Show HN: Refurb – A tool for refurbishing and modernizing Python codebases

Re:Show HN: Refurb – A tool for refurbishing and modernizing... by kevincox on Wednesday September 28, 2022 @09:37AM (#33007761)

My guess is that it is because for int `x == hash(x)`. This means that if your integers are smaller than the number of buckets in the hashtable they will fall into buckets in order and the iteration is likely just over the buckets.

For example you can see that by adding a large power of two to the numbers you can see that they are sorted by the remainder (as it seems that the default number of buckets is a power of two):

  > for x in {2048 + 1, 256 + 3, 1024 + 2, 512 + 4}: print(x)
  2049
  1026
  259
  516

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Re:Show HN: Refurb – A tool for refurbishing and modernizing... by cmcconomy on Wednesday September 28, 2022 @10:29AM (#33008388)

right, python caches ints from -5 to 256 as singletons, so id(x) for those numbers is constant.

whereas each time you run id() on a number outside that range you will get a different id as a new object is instantiated

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Re:Show HN: Refurb – A tool for refurbishing and modernizing... by kevincox on Wednesday September 28, 2022 @10:32AM (#33008431)

I don't think this is related. `set()` uses `hash(x)` it doesn't care what `id(x)` is. The example I provided works equally well with small numbers and large numbers.

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