tiissa <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
tiissa wrote:
If you know the number of characters to match can't you just compare slices?
If you don't, you can still do it by hand:
In [7]: def cmp(s1,s2):
....: diff_map=[chr(s1[i]!=s2[i]) for i in range(min(len(s1), len(s2)))]
....: diff_index=''.join(diff_map).find(chr(True))
....: if -1==diff_index:
....: return min(len(s1), len(s2))
....: else:
....: return diff_index
....:
I will look at that, although if i have 300 images i dont want to type all the comparisons (In [9]: cmp('ccc','cccap')) by hand, it would just be easier to sort them then :).
I didn't meant you had to type it by hand. I thought about writing a small script (as opposed to using some in the standard tools). It might look like:
In [22]: def make_group(L): ....: root,res='',[] ....: for i in range(1,len(L)): ....: if ''==root: ....: root=L[i][:cmp(L[i-1],L[i])] ....: if ''==root: ....: res.append((L[i-1],[L[i-1]])) ....: else: ....: res.append((root,[L[i-1],L[i]])) ....: elif len(root)==cmp(root,L[i]): ....: res[-1][1].append(L[i]) ....: else: ....: root='' ....: if ''==root: ....: res.append((L[-1],[L[-1]])) ....: return res ....:
In [23]: L=['cccat','cccap','cccan','dddfa','dddfg','dddfz']
In [24]: L.sort()
In [25]: make_group(L)
Out[25]: [('ccca', ['cccan', 'cccap', 'cccat']), ('dddf', ['dddfa', 'dddfg', 'dddfz'])]
However I guarantee no optimality in the number of classes (but, hey, that's when you don't specify the size of the prefix).
(Actually, I guarantee nothing at all ;p)
But in particular, you can have some file singled out:
In [26]: make_group(['cccan','cccap','cccat','cccb'])
Out[26]: [('ccca', ['cccan', 'cccap', 'cccat']), ('cccb', ['cccb'])]It is a matter of choice: either you want to specify by hand the size of the prefix and you'd rather look at itertools as pointed out by Kent, or you don't and a variation with the above code might do the job.
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