Matrix transposition
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You are encouraged to solve this task according to the task description, using any language you may know.
Transpose an arbitrarily sized rectangular Matrix.
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
Compiler: QuickBasic 4.5
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
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)
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)); } } }
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;
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
Python
#!/usr/bin/env python from pprint import pprint m=((1, 1, 1, 1), (2, 4, 8, 16), (3, 9, 27, 81), (4, 16, 64, 256), (5, 25,125, 625)); pprint(zip(*m))
Output:
[(1, 2, 3, 4, 5), (1, 4, 9, 16, 25), (1, 8, 27, 64, 125), (1, 16, 81, 256, 625)]