• tunetardis@lemmy.ca
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    6 months ago

    Can you use it to initialize vars outside the scope of the lambda?

    No, that’s not what it’s for. It lets you define a temporary local variable within an expression. This is useful in situations where you might want to use the same value more than once within the expression. In a regular function, you would just define a variable first and then use it as many times as you want. But until the walrus operator came along, you couldn’t define a variable within a lambda expression.

    Can you give an example?

    Ok, I’m trying to think of a simple example. Let’s say you had a database that maps student IDs to records contain their names. To keep things simple, I’ll just make it plain old dict. And then you have a list of student IDs. You want to sort these IDs using the student names in the form “last, first” as the key. So you could go:

    >>> student_recs = {1261456: {"first": "Harry", "last": "Potter"}, 532153: {"first": "Ron", "last": "Weasley"}, 632453: {"first": "Hermione", "last": "Granger"}}
    >>> student_ids = [1261456, 532153, 632453]
    >>> sorted(student_ids, key = lambda i: (rec := student_recs[i])['last'] + ', ' +  rec['first'])
    [632453, 1261456, 532153]
    

    The problem here is that student_ids doesn’t contain the student names. You need use the ID to look up the record that contains those. So let’s say the first ID i is 1261456. That would mean:

    rec := student_recs[i]
    

    evaluates to:

    {"first": "Harry", "last": "Potter"}
    

    Then we are effectively going:

     rec['last'] + ', ' + rec['first']
    

    which should give us:

     'Potter, Harry'
    

    Without the := you would either have to perform 2 student_recs[i] look-ups to get each name which would be wasteful or replace the lambda with a regular function where you can write rec = student_recs[i] on its own line and then use it.

    Am I making any sense?

    • tunetardis@lemmy.ca
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      6 months ago

      Actually, now that I think of it, there’s no reason you need to join the 2 names into a single str. You could just leave it as a tuple of last, first and Python will know what to do in comparing them.

      >>> sorted(student_ids, key = lambda i: ((rec := student_recs[i])['last'], rec['first']))
      [632453, 1261456, 532153]
      

      So the lambda would be returning ('Potter', 'Harry') rather than 'Potter, Harry'. But whatever. The := part is still the same.