Question

Parse string & return nested list

Asked by: HonorGod

I'm looking for an elegant solution for string parsing problem.
The string contains bracketed name/value pairs.

I want an elegant solution.
- Don't expect to get points for a quick and dirty solution
Especially one that "almost" works
I don't want, and won't accept a LEX/YACC solution.
I must be able to implement it in Python 2.1 (so, pyParsing won't work).
I will accept a complete pseudo-code algorithm that I can implement

Notes:
- Name & value are separated by a single space
- Value may be empty (e.g., "[name ]")
- Values may contain other bracketed name/value pairs.
- Nested result may be arbitrarily deep
- solution must be able to handle nesting at least 7 deep

Sample input:

[[XXXXXXXXXXXXXXXXXXXX XXXX] [XXXXXXXXXXXXXXXXXX XXXX] [XXXXXXXXXXXXXXXXXXX XXXX] [XXXXXXXXXXXXXXXXXXXXXX XXXXXXXX] [XXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXX] [XXXXXXXXXXXX 999] [XXXXXXXXXXXXXXXXXXXXXX XXXXX] [XXXXXXXXXXX9XXXXXXXX XXXXX] [XXXXXXXXXXXXXXXXXX XXXXX] [XXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXX] [XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXX] [XXXXXXXXXX [[[XXXX XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXX] [XXXXX ] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXXXX.XXX_XXXXXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXXXX.XXX_XXXXXXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXXXX.XXX_XXXXXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXX] [XXXXX XXXXXX.XXXXXXX] ][[XXXX XXX.XXX.XXXXX.XXXXXXXX.XXXX.XXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXXXXXXXXX|XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXX9XXXXXXX|XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XXXXX.XXXXXXXX.XXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XXXXX.XXXXXXXX.XXXXX.XXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XXXXX.XXXXXXXX.XXXXX.XXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XXXXX.XXXXXXXX.XXXXX.XXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXX.XXXXXXXXX9XXXXXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXXXXXXX.XXXXXXX] [XXXXX XXXXX] ][[XXXX XXX.XXX.XXXXXXXXX.XXXXXXXX.XXXXXXXXXXXXX] [XXXXX XXX.XXX.XX.XXXXXXXX.XXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XX.XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXX] [XXXXX XXX.XXX.XXXXXXXX.XXXX.*:XXXXX.XXXXXXXX.XXXX.XXXXXXXX.XXXXXXXXXXX:XXXXX.XXXXXXXX.XXXX.XXXXXXXX.XXXXXXXXXXXXXX] ][[XXXX XXX.XXX.XXXXXXXXX.XXXXXXXX.XXX.XXXXXXXXXXXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXXXXXXXX.XXXXXXXX.XXX.XXXXXXXXX_XXXX] [XXXXX XXXX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXXXXXXX] [XXXXX XX=XXXXXX99.XXX.XXXXXXX.XXX.XXX,XX=XXXXXX99XXXX99XXXX,XX=XXXXXX99XXXX99,X=XXX,X=XX] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXX] [XXXXX [XX=XXXXXX99.XXX.XXXXXXX.XXX.XXX,XX=XXXX XXXXXXXXXXX,XX=XXXXXX99XXXX99XXXX,XX=XXXXXX99XXXX99,X=XXX,X=XX]] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXX] [XXXXX 9999] ][[XXXX XXX.XXX.XXX.XXXXXXXXXXXXXXXXXX] [XXXXX 999] ]]] ]

1:

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Asked On: 2009-08-16 at 18:21:02ID24657015
Topics: Python Scripting Language
,; Algorithms
,; Parsers
Participating Experts: 2
Points: 500
Comments: 19

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Answers

by: peprPosted on 2009-08-17 at 01:53:36ID: 25112461

There is a sequence like (spaces replaced by _ to visualize them)

[[XXXX_XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX]_[XXXXX XXXX]_]

How this should be interpreted? To simplify further, the X'es shortened to A, B, C, D respectively:

[[A_B]_[C_D]_]

If the name-value are paired with possibly missing value, what is the name for the last empty value. In other words, is it only a syntax feature and the above should be considered the same as

[[A_B]_[C_D]]

I.e. what to do with the space in front of the last ] ?

by: peprPosted on 2009-08-17 at 01:55:32ID: 25112472

Should also be the simple name-value pairs be represented as lists with two elements?

by: HonorGodPosted on 2009-08-17 at 03:58:42ID: 25113066

Unfortunately, not everything is a name value pair...
Sorry for not being more clear.

