inserting bracketings into a string

[Michael Loritsch]

Steven Bethard <steven.bethard at gmail.com> wrote in message news:<mailman.6460.1100648300.5135.python-list at python.org>...
> I'm trying to insert some bracketings in a string based on a set of
> labels and associated start and end indices.  For example, I'd like to
> do something like:
> 
> >>> text = 'abcde fgh ijklmnop qrstu vw xyz'
> >>> spans = [('A', 0, 9), ('B', 6, 9), ('C', 25, 31)]
> >>> insert_bracketings(text, spans)
> '[A abcde [B fgh]] ijklmnop qrstu [C vw xyz]'
> 
> My current implementation looks like:
> 
> >>> def insert_bracketings(text, spans):
> ... 	starts = [start for _, start, _ in spans]
> ... 	ends = [end for _, _, end in spans]
> ... 	indices = sorted(set(starts + ends))
> ... 	splits = [(text[start:end], start, end)
> ... 	          for start, end in zip([None] + indices, indices + [None])]
> ... 	start_map, end_map = {}, {}
> ... 	for label, start, end in spans:
> ... 		start_map.setdefault(start, []).append('[%s ' % label)
> ... 		end_map.setdefault(end, []).append(']')
> ... 	result = []	
> ... 	for string, start, end in splits:
> ... 		if start in start_map:
> ... 			result.extend(start_map[start])
> ... 		result.append(string)
> ... 		if end in end_map:
> ... 			result.extend(end_map[end])
> ... 	return ''.join(result)
> ... 
> 
> but it seems like there ought to be an easier way.  Can anyone help me?
> 
> Thanks in advance,
> 
> Steve


Below is a little more readable and compact implementation that
produces the same result.  I'm not entirely sure if it qualifies as
'better', but I do believe it is ultimately more readable.

def insert_brackets(text, spans):
    brackets = []
    for span in spans:
        brackets.append((span[1], ("".join(('[', span[0], " ")))))
        brackets.append((span[2], ']'))
    brackets.sort()                 #Note: (n, '[X ') < (n, ']')
    answer = []
    lastIndex = 0
    for bracket in brackets:
        if lastIndex == bracket[0]: #Repeated index
            answer.append(bracket[1])
        else:                       #Non repeated index
            answer.extend((text[lastIndex:bracket[0]], bracket[1]))
            lastIndex = bracket[0]
    return "".join(answer)

Regards,

Michael Loritsch