parsing a dictionary from a string

[Benjamin Georgi]

Hello list,

I could use some help extracting the keys/values of a list of
dictionaries from a string that is just the str() representation of the
list (the problem is related to some flat file format I'm using for file
IO).

Example:

>>s = str(dict_list)
s

'[{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]'

Then, what I want to do is to reconstruct dict_list given s.
Now, one possible solution would be

>>dict_list = eval(s)

but since the content of s cannot be blindly trusted I`d rather not do
it that way. Basically my question is whether there is another solution
which is simpler than using regular expressions.

Best,
Benjamin

[Paul McGuire]

"Benjamin Georgi" <ge****@molgen. mpg.dewrote in message
news:ma******** *************** *************** *@python.org...

Hello list,

I could use some help extracting the keys/values of a list of dictionaries
from a string that is just the str() representation of the list (the
problem is related to some flat file format I'm using for file IO).

Example:

>s = str(dict_list)
s

'[{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]'

Pyparsing comes with a working example that will parse strings representing
lists, even if they are nested. This is actually more complex than the
example you've given - none of your lists is nested. Here is that example,
adapted to handle dict elements, too.

-- Paul
(The pyparsing home page/wiki is at pyparsing.wikis paces.com.)
from pyparsing import *

cvtInt = lambda toks: int(toks[0])
cvtReal = lambda toks: float(toks[0])
cvtTuple = lambda toks : tuple(toks.asLi st())
cvtDict = lambda toks: dict(toks.asLis t())

# define punctuation as suppressed literals
lparen,rparen,l brack,rbrack,lb race,rbrace,col on = \
map(Suppress,"( )[]{}:")

integer = Combine(Optiona l(oneOf("+ -")) + Word(nums))\
.setName("integ er")\
.setParseAction ( cvtInt )
real = Combine(Optiona l(oneOf("+ -")) + Word(nums) + "." +
Optional(Word(n ums)))\
.setName("real" )\
.setParseAction ( cvtReal )
tupleStr = Forward()
listStr = Forward()
dictStr = Forward()

listItem = real|integer|qu otedString.setP arseAction(remo veQuotes)| \
Group(listStr) | tupleStr | dictStr

tupleStr << ( Suppress("(") + Optional(delimi tedList(listIte m)) +
Optional(Suppre ss(",")) + Suppress(")") )
tupleStr.setPar seAction( cvtTuple )

listStr << (lbrack + Optional(delimi tedList(listIte m) +
Optional(Suppre ss(","))) + rbrack)

dictEntry = Group( listItem + colon + listItem )
dictStr << (lbrace + Optional(delimi tedList(dictEnt ry) + \
Optional(Suppre ss(","))) + rbrace)
dictStr.setPars eAction( cvtDict )

tests = """['a', 100, ('A', [101,102]), 3.14, [ +2.718, 'xyzzy', -1.414] ]
[{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]"""

for test in tests.split("\n "):
print "Test:", test.strip()
result = listStr.parseSt ring(test)
print "Result:", result
for dd in result:
if isinstance(dd,d ict): print dd.items()
print

Prints:
Test: ['a', 100, ('A', [101,102]), 3.14, [ +2.718, 'xyzzy', -1.414] ]
Result: ['a', 100, ('A', [101, 102]), 3.1400000000000 001, [2.718,
'xyzzy', -1.4139999999999 999]]

Test: [{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]
Result: [{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]
[(0, [2]), (1, [])]
[(0, []), (1, []), (2, [])]
[(0, [1, 2])]

[Peter Otten]

Benjamin Georgi wrote:

I could use some help extracting the keys/values of a list of
dictionaries from a string that is just the str() representation of the
list (the problem is related to some flat file format I'm using for file
IO).

Example:

>>s = str(dict_list)
>>s

'[{0: [2], 1: []}, {0: [], 1: [], 2: []}, {0: [1, 2]}]'

Then, what I want to do is to reconstruct dict_list given s.
Now, one possible solution would be

>>dict_list = eval(s)

but since the content of s cannot be blindly trusted I`d rather not do
it that way. Basically my question is whether there is another solution
which is simpler than using regular expressions.

Michael Spencer has published a simple eval() that only allows constant
expressions:

http://aspn.activestate.com/ASPN/Coo.../Recipe/364469

Peter