# 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

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))
``````

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
``````

@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
``````
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