sort
.SortFields.Clear
.SortFields.Add Key:=IIf(a<0,ActiveSheet.Columns(Abs(a)),ActiveSheet.Rows(Abs(a)))
.SetRange ActiveSheet.UsedRange
.Orientation=IIf(a<0,1,2)
.Apply
End With
End Sub


This sorts all the data on the open (active) worksheet using UsedRange... which can be buggy, but should only contain cells that have been edited.

UnGolfed:

Sub d(a)
'Clear any Sort preferences that already exists
ActiveSheet.Sort.SortFields.Clear
'Use the column if A is negative, the row if A is positive
ActiveSheet.Sort.SortFields.Add Key:=IIf(a < 0, ActiveSheet.Columns(Abs(a)), ActiveSheet.Rows(Abs(a)))
'Set the area to sort
ActiveSheet.Sort.SetRange ActiveSheet.UsedRange
'Orient sideways if sorting by row, vertical if by column
ActiveSheet.Sort.Orientation = IIf(a < 0, xlTopToBottom, xlLeftToRight)
'Actually sort it now
ActiveSheet.Sort.Apply
End Sub


Perl 6, 43 bytes

($!=$_>0??&[Z]!!*)o*.sort(*[.abs-1])o$!


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Curried function.

# Block returning function composed of
o$! # 1. Apply $! (transpose or not)
o*.sort(*[.abs-1]) # 2. Sort rows by column abs(i)-1
$_>0??&[Z] # 3. If i > 0 transpose matrix
!!* # Else identity function
($!= ) # Store in $!


Physica, 45 bytes

Very similar to Arnauld's JS answer.

F=>n;m:n<0&&Sort[->u:u~n;m]||Zip@F#Zip@m#-n


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A more elaborate and visual explanation can be found in the linked answer.

F=>n;m: // Create a function F that takes two arguments, n and m.
n<0&& // If n < 0 (i.e. is negative)
Sort[->u~n;m] // Sort the rows u of m by the result of the function u[~n].
// In short, sort by indexing from the end with n.
|| F#Zip@m#-n // Else, apply F to Zip[m] and -n. Uses a new feature, binding.
Zip@ // And transpose the result.


Red, 190 bytes

func[b n][t: func[a][c: length? a/1 a: to-block form a
d: copyloop c[append/only d extract a c take a]d]d: does[if n > 0[b: t b]]d
m: absolute n sort/compare b func[x y][x/(m) < y/(m)]d b]


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f: func [ b n ] [
t: func [ a ] [ ; helper transpose function
c: length? a/1 ; c is the length of the rows
a: to-block form a ; flatten the list
d: copy ; an empty block (list)
loop c [ ; do as many times as the number of columns
append/only d extract a c ; extract each c-th element (an entire column)
; and append it as a sublist to d
take a ; drop the first element
]
d ; return the transposed block (list of lists)
]
d: does [ if n > 0 [ b: t b ] ] ; a helper function (parameterless) to transpose
; the array if positive n
d ; call the function
m: absolute n ; absolute n
sort/compare b func[ x y ] [ x/(m) < y/(m) ]; sort the array according to the chosen column
d ; transpose if positive n
b ; return the array
]


My actual solution is 175 bytes long, but it doesn't work in TIO. Here it is, working normalyl in the Red console:

func[b n][d: does[if n > 0[c: length? b/1 a: to-block form b
t: copyloop c[append/only t extract a c take a]b: t]]d
m: absolute n sort/compare b func[x y][x/(m) < y/(m)]d b]


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