Transpose this matrix(python3)

Question

I am trying to find a way to transpose a matrix, for example for:

[[1, 2, 3],
 [4, 5, 6],
 [7, 8, 9]]

it would change the matrix to:

[[1, 4, 7],
 [2, 5, 8],
 [3, 6, 9]]

So far, I tried several things but it never worked. I tried:

def transpose_matrix(matrix): # this one doesn't change the matrix at all
    zip(*matrix)
    return matrix

or

def transpose_matrix(matrix):
   map(list, zip(*matrix))
   return matrix

or

 def transpose_matrix(matrix): # this one returns an empty list []
    print(list(zip(*matrix)))

I want to code it without using external library such as numpy.

After that, it returns the result to this menu and use the transposed matrix for upcoming option (not shown):

def menu(matrix):
    print('choose option')
    loop = True
    while loop:
        print('''
        1-display matrix
        7-transpose it
        8-other option
        0-back
        ''')
        choi = input('cchoice:')
        if choi =='1':
            print('\n'.join([' '.join(map(str, row)) for row in matrix]))

        elif choix == '7':
            matrix = transpose_matrix(matrix)

        else:
            print('invalid choice')

jonrsharpe · Accepted Answer · 2014-05-04T23:30:30.737

2

You aren't actually assigning the result of your manipulations, and therefore return the original matrix unchanged. For example:

def transpose_matrix(matrix):
    zip(*matrix) # ignore result of computation
    return matrix # return argument unchanged

Instead, try:

def transpose_matrix(matrix):
    matrix = list(map(list, zip(*matrix))) # assign result 
    return matrix # return transposed matrix

or simply:

def transpose_matrix(matrix):
    return list(map(list, zip(*matrix)))

Bear in mind that you will need to assign the return from the function in the calling function, too:

matrix = transpose_matrix(matrix)

In-place modification

Alternatively, you could transpose in-place and implicitly return None:

def transpose_matrix(matrix):
    matrix[:] = list(map(list, zip(*matrix)))

This way you don't have to assign back to matrix when you call the function.

Examples

Examples of the functions in action:

>>> def transpose_matrix(matrix):
    return list(map(list, zip(*matrix)))

>>> transpose_matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
>>> def in_place_transpose(matrix):
    matrix[:] = list(map(list, zip(*matrix)))


>>> m = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> in_place_transpose(m)
>>> m
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

zip behaviour

It is important to note that zip will truncate its output to the shortest argument iterable, e.g.:

>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> c = [7, 8]
>>> d = [9]
>>> e = []
>>> for l in (b, c, d, e):
    print(list(zip(a, l)))


[(1, 4), (2, 5), (3, 6)]
[(1, 7), (2, 8)]
[(1, 9)]
[]

Therefore if any of the rows of your matrix are empty lists, your output will be an empty list. You can use zip_longest from itertools to insert dummy fillvalues (None by default) instead:

>>> from itertools import zip_longest
>>> list(zip_longest([1, 2, 3], [4, 5]))
[(1, 4), (2, 5), (3, None)]
>>> m = [[1, 2, 3], [4, 5, 6], []]
>>> list(zip_longest(*m, fillvalue=0))
[(1, 4, 0), (2, 5, 0), (3, 6, 0)]

You could implement this as:

def transpose_matrix(matrix):
    return list(map(list, zip_longest(*matrix, fillvalue=0)))

with from itertools import zip_longest at the top of your script.

edited May 04 '14 at 23:30

answered May 04 '14 at 21:45

jonrsharpe

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I tried all the options,and each time the matrix remain the same – Program_coder May 04 '14 at 21:56
@Program_coder I have added more - you need to assign the result of the function or modify the `matrix` in-place – jonrsharpe May 04 '14 at 21:58
I tried all and nothing works! I dont get it; either it returns an empty list,or doesnt change the matrix at all; im using print('\n'.join([' '.join(map(str, row)) for row in matrix]))to display the matrix in my menu module and it display the same matrix after passing in the transposing module – Program_coder May 04 '14 at 22:27
@Program_coder are you certain that you are are assigning the result of the computation back to the name `matrix`, or modifying it in-place? This code works perfectly for me. You could edit your question to show precisely what code you have now, what inputs it takes and what outputs you get - you can even paste directly from the interpreter, as I have just done. – jonrsharpe May 04 '14 at 22:30
@ jonrsharpe I edited my question for the complete code – Program_coder May 04 '14 at 22:51
@Program_coder yes, and you haven't assigned the result of `transpose_matrix` to anything. Make it `matrix = transpose_matrix(matrix)`, or use the in-place version of the function. It already tells you this in my answer. – jonrsharpe May 04 '14 at 22:53
Sorry,im new in programming.I edited my question,is it what you mean? Under option 7.I tried it,and when I select option 1 to print the matrix, it print [] ,while the matrix before was [[1,2,3],[4,5,6],[]] – Program_coder May 04 '14 at 22:58
@Program_coder ah, finally I see the problem; your example matrices aren't quite representative. I have edited my answer accordingly. Also, note that `'choix' != 'choi'`. – jonrsharpe May 04 '14 at 23:03
it says zip_longest global name not defined(error) when I write list(zip_longest(*matrix, fillvalue=0)) – Program_coder May 04 '14 at 23:21
2

@Program_coder it's part of `itertools`; you will need to `from itertools import zip_longest` – jonrsharpe May 04 '14 at 23:24
from itertools import zip_longest list(zip_longest(*matrix, fillvalue=0)) return matrix Then when I choose option 1 from my menu above,it display the same matrix – Program_coder May 04 '14 at 23:28
2

@Program_coder that code is difficult to read in the comments. However, it looks like you **still aren't actually assigning the result of the computation back to `matrix`**. I give up: put the `import` at the top of your script, and copy and paste the function I just added to the end of my question. – jonrsharpe May 04 '14 at 23:31
It finally worked.Finally.Yes! My last question would be,would it be possible to display nothing instead of the 0 value? when I display the list.Thanks a lot for all the support. – Program_coder May 04 '14 at 23:36
@Program_coder you could use `fillvalue=""` instead, but might get into trouble if later code expects all values to be numbers. Is `0` a valid value for the matrix? You could use the default `fillvalue=None` then `print("\n".join(" ".join(str(i) if i is not None else "" for i in row) for row in matrix))`. – jonrsharpe May 04 '14 at 23:39

poke · Answer 2 · 2014-05-04T22:30:52.483

1

There are some problems with the solutions you tried:

def transpose_matrix(matrix):
    zip(*matrix)
    return matrix

The result from zip(…) is thrown away, and you return the original matrix instead. Write return zip(*matrix).

def transpose_matrix(matrix):
   map(list, zip(*matrix))
   return matrix

Same as above, the result from map(…) is thrown away. Write return map(list, zip(*matrix)) instead.

def transpose_matrix(matrix):
    print(list(zip(*m)))

The argument is named matrix, but you are working on m. So either rename the parameter m, or change the zipped value to *matrix.

Examples:

>>> m = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> list(zip(*m))
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]
>>> list(map(list, zip(*m)))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

edited May 04 '14 at 22:30

answered May 04 '14 at 21:54

poke

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print(list(zip(*matrix))) print an empty list [],I just tried it – Program_coder May 04 '14 at 22:16
map(list, zip(*matrix)) return matrix leave the matrix same as it was before – Program_coder May 04 '14 at 22:18

Transpose this matrix(python3)

2 Answers2