### Oz: Translating Python into Oz

I've been working my way through Concepts, Techniques, and Models of Computer Programming. In chapter 3, one of the exercises was to transpose a matrix.

I immediately felt completely overwhelmed. I couldn't even remember what it meant to transpose a matrix, and I definitely hadn't seen anything matrix-related in the book. I went to Wikipedia and figured out that transposing a matrix just means swapping the rows and the columns. That seemed more possible, but I still didn't think I had the necessary tools to code it in Oz yet.

I decided to try solving the problem in Python. Python is sometimes called "executable pseudocode", and sure enough, the solution came to me pretty easily:
`m = [[1, 2, 3],     [4, 5, 6]]def transpose(m):    transposed = []    if len(m) >= 1:        len_cols = len(m[0])        for col in xrange(len_cols):            transposed_row = []            for row in m:                transposed_row.append(row[col])            transposed.append(transposed_row)    return transposedprint transpose(m)`
Now, in real life, I might refactor that into the following:
`m = [[1, 2, 3],     [4, 5, 6]]def transpose(m):    if not m:        return []    return [[row[col] for row in m]            for col in xrange(len(m[0]))]print transpose(m)`
However, the goal was to write the solution in Oz, not Python. Hence, I started refactoring the code into something I could translate to Oz. Oz uses Lisp-like lists, not arrays, and at least for the chapter I'm in, it uses recursion, not foreach loops. Hence I ended up with:
`m = [[1, 2, 3],     [4, 5, 6]]def transpose(m):    def transpose_row(m, i, transposed_row):        if not m:            transposed_row.reverse()            return transposed_row        car, cdr = m[0], m[1:]        transposed_row.insert(0, car[i])        return transpose_row(cdr, i, transposed_row)    def iter_row_len(i, transposed):        if i == row_len:            transposed.reverse()            return transposed        transposed.insert(0, transpose_row(m, i, []))        return iter_row_len(i + 1, transposed)    if not m:        return m    row_len = len(m[0])    transposed = iter_row_len(0, [])    return transposedprint transpose(m)`
At this point, it was fairly straightforward to translate that into Oz:
`declare MM = [[1 2 3]     [4 5 6]]declarefun {Transpose M}   fun {TransposeRow M I TransposedRow}      case M      of nil then  {Reverse TransposedRow}      [] M1|Mr then  {TransposeRow Mr I {List.nth M1 I}|TransposedRow}      end   end   fun {IterRowLen I Transposed}      if I==RowLen+1 then  {Reverse Transposed}      else  {IterRowLen I+1 {TransposeRow M I nil}|Transposed}      end   end   RowLenin   case M   of nil then nil   [] M1|Mr then      RowLen={Length M1}      {IterRowLen 1 nil}   endend{Browse {Transpose M}}  % Prints [[1 4] [2 5] [3 6]]`
Viola! Problem solved!

That solution is pretty long, which just goes to show how bad an Oz programmer I am. I found someone else's solution online:
`fun {Transpose Matrix}   {List.foldR Matrix    fun {\$ R1 R2} {List.zip R1 R2 fun {\$ E1 E2} E1|E2 end} end    for E in Matrix.1 collect:C do {C nil} end}end`
Obviously, he knows Oz and functional programming in general a heck of a lot better than I do.

The moral of the story is that Python can indeed serve as executable pseudocode, but it won't help me to think like an Oz programmer.

A cute way of transposing
m = [[1, 2, 3], [4, 5, 6]]
is
zip(*m)

thats it..done
jjinux said…
Bravo! I'd add the following, though, to make sure the lists stay lists:

map(list, zip(*m))

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