Matrix transposition
You are encouraged to solve this task according to the task description, using any language you may know.
Transpose an arbitrarily sized rectangular Matrix.
Ada
Transpose is a function of the standard Ada library provided for matrices built upon any floating-point or complex type. The relevant packages are Ada.Numerics.Generic_Real_Arrays and Ada.Numerics.Generic_Complex_Arrays, correspondingly.
This example illustrates use of Transpose for the matrices built upon the standard type Float: <ada> with Ada.Numerics.Real_Arrays; use Ada.Numerics.Real_Arrays; with Ada.Text_IO; use Ada.Text_IO;
procedure Matrix_Transpose is
procedure Put (X : Real_Matrix) is
type Fixed is delta 0.01 range -500.0..500.0;
begin
for I in X'Range (1) loop
for J in X'Range (2) loop
Put (Fixed'Image (Fixed (X (I, J))));
end loop;
New_Line;
end loop;
end Put;
Matrix : constant Real_Matrix :=
( (0.0, 0.1, 0.2, 0.3),
(0.4, 0.5, 0.6, 0.7),
(0.8, 0.9, 1.0, 1.1)
);
begin
Put_Line ("Before Transposition:");
Put (Matrix);
New_Line;
Put_Line ("After Transposition:");
Put (Transpose (Matrix));
end Matrix_Transpose; </ada> Sample output:
Before Transposition: 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 After Transposition: 0.00 0.40 0.80 0.10 0.50 0.90 0.20 0.60 1.00 0.30 0.70 1.10
ALGOL 68
main:(
[,]REAL m=((1, 1, 1, 1),
(2, 4, 8, 16),
(3, 9, 27, 81),
(4, 16, 64, 256),
(5, 25,125, 625));
OP ZIP = ([,]REAL in)[,]REAL:(
[2 LWB in:2 UPB in,1 LWB in:1UPB in]REAL out;
FOR i FROM LWB in TO UPB in DO
out[,i]:=in[i,]
OD;
out
);
PROC pprint = ([,]REAL m)VOID:(
FORMAT real fmt = $g(-6,2)$; # width of 6, with no '+' sign, 2 decimals #
FORMAT vec fmt = $"("n(2 UPB m-1)(f(real fmt)",")f(real fmt)")"$;
FORMAT matrix fmt = $x"("n(UPB m-1)(f(vec fmt)","lxx)f(vec fmt)");"$;
# finally print the result #
printf((matrix fmt,m))
);
printf(($x"Transpose:"l$));
pprint((ZIP m))
)
Output:
Transpose: (( 1.00, 2.00, 3.00, 4.00, 5.00), ( 1.00, 4.00, 9.00, 16.00, 25.00), ( 1.00, 8.00, 27.00, 64.00,125.00), ( 1.00, 16.00, 81.00,256.00,625.00));
BASIC
CLS
DIM m(1 TO 5, 1 TO 4) 'any dimensions you want
'set up the values in the array
FOR rows = LBOUND(m, 1) TO UBOUND(m, 1) 'LBOUND and UBOUND can take a dimension as their second argument
FOR cols = LBOUND(m, 2) TO UBOUND(m, 2)
m(rows, cols) = rows ^ cols 'any formula you want
NEXT cols
NEXT rows
'declare the new matrix
DIM trans(LBOUND(m, 2) TO UBOUND(m, 2), LBOUND(m, 1) TO UBOUND(m, 1))
'copy the values
FOR rows = LBOUND(m, 1) TO UBOUND(m, 1)
FOR cols = LBOUND(m, 2) TO UBOUND(m, 2)
trans(cols, rows) = m(rows, cols)
NEXT cols
NEXT rows
'print the new matrix
FOR rows = LBOUND(trans, 1) TO UBOUND(trans, 1)
FOR cols = LBOUND(trans, 2) TO UBOUND(trans, 2)
PRINT trans(rows, cols);
NEXT cols
PRINT
NEXT rows
Common Lisp
<lisp>(defun transpose (m)
(apply #'map #'list m))</lisp>
D
<d>module mxtrans ; import std.stdio ; import std.string ;
bool isRectangular(T)(T[][] a){
if(a.length < 1 || a[0].length < 1)
return false ;
for(int i = 1 ; i < a.length; i++)
if(a[i].length != a[i-1].length)
return false ;
return true ;
}
T[][] transpose(T=real)(T[][] a) {
T[][] res ;
if(isRectangular(a)){
res = new T[][](a[0].length, a.length) ;
for(int i = 0 ; i < a.length ; i++)
