turn list of letters into an array of integers

[wxjmfauth at gmail.com]

Le mercredi 24 octobre 2012 07:23:11 UTC+2, seektime a écrit :
> 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://gomputor.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'
> 
> 
> 
> My question is how can I turn "seq" into a python array?
> 
> 
> 
> Thanks
> 
> Michael

Not so sure what you mean by an "array of integers".

>>> def z(s):
...     a = s.splitlines()
...     b = [e.split() for e in a]
...     for row in range(len(b)):
...         for col in range(len(b[row])):
...             b[row][col] = ord(b[row][col]) - ord('a')
...     return b
...     
>>> z('a b a\n b b a')
[[0, 1, 0], [1, 1, 0]]
>>> 
>>> # or
>>> table = {'a': 111, 'b': 222}
>>> 
>>> def z2(s, table):
...     a = s.splitlines()
...     b = [e.split() for e in a]
...     for row in range(len(b)):
...         for col in range(len(b[row])):
...             b[row][col] = table[b[row][col]]
...     return b
...     
>>> z2('a b a\n b b a', table)
[[111, 222, 111], [222, 222, 111]]
>>> 
>>> # note
>>> z('a\n b b b b b\n a a')
[[0], [1, 1, 1, 1, 1], [0, 0]]

jmf