Python parse issue

Question

I need to do a sort of reverse .format() to a string like

a = "01AA12345AB12345AABBCCDDEE".reverseformat({id:2d}{type:2s}{a:3d}{b:4s}{c:5d}{d:2s})
print a

>>>> {'id':1, 'type':'aa', 'a':'123', 'b':'45AB', 'c':'12345', 'd':'AA'}

I found this lib that makes almost what i need, the problem is that it gives me this result

msg = parse.parse("{id:2d}{type:3S}{n:5S}", "01D1dddffffffff")

print msg.named

>>>>{'type': 'D1dddfffffff', 'id': 1, 'n': 'f'}

and not

{'id':1, 'type':'D1d', 'n':'ddfffff'}

Does another lib/method/wathever that can "unpack" a string to a dict exists?

EDIT: Just for clarify, i already tryed the w and D format specification for string

Matt Taylor · Accepted Answer · 2015-11-26 11:46:41Z

4

Is there any reason you can't just slice it like a normal string if your format is always the same?

s = "01D1dddffffffff"
id = s[:2]
type = s[2:5]
n = s[5:]

Which gives id, type, and n as:

01
D1d
ddffffffff

And it's trivial to convert this into a dictionary from there if that's your need. If your parsing doesn't need to be dynamic (which it doesn't seem to be from your question in it's current state) then it's easy enough to wrap the slicing in a function which will extract all of the values.

This also has the advantage that from the slice it's clear how many characters and what position in the string you're extracting, but in the parse formatter the positions are all relative (i.e. finding which characters n extracts means counting how many characters id and type consume).

edited Nov 26, 2015 at 11:46

answered Nov 26, 2015 at 10:40

Matt Taylor

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2 Comments

barbaanto Over a year ago

I know, before the parse lib it was like that, but it would have been more readable with that string format

Matt Taylor Over a year ago

I've responded to you as an extra edit to my answer. I personally think that the slice is clearer than the format specifier.

Adam Smith · Accepted Answer · 2015-11-26 11:42:38Z

2

You can use regular expressions to do what you want here.

import re

a = "01AA12345AB12345AABBCCDDEE"
expr = re.compile(r"""
    (?P<id>.{2})          # id:2d
    (?P<type>.{2})        # type:2s
    (?P<a>.{3})           # a:3d
    (?P<b>.{4})           # b:4s
    (?P<c>.{5})           # c:5d
    (?P<d>.{2})           # d:2s""", re.X)

expr.match(a).groupdict()
# {'id': '01', 'b': '45AB', 'c': '12345', 'd': 'AA', 'a': '123', 'type': 'AA'}

You could even make a function that does this.

def unformat(s, formatting_str):
    typingdict = {'s': str, 'f': float, 'd':int}  # are there any more?
    name_to_type = {}
    groups = re.findall(r"{([^}]*)}", formatting_str)
    expr_str = ""
    for group in groups:
        name, formatspec = group.split(":")
        length, type_ = formatspec[:-1], typingdict.get(formatspec[-1], str)
        expr_str += "(?P<{name}>.{{{length}}})".format(name=name, length=length)
        name_to_type[name] = type_
    g = re.match(expr_str, s).groupdict()
    for k,v in g.items():
        g[k] = name_to_type[k](v)

    return g

Then calling like...

>>> a
'01AA12345AB12345AABBCCDDEE'
>>> result = unformat(a, "{id:2d}{type:2s}{a:3d}{b:4s}{c:5d}{d:2s}")
>>> result
{'id': 1, 'b': '45AB', 'c': 12345, 'd': 'AA', 'a': 123, 'type': 'AA'}

However I hope you can see how incredibly ugly this is. Don't do this -- just use string slicing.

edited Nov 26, 2015 at 11:42

answered Nov 26, 2015 at 11:35

Adam Smith

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3 Comments

Matt Taylor Over a year ago

Upvoted just for doing a regex version. Regex always makes me cringe slightly though.

Adam Smith Over a year ago

@MattTaylor yeah it's a terrible solution, but it was fun to bite into. Building dynamic regexes seems like a task better suited for the damned or the mad, however! :)

barbaanto Over a year ago

Yay, exactly what i needed, i know that it's quite ugly, but it's easier to implement in my program since it's easier than the abnormous string slicing that i should have done

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