Re-sort string based on precedence of separator

Question

I have a string with a certain meaning for example "a,b;X1" or e&r1. In total there are 3 possible separators between values: ,;& where ; has low precedence.

Also "a,b;X1" and "b,a;X1"are the same but to be able to compare they are the same, I want to predictably resort the string so that indeed the 2 can be compared to be equal. In essence "b,a;X1" must be "sorted" to become "a,b;X1" and this is rather a simple example. The expression can be more complex.

The precedence is of importance as "a,b;X1" is not the same as "a;b,X1".

In general I would need to split into "groups by precedence and then sort the groups and merge things together again but unclear how to achieve this.

So far I have:

example = "b,a;X1"
ls = example.split(';')
ls2 = [x.split(",") for x in ls]
ls3 = [[y.split("&") for y in x] for x in ls2]      
ls3.sort()
print(ls3)
# [[['X1']], [['b'], ['a']]]

Sorting doesn’t yet work as a should be before b and then I’m not sure how to "stitch" the result back together again.

For clarification:

, means OR
& means AND (high precedence)
; means AND (low precedence)

"a,b;X1" therefore means (a OR b) AND X1

"b,a;X1" therefore means (b OR a) AND X1 i.e. the same

Asked By: beginner_

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Source

Answer 1

I would suggest writing a function which recursively sorts each list. Here’s an example of how to do that:

delim_precedence = (';', ',', '&')


def recursive_split(s, delim):
    if isinstance(s, str):
        return s.split(delim)
    elif isinstance(s, list):
        return [recursive_split(i, delim) for i in s]
    else:
        raise Exception("unknown type")


def split_by_precedence(s):
    for delim in delim_precedence:
        s = recursive_split(s, delim)
    return s


def recursive_sort(s):
    if isinstance(s, str):
        return s
    elif isinstance(s, list):
        return sorted([recursive_sort(i) for i in s])
    else:
        raise Exception("unknown type")


def rejoin(s, delims=delim_precedence):
    if len(delims) == 0:
        return s
    return delims[0].join(rejoin(i, delims[1:]) for i in s)
    

def canonicalize(s):
    return rejoin(recursive_sort(split_by_precedence(s)))


print(canonicalize(example))

Answered By: Nick ODell

Answer 2

You could use split, sort as you did (combined with join) , but it should happen at every level of operator:

def normalize(s):
    return ";".join(sorted(
        ",".join(sorted(
            "&".join(sorted(factor.split("&")))
            for factor in term.split(",")
        ))
        for term in s.split(";")
    ))


example = "b,a&z&x;x1;m&f,q&c"
print(normalize(example))  # a&x&z,b;c&q,f&m;x1

Answered By: trincot

Answer 3

@trincot’s answer works but is a maintenance nightmare since it hard-codes the separators for nested splits and joins, so if there is a need for a change of a separator, both the corresponding split and join need to be modified, and if there is a need for an additional separator, an additional layer of split and join needs to be nested. .

A more general approach would be to use a recursive function to split by and join with one separator at a time, and pass the rest of the separators to the next level of recursive call:

def sort(string, separators):
    sep, *rest = separators
    pieces = string.split(sep)
    return sep.join(sorted((sort(p, rest) for p in pieces) if rest else pieces))

ieces))

so that (using @trincot’s test case):

example = "b,a&z&x;x1;m&f,q&c"
separators = ';,&'
print(sort(example, separators))

would output:

a&x&z,b;c&q,f&m;x1

Answered By: blhsing

Re-sort string based on precedence of separator

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