for(int j = 0 ; j < a[0].length ; j++) {
res[j][i] = a[i][j] ;
}
} else
throw new Exception("Transpose Error") ;
return res ;
}
string toString(T=real)(T[][] a){ // for pretty print
string[] s;
foreach(e ; a)
s ~= format("%8s", e)[1..$-1] ;
return "\n<" ~ join(s,"\n ") ~ ">" ;
}
void main() {
float[][] n, m = [[0.5,0,0,1],[0.0,0.5,0,0],[0.0,0,0.5,-1]] ;
writefln(" m = ", m.toString()) ;
n = m.transpose() ;
writefln(" n (m's transpose) = ", n.toString()) ;
}</d>
Fortran
In ISO Fortran 90 or later, use the TRANSPOSE intrinsic function:
integer, parameter :: n = 3, m = 5
real, dimension(n,m) :: a = reshape( (/ (i,i=1,n*m) /), (/ n, m /) )
real, dimension(m,n) :: b
b = transpose(a)
do i = 1, n
print *, a(i,:)
end do
do j = 1, m
print *, b(j,:)
end do
In ANSI FORTRAN 77 with MIL-STD-1753 extensions or later, use nested structured DO loops:
REAL A(3,5), B(5,3)
DATA ((A(I,J),I=1,3),J=1,5) /1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15/
DO I = 1, 3
DO J = 1, 5
B(J,I) = A(I,J)
END DO
END DO
In ANSI FORTRAN 66 or later, use nested labeled DO loops:
REAL A(3,5), B(5,3)
DATA ((A(I,J),I=1,3),J=1,5) /1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15/
DO 10 I = 1, 3
DO 20 J = 1, 5
B(J,I) = A(I,J)
20 CONTINUE
10 CONTINUE
GAP
originalMatrix := [[1, 1, 1, 1],
[2, 4, 8, 16],
[3, 9, 27, 81],
[4, 16, 64, 256],
[5, 25, 125, 625]];
transposedMatrix := TransposedMat(originalMatrix);
Haskell
For matrices represented as lists, there's transpose:
*Main> transpose [[1,2],[3,4],[5,6]] [[1,3,5],[2,4,6]]
For matrices in arrays, one can use ixmap:
import Data.Array swap (x,y) = (y,x) transpArray :: (Ix a, Ix b) => Array (a,b) e -> Array (b,a) e transpArray a = ixmap (swap l, swap u) swap a where (l,u) = bounds a
IDL
Standard IDL function transpose()
m=[[1,1,1,1],[2, 4, 8, 16],[3, 9,27, 81],[5, 25,125, 625]] print,transpose(m)
J
Transpose is the monadic primary verb |:
]matrix=: 3 5 $ 10+ i. 15 NB. make and show example matrix 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |: matrix 10 15 20 11 16 21 12 17 22 13 18 23 14 19 24
Java
<java>import java.util.Arrays; public class Transpose{
public static void main(String[] args){
double[][] m = {{1, 1, 1, 1},
{2, 4, 8, 16},
{3, 9, 27, 81},
{4, 16, 64, 256},
{5, 25, 125, 625}};
double[][] ans = new double[m[0].length][m.length];
for(int rows = 0; rows < m.length; rows++){
for(int cols = 0; cols < m[0].length; cols++){
ans[cols][rows] = m[rows][cols];
}
}
for(double[] i:ans){//2D arrays are arrays of arrays
System.out.println(Arrays.toString(i));
}
}
}</java>
Mathematica
originalMatrix = {{1, 1, 1, 1},
{2, 4, 8, 16},
{3, 9, 27, 81},
{4, 16, 64, 256},
{5, 25, 125, 625}}
transposedMatrix = Transpose[originalMatrix]
Maxima
originalMatrix : matrix([1, 1, 1, 1],
[2, 4, 8, 16],
[3, 9, 27, 81],
[4, 16, 64, 256],
[5, 25, 125, 625]);
transposedMatrix : transpose(originalMatrix);
MAXScript
Uses the built in transpose() function
m = bigMatrix 5 4 for i in 1 to 5 do for j in 1 to 4 do m[i][j] = pow i j m = transpose m
Nial
make an array
|a := 2 3 reshape count 6 =1 2 3 =4 5 6
transpose it
|transpose a =1 4 =2 5 =3 6
OCaml
Matrices can be represented in OCaml as a type 'a array array, or using the module Bigarray.
