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pythonjsonpython-3.xpattern-matchingpython-3.10

Python 3.10 pattern matching (PEP 634) - wildcard in string

I got a large list of JSON objects that I want to parse depending on the start of one of the keys, and just wildcard the rest. A lot of the keys are similar, like "matchme-foo" and "matchme-bar". There is a builtin wildcard, but it is only used for whole values, kinda like an else.

I might be overlooking something but I can't find a solution anywhere in the proposal:

https://docs.python.org/3/whatsnew/3.10.html#pep-634-structural-pattern-matching

Also a bit more about it in PEP-636:

https://www.python.org/dev/peps/pep-0636/#going-to-the-cloud-mappings

My data looks like this:

data = [{
          "id"     : "matchme-foo",
          "message": "hallo this is a message",
      },{
          "id"     : "matchme-bar",
          "message": "goodbye",
      },{
          "id"     : "anotherid",
          "message": "completely diffrent event"
      }, ...]

I want to do something that can match the id without having to make a long list of |'s.

Something like this:

for event in data:
    match event:
        case {'id':'matchme-*'}: # Match all 'matchme-' no matter what comes next
            log.INFO(event['message'])
        case {'id':'anotherid'}:
            log.ERROR(event['message'])

It's a relatively new addition to Python so there aren't many guides on how to use it yet.

Solution

  • You can use a guard.

    for event in data:
    match event:
        case {'id': x} if x.startswith("matchme"): # guard
            print(event["message"])
        case {'id':'anotherid'}:
            print(event["message"])

    Quoting from the official documentation:

    Guard

    We can add an if clause to a pattern, known as a “guard”. If the guard is false, match goes on to try the next case block. Note that value capture happens before the guard is evaluated:

    match point:
     case Point(x, y) if x == y:
         print(f"The point is located on the diagonal Y=X at {x}.")
     case Point(x, y):
         print(f"Point is not on the diagonal.")

    See also: