This may work:
from pprint import pprint
plan = """xooooo
xooooo
oxxxxx
oooooo
ooooox
ooooox
"""
width = plan.index("\n")
height = plan.count("\n")
a = [[w, h] for h in xrange(height) for w in xrange(width)]
for (xy, c) in zip(a, plan.replace("\n", "")):
xy.append(c)
pprint(a)
|46>%run test.py
[[0, 0, 'x'],
[1, 0, 'o'],
[2, 0, 'o'],
[3, 0, 'o'],
[4, 0, 'o'],
[5, 0, 'o'],
[0, 1, 'x'],
[1, 1, 'o'],
[2, 1, 'o'],
[3, 1, 'o'],
[4, 1, 'o'],
[5, 1, 'o'],
[0, 2, 'o'],
[1, 2, 'x'],
[2, 2, 'x'],
[3, 2, 'x'],
[4, 2, 'x'],
[5, 2, 'x'],
[0, 3, 'o'],
[1, 3, 'o'],
[2, 3, 'o'],
[3, 3, 'o'],
[4, 3, 'o'],
[5, 3, 'o'],
[0, 4, 'o'],
[1, 4, 'o'],
[2, 4, 'o'],
[3, 4, 'o'],
[4, 4, 'o'],
[5, 4, 'x'],
[0, 5, 'o'],
[1, 5, 'o'],
[2, 5, 'o'],
[3, 5, 'o'],
[4, 5, 'o'],
[5, 5, 'x']]
* David Crisp <david.cr...@gmail.com> [2011-04-04 13:09:35 +1000]:
> On Fri, Apr 1, 2011 at 2:04 AM, Peter Otten <__pete...@web.de> wrote:
> > David Crisp wrote:
> >
> >> Hello,
> >>
> >> I have a very simple question / problem I need answered. The problem
> >> is imnot entirely sure of the correct terminology and langauge to use
> >> to describe it. (One of the reasons im using this miling list)
> >>
> >> I have a 2d matrix representing the X the Y and the Z value of a
> >> point. I wish to convert that matrix to an array. What is a good
> >> way of doing so?
> >>
> >> Eg:
> >> Matrix
> >> 012345
> >> 0xooooo
> >> 1xooooo
> >> 2oxxxxx
> >> 3oooooo
> >> 4ooooox
> >> 5ooooox
> >>
> >>
> >> I want to convert that to a 2d array which looks like:
> >> 0,0,x
> >> 0,1,o
> >> 0,2,o
> >> 0,3,o
> >> 0,4,o
> >> 0,5,o
> >> .......
> >> 5,4,o
> >> 5,5,o
> >>
> >> I am pretty sure it is simple. I'm just having a brain fade.
> >
> > Using basic numpy:
> >
> >>>> import numpy as np
> >>>> a = np.array(list("xoo"
> > ... "oxx"
> > ... "oxo")).reshape(3,3)
> >>>> a
> > array([['x', 'o', 'o'],
> > ['o', 'x', 'x'],
> > ['o', 'x', 'o']],
> > dtype='|S1')
> >>>> np.array([np.arange(9)//3, np.arange(9)%3, a.flatten()]).transpose()
> > array([['0', '0', 'x'],
> > ['0', '1', 'o'],
> > ['0', '2', 'o'],
> > ['1', '0', 'o'],
> > ['1', '1', 'x'],
> > ['1', '2', 'x'],
> > ['2', '0', 'o'],
> > ['2', '1', 'x'],
> > ['2', '2', 'o']],
> > dtype='|S8')
> >>>> np.array([np.arange(9)//3, np.arange(9)%3,
> > (a=="x").flatten()]).transpose()
> > array([[0, 0, 1],
> > [0, 1, 0],
> > [0, 2, 0],
> > [1, 0, 0],
> > [1, 1, 1],
> > [1, 2, 1],
> > [2, 0, 0],
> > [2, 1, 1],
> > [2, 2, 0]])
> >>>> np.array([np.arange(9)//3, np.arange(9)%3, a.flatten()],
> > dtype=object).transpose()
> > array([[0, 0, x],
> > [0, 1, o],
> > [0, 2, o],
> > [1, 0, o],
> > [1, 1, x],
> > [1, 2, x],
> > [2, 0, o],
> > [2, 1, x],
> > [2, 2, o]], dtype=object)
> >
> > If that's not good enough you may also ask on the numpy mailing list.
>
> Thanks Peter,
>
> That appears to do what I want, in a way. How does this work if you
> have a matrix which is of variable size? For instance, some of my
> data will create a 10 by 10 matrix but some will create a 40 by 40
> matrix, Or for that matter any size. I notice your example
> specifically states there will be 9 outputs ( tupples? ) what if I
> want to say "just create as many tuples as you need to use to
> transpose the data"
>
> Regards,
> David
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