The implementation below uses a bigarray:
<ocaml>open Bigarray
let transpose b =
let dim1 = Array2.dim1 b
and dim2 = Array2.dim2 b in
let kind = Array2.kind b
and layout = Array2.layout b in
let b' = Array2.create kind layout dim2 dim1 in
for i=0 to pred dim1 do
for j=0 to pred dim2 do
b'.{j,i} <- b.{i,j}
done;
done;
(b')
let array2_display print newline b =
for i=0 to Array2.dim1 b - 1 do
for j=0 to Array2.dim2 b - 1 do
print b.{i,j}
done;
newline();
done;
let a = Array2.of_array int c_layout [|
[| 1; 2; 3; 4 |]; [| 5; 6; 7; 8 |];
|]
array2_display (Printf.printf " %d") print_newline (transpose a) ;;</ocaml>
This will output:
1 5 2 6 3 7 4 8
A version for lists: <ocaml>let rec transpose m =
assert (m <> []); if List.mem [] m then [] else List.map List.hd m :: transpose (List.map List.tl m)</ocaml>
Example:
# transpose [[1;2;3;4];
[5;6;7;8]];;
- : int list list = [[1; 5]; [2; 6]; [3; 7]; [4; 8]]
Perl
<perl>use Math::Matrix;
$m = Math::Matrix->new(
[1, 1, 1, 1], [2, 4, 8, 16], [3, 9, 27, 81], [4, 16, 64, 256], [5, 25, 125, 625],
);
$m->transpose->print;</perl>
Output:
1.00000 2.00000 3.00000 4.00000 5.00000 1.00000 4.00000 9.00000 16.00000 25.00000 1.00000 8.00000 27.00000 64.00000 125.00000 1.00000 16.00000 81.00000 256.00000 625.00000
Pop11
define transpose(m) -> res;
lvars bl = boundslist(m);
if length(bl) /= 4 then
throw([need_2d_array ^a])
endif;
lvars i, i0 = bl(1), i1 = bl(2);
lvars j, j0 = bl(3), j1 = bl(4);
newarray([^j0 ^j1 ^i0 ^i1], 0) -> res;
for i from i0 to i1 do
for j from j0 to j1 do
m(i, j) -> res(j, i);
endfor;
endfor;
enddefine;
Python
<python>m=((1, 1, 1, 1),
(2, 4, 8, 16), (3, 9, 27, 81), (4, 16, 64, 256), (5, 25,125, 625))
print(zip(*m))</python> Output:
[(1, 2, 3, 4, 5), (1, 4, 9, 16, 25), (1, 8, 27, 64, 125), (1, 16, 81, 256, 625)]
Ruby
m=[[1, 1, 1, 1],
[2, 4, 8, 16], [3, 9, 27, 81], [4, 16, 64, 256], [5, 25,125, 625]]
puts m.transpose Output:
[[1, 2, 3, 4, 5], [1, 4, 9, 16, 25], [1, 8, 27, 64, 125], [1, 16, 81, 256, 625]]
Scheme
<scheme>(define (transpose m)
(apply map list m))</scheme>
TI-83 BASIC
Assuming the original matrix is in [A], place its transpose in [B]:
[A]T->[B]
The T operator can be found in the matrix math menu.