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Comment by kazinator | Hacker News Reader

This rewrite is more performant than the original:

  query_params = s.split('?')[1].split('&')[-3:]
  map(lambda x: x.split('=')[1], query_params)

The calculation of query_params, having no dependency on the lambda parameters or anything being mutated, has been lifted out of the lambda, and thus spared from repeated execution by map. The compiler for that language won't do this automatically.

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What? No it isn't! You didn't parse that correctly.

The query params were never in the lambda to begin with. Python function calls have strict (not lazy) semantics, i.e. "applicative order", i.e. both expressions passed as arguments to map() are evaluated before the map body gets them as parameters, thus the query params would only be evaluated once, even when inlined as they were originally.

Same with the lambda definition: it's evaluated only once. It's just the lambda body that gets reevaluated each loop, and only evaluated for the first time on the first loop.

Sorry; it looked to me like x.split('=')[1], query_params is a tuple being returned by the lambda. But of course that leaves map without the needed argument.

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