How to form a dict out of a string by doing regex ?

[Mel]

Satyajit Sarangi wrote:

> 
> 
> data = "GEOMETRYCOLLECTION (POINT (-8.9648437500000000
> -4.1308593750000000), POINT (2.0214843750000000 -2.6367187500000000),
> POINT (-1.4062500000000000 -11.1621093750000000), POINT
> (-11.9531250000000000,-10.8984375000000000), POLYGON
> ((-21.6210937500000000 1.8457031250000000,2.4609375000000000
> 2.1972656250000000, -18.9843750000000000 -3.6914062500000000,
> -22.6757812500000000 -3.3398437500000000, -22.1484375000000000
> -2.6367187500000000, -21.6210937500000000
> 1.8457031250000000)),LINESTRING (-11.9531250000000000
> 11.3378906250000000, 7.7343750000000000 11.5136718750000000,
> 12.3046875000000000 2.5488281250000000, 12.2167968750000000
> 1.6699218750000000, 14.5019531250000000 3.9550781250000000))"
> 
> This is my string .
> How do I traverse through it and form 3 dicts of Point , Polygon and
> Linestring containing the co-ordinates ?

Except for those space-separated number pairs, it could be a job for some 
well-crafted classes (e.g. `class GEOMETRYCOLLECTION ...`, `class POINT 
...`) and eval.

My approach would be to use a loop with regexes to recognize the leading 
element and pick out its arguments, then use the string split and strip 
methods beyond that point.  Like (untested):

recognizer = re.compile (r'(?(POINT|POLYGON|LINESTRING)\s*\(+(.*?)\)+,(.*)')
# regex is not good with nested brackets, 
# so kill off outer nested brackets..
s1 = 'GEOMETRYCOLLECTION ('
if data.startswith (s1):
    data = data (len (s1):-1)

while data:
    match = recognizer.match (data)
    if not match:
        break	# nothing usable in data
    ## now the matched groups will be:
    ## 1: the keyword
    ## 2: the arguments inside the smallest bracketed sequence
    ## 3: the rest of data
    ##  so use str.split and str.match to pull out the individual arguments,
    ## and lastly
    data = match.group (3)

This is all from memory.  I might have got some details wrong in recognizer.

	Mel.