matrix

[kur3k]

hey gues

i search 'replace' matrix?

123
456
789

i need in output

147
258
369

[vegaseat]

You need to give us more info!

[kur3k]

my code

def y():

    matrix = [ "12.34, 67.89, 56.21", 
                "22.01, 11.0,  34.56",
                "20.0,  56,    0"     ]

    a = [x.split(", ") for x in matrix]

In polsih this word name is 'tranpozycja' ( in math ).

[vegaseat]

An excellent use of module numpy ...

# transpose a Python list of lists (matrix)
# using module numpy from http://sourceforge.net/projects/numpy
# tested with Python 2.6.5

import numpy as np

# a Python list of lists (made to look pretty)
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]

# change to a numpy matrix
matrix_np = np.array(matrix)

print( matrix_np )

"""my result -->
[[1 2 3]
 [4 5 6]
 [7 8 9]]
"""

matrix_np_transposed = np.transpose(matrix_np)

print( matrix_np_transposed )

"""my result -->
[[1 4 7]
 [2 5 8]
 [3 6 9]]
"""

# change back into a Python list of lists (matrix)
matrix_transposed = matrix_np_transposed.tolist()

print( matrix_transposed )

"""my result (made pretty) -->
[
[1, 4, 7], 
[2, 5, 8], 
[3, 6, 9]
]
"""

[kur3k]

i don't want libary, i want new function! ;p

[Gribouillis]

There is the classical zip(*list) :

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

[TrustyTony]

Also

>>> n=10
>>> a=[list(range(n)) for y in range(n)]
>>> from pprint import pprint as pp
>>> pp(a)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]
>>> pp(map(None,*a))
[(0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1),
 (2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
 (3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
 (4, 4, 4, 4, 4, 4, 4, 4, 4, 4),
 (5, 5, 5, 5, 5, 5, 5, 5, 5, 5),
 (6, 6, 6, 6, 6, 6, 6, 6, 6, 6),
 (7, 7, 7, 7, 7, 7, 7, 7, 7, 7),
 (8, 8, 8, 8, 8, 8, 8, 8, 8, 8),
 (9, 9, 9, 9, 9, 9, 9, 9, 9, 9)]
>>>

(zip stops in shortest iterable)
And if you happen to need to make the shorter lines longer by filling by None. You can do howerver map(None,*matrix)

>>> a=[list(range(y)) for y in range(n)]
>>> pp(a)
[[],
 [0],
 [0, 1],
 [0, 1, 2],
 [0, 1, 2, 3],
 [0, 1, 2, 3, 4],
 [0, 1, 2, 3, 4, 5],
 [0, 1, 2, 3, 4, 5, 6],
 [0, 1, 2, 3, 4, 5, 6, 7],
 [0, 1, 2, 3, 4, 5, 6, 7, 8]]
>>> pp(zip(*a))
[]
>>> pp(map(None,*a))
[(None, 0, 0, 0, 0, 0, 0, 0, 0, 0),
 (None, None, 1, 1, 1, 1, 1, 1, 1, 1),
 (None, None, None, 2, 2, 2, 2, 2, 2, 2),
 (None, None, None, None, 3, 3, 3, 3, 3, 3),
 (None, None, None, None, None, 4, 4, 4, 4, 4),
 (None, None, None, None, None, None, 5, 5, 5, 5),
 (None, None, None, None, None, None, None, 6, 6, 6),
 (None, None, None, None, None, None, None, None, 7, 7),
 (None, None, None, None, None, None, None, None, None, 8)]
>>>

[vegaseat]

One problem is that
map(None, *matrix)
will not work with Python3

However
list(zip(*matrix))
does work with Python3

[TrustyTony]

And with few iterations of interactive testing the triangle matrix can be got back by :icon_wink:

pp(map(lambda x: filter(None,x),zip(*map(None,*a))))

Surely not most straigt forward thing but I got just curious.
If you got confused:
Do transform Map(None way
Undo it zip(* way
Filter Nones by map filter with lambda function for standard parameter None

vegaseat:

In python2.6 zip(*matrix) seem to work without list i.e list(zip(*matrix))==zip(*matrix)

Tony

[kur3k]

I want new function, i dont want use this function ...

New function == algorythims tranpose matrix

[hondros]

Look, if you want a new function, and you're sure that it'll always be a 3x3 matrix, think about what you need to do. Outline it. Do it by hand a few times, and then list EVERY step needed to do it. Then translate it into the python code. So let me show you something:

# Our 3x3 matrix
matrix = \
[
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
]
# What we want it as:
new_matrix = \
[
    [1, 4, 7],
    [2, 5, 8],
    [3, 6, 9]
]
# What steps do we need?
# If we take a look at the number placements, we get this:
# matrix[0][0] = 1
# matrix[0][1] = 2
# matrix[0][2] = 3
# matrix[1][0] = 4
# matrix[1][1] = 5
# matrix[1][2] = 6
# matrix[2][0] = 7
# matrix[2][1] = 8
# matrix[2][2] = 9

# We want it to look like this:
# new_matrix[0][0] = 1
# new_matrix[0][1] = 4
# new_matrix[0][2] = 7
# new_matrix[1][0] = 2
# new_matrix[1][1] = 5
# new_matrix[1][2] = 8
# new_matrix[2][0] = 3
# new_matrix[2][1] = 6
# new_matrix[2][2] = 9

# What steps can we take to make this possible?

If you need further help, don't be hesitant to ask

[TrustyTony]

Why use new function when it is already built in the language zip(*matrix) or map(None,*matrix) You want to do format checking or something?