Request: inspect: signature & getfullargspec & getcallargs
Hello, I have a request. Would it be possible to include `follow_wrapper_chains` and `skip_bound_arg` arguments to higher level functions of `inspect` module? Would exposing them, but setting defaults to what they currently are, be possible? I sometimes need: * `getcallargs`, but without `bound_arg` * `getfullargspec` to `follow_wrapper_chains` * `signature` to include `bound_arg`. Regards, DG -- https://mail.python.org/mailman/listinfo/python-list
Re: on writing a number as 2^s * q, where q is odd
On Sun, 3 Dec 2023 at 10:25, Julieta Shem via Python-list wrote: > > Alan Bawden writes: > > > > def powers_of_2_in(n): > > bc = (n ^ (n - 1)).bit_count() - 1 > > return bc, n >> bc > > That's pretty fancy and likely the fastest. It might be the fastest but it depends how big you expect n to be and how many trailing zeros you expect. If n is a very large integer then this does three large integer operations in (n^(n-1)).bit_count(). They are relatively fast operations but all linear in bit size. By contrast a check like n & 1 is O(1) and half the time proves that no further steps are necessary. The mpmath library needs this exact operation and is generally intended for the large n case so I checked how it is implemented there: https://github.com/mpmath/mpmath/blob/f13ea4dc925d522062ac734bd19a0a3cc23f9c04/mpmath/libmp/libmpf.py#L160-L177 That code is: # Strip trailing bits if not man & 1: t = trailtable[man & 255] if not t: while not man & 255: man >>= 8 exp += 8 bc -= 8 t = trailtable[man & 255] man >>= t exp += t bc -= t The trailtable variable is a pre-initialised list of shifts needed to remove zeros from an 8-bit integer. The bc variable here is just bc=man.bit_length() which is redundant but this code predates the addition of the int.bit_length() method. In principle this could use a large number of man>>=8 shifts which would potentially be quadratic in the bit size of man. In practice the probability of hitting the worst case is very low so the code is instead optimised for the expected common case of large man with few trailing zeros. -- Oscar -- https://mail.python.org/mailman/listinfo/python-list
Re: on writing a number as 2^s * q, where q is odd
Julieta Shem ha scritto: jak writes: [...] --8<---cut here---start->8--- def powers_of_2_in(n): if remainder(n, 2) != 0: return 0, n else: s, r = powers_of_2_in(n // 2) return 1 + s, r --8<---cut here---end--->8--- for n = 0 your function get stack overflow That's right. Let's say ``assert n > 0'' before we say ``if''. ...or just: ... if n == 0 or remainder(n, 2) != 0: ... -- https://mail.python.org/mailman/listinfo/python-list
Re: on writing a number as 2^s * q, where q is odd
jak writes: [...] >> --8<---cut here---start->8--- >> def powers_of_2_in(n): >>if remainder(n, 2) != 0: >> return 0, n >>else: >> s, r = powers_of_2_in(n // 2) >> return 1 + s, r >> --8<---cut here---end--->8--- > > for n = 0 your function get stack overflow That's right. Let's say ``assert n > 0'' before we say ``if''. -- https://mail.python.org/mailman/listinfo/python-list
Re: on writing a number as 2^s * q, where q is odd
Julieta Shem ha scritto: Alan Bawden writes: jak writes: Alan Bawden ha scritto: > Julieta Shem writes: > > How would you write this procedure? > def powers_of_2_in(n): > ... > > def powers_of_2_in(n): > return (n ^ (n - 1)).bit_count() - 1 > Great solution, unfortunately the return value is not a tuple as in the OP version. Maybe in this way? def powers_of_2_inB(n): bc = (n ^ (n - 1)).bit_count() - 1 return bc, int(n / (1 << bc)) Good point. I overlooked that. I should have written: def powers_of_2_in(n): bc = (n ^ (n - 1)).bit_count() - 1 return bc, n >> bc That's pretty fancy and likely the fastest. I was pretty happy with a recursive version. If I'm not mistaken, nobody has offered a recursive version so far. It's my favorite actually because it seems to be the clearest one. --8<---cut here---start->8--- def powers_of_2_in(n): if remainder(n, 2) != 0: return 0, n else: s, r = powers_of_2_in(n // 2) return 1 + s, r --8<---cut here---end--->8--- for n = 0 your function get stack overflow -- https://mail.python.org/mailman/listinfo/python-list
Re: on writing a number as 2^s * q, where q is odd
Alan Bawden writes: > jak writes: > >Alan Bawden ha scritto: >> Julieta Shem writes: >> >> How would you write this procedure? >> def powers_of_2_in(n): >> ... >> >> def powers_of_2_in(n): >> return (n ^ (n - 1)).bit_count() - 1 >> > >Great solution, unfortunately the return value is not a tuple as in the >OP version. Maybe in this way? > >def powers_of_2_inB(n): >bc = (n ^ (n - 1)).bit_count() - 1 >return bc, int(n / (1 << bc)) > > Good point. I overlooked that. I should have written: > > def powers_of_2_in(n): > bc = (n ^ (n - 1)).bit_count() - 1 > return bc, n >> bc That's pretty fancy and likely the fastest. I was pretty happy with a recursive version. If I'm not mistaken, nobody has offered a recursive version so far. It's my favorite actually because it seems to be the clearest one. --8<---cut here---start->8--- def powers_of_2_in(n): if remainder(n, 2) != 0: return 0, n else: s, r = powers_of_2_in(n // 2) return 1 + s, r --8<---cut here---end--->8--- -- https://mail.python.org/mailman/listinfo/python-list