I see that you put underscores where the blanks were.

Q: How should this be interpreted?
[[XXXX_XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX]_[XXXXX XXXX]_]

A: I would like to have a list returned with 2 name pair elements,
the first would contain:
[ 'XXXX', 'XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX' ]
and the second would contain:
['XXXXX', 'XXXX']

So, that specific string should return the following list:

[ [ 'XXXX', XXXXXXXX.XXXXXXXXXXXXXXXXXXXXXXXXXXXXX ], [ 'XXXXX', 'XXXX' ] ]

The think with which I have been having real trouble is when a value contains nested brackets...

For example (with single character "items"):

'[ A [ [ [ B C ] [ D E ] [ [ F G ] [ H I ] ] ] [ [ J K ] [ L M ] [ [ N O ] [ P Q ] ] ] ] ]'

by: HonorGodPosted on 2009-08-17 at 04:00:19ID: 25113071

Interesting problem... eh? ;-)

Thanks for taking a look at this. I have been trying to force fit a few different things over the weekend, without success.

I have something that I thought of during my sleep last night that I want to try.

I'll let you know.

by: t0t0Posted on 2009-08-17 at 05:48:00ID: 25113748

Okay, to make things clearer.... what would be the output for these (given in the format you want the output to be in):

1) [adam amanda]

2) [[adam amanda][brian brenda]]

3) [[adam amanda][brian brenda][craig ]]

4) [adam ]

by: t0t0Posted on 2009-08-17 at 05:50:03ID: 25113774

and these:

5) [adam amanda][brian brenda]

6) [adam [brian [craig cassy]]]

by: t0t0Posted on 2009-08-17 at 05:50:38ID: 25113778

or am i completely off the track there?

by: HonorGodPosted on 2009-08-17 at 06:13:12ID: 25113982

Here is what I would like to see:

1) [adam amanda]
['adam', 'amanda']

2) [[adam amanda][brian brenda]]
[['adam', 'amanda'], ['brian', 'brenda'] ]

3) [[adam amanda][brian brenda][craig ]]
[ ['adam', 'amanda'], ['brian', 'brenda'], ['craig', ''] ]

4) [adam ]
['adam', '' ]

5) [adam amanda][brian brenda]
This would not occur. The data is guaranteed to be well formed.
go ahead and throw an exception

6) [adam [brian [craig cassy]]]
['adam', [ 'brian', ['craig', 'cassy'] ] ]

>> or am i completely off the track there?

Not at all. You seem to understand my challenge... ;-)

by: peprPosted on 2009-08-17 at 06:16:23ID: 25114013

OK. Let's try the snippet below first to clarify how it should work. The code is based on finite automaton (processing single characters and changing the status to capture the situation). However, the finite automaton apparatus (theoretically of the same power as regular expressions) is not capable to capture nested pair structures. This way it could be done recursively (but not that easy or efficient because of the stream of processed characters) or using a stack that captures the not-finished levels. I have chosen the second approach. The resulting list can be accessed as stack[0]. The code should be cleaned, yet. It also depends on your clarification.

# Get the string from somewhere (here stored in the file).
f = open('a.txt')
s = f.read()
f.close()
 
# Introduce a stack for avoiding the need for a recursive algorithm.
# The stack is implemented using the built in list.
stack = []  # init as empty list
 
# Initialize status of the automaton.
status = 0
 
lst = None         # init -- current list
lstStr = None      # init -- collected string
 
# Process all characters in the loop.
for c in s:
 
    if status == 0:     # not collecting a string
    
        if c == '[':        # start the new list
            lst = []            # list of the next level
            stack.append(lst)   # push to the stack
            # do not change the status
            
        elif c == ' ':      
            pass            # ignore the extra separator
            
        elif c == ']':
            # If the list contains a single element, then append
            # the missing empty value.
            if len(lst) == 1:
                lst.append('')
            
