[Python-Dev] PEP 572: Assignment Expressions (original) (raw)

Glenn Linderman v+python at g.nevcal.com
Mon Apr 23 18:45:59 EDT 2018

On 4/23/2018 1:01 PM, Ethan Furman wrote:

On 04/22/2018 10:44 PM, Tim Peters wrote:

[Guido]

In reality there often are other conditions being applied to the match for which if expr as name is inadequate. The simplest would be something like

if ...: elif (m := re.match('(.):(.)', line)) and m.group(1) == m.group(2): And the match() call may not even be the first thing to check -- e.g. we could have elif line is not None and (m := re.match('(.):(.)', line)) and m.group(1) == m.group(2): I find myself warming more to binding expressions the more I keep them in mind while writing new code.  And I think it may be helpful to continue showing real examples where they would help. Today's example:  I happened to code this a few hours ago: diff = x - xbase if diff: g = gcd(diff, n) if g > 1: return g It's not really hard to follow, but two levels of nesting "feels excessive", as does using the names "diff" and "g" three times each. It's really an "and" test:  if the diff isn't 0 and gcd(diff, n) > 1, return the gcd.  That's how I thought of it from the start. Which this alternative expresses directly: if (diff := x - xbase) and (g := gcd(diff, n)) > 1: return g So I really like being able to make the assignment in the expression, but I have a really hard time parsing it with the name first. Attempting to read just the names first: if diff scan for ending right paren found and g scan for ending right paren oops, found opening left paren scan for ending right paren found resume scanning for final right paren found > 1: return g Attempting to read expressions first: if x - xbase and gcd(diff, n) what's diff? scan backwards diff is x - xbase > 1: return g what's g? scan up and backwards g is gcd(diff, n) Attempting to read interleaved: if skip diff x - xbase back to diff as diff and skip g gcd(diff, n) back to g as g > 1: return g On the other hand, if it were using the "as" keyword: if (x - xbase as diff) and (gcd(diff, n) as g) > 1: return g I would parse as: if x - xbase as diff and gcd(diff, n) as g > 1: return g For me at least, the last is much more readable.  Thinking about it some more, the problem (or maybe just my problem) is that I see an "if" or "while" and the I look for the thing that is True or False, and using the ":=" syntax the first thing I see is a placeholder for a result that doesn't exist yet, making me constantly scan backwards and forwards to put all the pieces in the correct place. With "as", it just flows forwards.

You need to borrow the time machine, and get with those early mathematicians that first said:

Let x be the sum of y and z

and convince them that what they should have said was:

Let the sum of y and z be called x.

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