David Hutto wrote:
> On Wed, Oct 24, 2012 at 1:23 AM, seektime <michael.j.kra...@gmail.com> wrote:
> > Here's some example code. The input is a list which is a "matrix" of
> > letters:
> > a b a
> > b b a
> >
> > and I'd like to turn this into a Python array:
> >
> > 1 2 1
> > 2 2 1
> >
> > so 1 replaces a, and 2 replaces b. Here's the code I have so far:
> >
> >>>> L=['a b a\n','b b a\n']
> >>>> s=' '.join(L)
> >>>> seq1=('a','b')
> >>>> seq2=('1','2')
> >>>> d = dict(zip(seq1,seq2))
> >>>> # Define method to replace letters according to dictionary (got this from
> http://gommeitputor.wordpress.com/2008/09/27/search-replace-multiple-words-or-characters-with-python/).
> > ... def replace_all(text, dic):
> > ... for i, j in dic.iteritems():
> > ... text = text.replace(i, j)
> > ... return text
> > ...
> >
> >>>> seq = replace_all(s,d)
> >>>> print seq
> > 1 2 1
> > 2 2 1
> >
> >>>> seq
> > '1 2 1\n 2 2 1\n'
> >
> I'd suggest, if this is what you're referring to:
>
> x = seq.split('\n ')
> array_list = [ ]
> next_3_d_array = []
> range_of_seq = len(seq)
> for num in range(0,range_of_seq):
> if num % 3 != 0:
> next_3_d_array.append(num)
> if num % 3 == 0:
> array_list.append(next_3_d_array)
> next_3_d_array = [ ]
>
Wow, that looks complicated. Why hardcode to 3 instead of where ever
the newline is?
>>> [ int(x.strip()) for subseq in seq.split('\n') for x in subseq.split() ]
[1, 2, 1, 2, 2, 1]
>>> lst = []
# OR
>>> for subseq in seq.split('\n'):
... for x in subseq.split():
... lst.append( int(x.strip()))
...
>>>
Ramit Prasad
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