# 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

## Answers:

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
```