            # Get the just-built list of the level (lst) and append it 
            # to the upper level list.
            stack.pop()          # remove the deeper-level list...
            if len(stack) > 0:
                stack[-1].append(lst) # append to the upper level
                lst = stack[-1]       # the upper level is the current list from now on
                status = 0            # keep the status
            else:
                stack.append(lst)     # should finish now
                status = 888          # final state
            
        else:
            assert lstStr is None
            lstStr = [c]    # first character to be collected
            status = 1      # change the status
 
            
    elif status == 1:   # collecting a string            
        
        if c == ' ':        # separate the just-collected
            s = ''.join(lstStr)  # collected string
            lstStr = None        # init
            lst.append(s)        # add to the current list
            status = 0           # change the status
 
        elif c == ']':
            s = ''.join(lstStr)  # collected string
            lstStr = None        # init
            lst.append(s)        # add to the current list
            
            # Get the just-built list of the level (lst) and append it 
            # to the upper level list.
            stack.pop()          # remove the deeper-level list...
            if len(stack) > 0:
                stack[-1].append(lst) # append to the upper level
                lst = stack[-1]       # the upper level is the current list from now on
                status = 0            # change the status
            else:
                stack.append(lst)     # should finish now
                status = 888          # final state
                
        else:
            lstStr.append(c)    # collect the next character
 
    
    elif status == 888:   # stop it
        break
    
    else:
        print 'Unknown status:', status
        print 'stack:', repr(stack)
        print 'lst:', repr(lst)
        print 'lstStr:', repr(lstStr)
    
assert len(stack) == 1
print repr(stack[0])

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by: HonorGodPosted on 2009-08-17 at 06:17:19ID: 25114026

And here's an ugly one...

7) [adam [[[brian brenda][craig christina]][[doug debbie][eric ethal]]]]
['adam', [ [ ['brian', 'brenda'],['craig', 'christina']],[['doug', 'debbie'],['eric', 'ethal']]]]

by: HonorGodPosted on 2009-08-17 at 06:19:01ID: 25114052

Hopefully this make sense.

Thank you for your continued interest.

by: HonorGodPosted on 2009-08-17 at 06:20:39ID: 25114063

Studying http:#a25114013 ...

by: HonorGodPosted on 2009-08-17 at 06:35:12ID: 31616375

Outstanding. Thank you ever so much.

by: HonorGodPosted on 2009-08-17 at 06:50:59ID: 25114384

pepr: Thanks again.

However, I did find one little flaw in the code.

lines ( 55-57 & 61-63) - you "reuse" s which is the same variable name as the string over which the for loop is iterating.

Fortunately, it is easily corrected.

by: peprPosted on 2009-08-17 at 13:44:48ID: 25118218

Yes. Sorry for the "mistake". This was not intentional. You can also completely get rid of it by replacing

s = ''.join(lstStr) # collected string
lstStr = None # init
lst.append(s) # add to the current list

by

lst.append(''.join(lstStr)) # add collected string to the current list
lstStr = None # init

I am not sure, but even the original code could be correct. The s is reused while the reference to the original string can be remembered by the code implementing the for loop. Anyway, even it it was the case, I do not like tricks like that. The code should be as understandable as possible.

by: peprPosted on 2009-08-17 at 13:50:28ID: 25118282

The doc says (http://docs.python.org/reference/compound_stmts.html#the-for-statement)

"The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator..."

This means that you should not observe an error in the produced result. Anyway, you are right.

by: HonorGodPosted on 2009-08-17 at 14:51:34ID: 25118740

Yes, I did exactly that (http:#a25118218)

>> "The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator..."

I agree completely. However, as I pointed out, this is for Python 2.1, which preceded iterators... :-)

by: peprPosted on 2009-08-17 at 23:30:13ID: 25120537

You are definitely good in programming, so you know what to do ;) The code can be simplified. For example, you can remove the

lstStr = None # init
and
assert lstStr is None

It is there only to make the potential errors more visible. You know. However, other people may be reading this...

Have a nice day ;)

by: HonorGodPosted on 2009-08-18 at 04:39:58ID: 25122017

You too. Thanks again for all of your help with this question, and all of the questions on this site in which you participate. Have a super day.

20120131-EE-VQP-002

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Parse string & return nested list

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