🌐 Mailman 3 March 2024 - Python-list - python.org

Trying to use pyinstaller under python 3.11, and, recently started receiving error message about specific module/distribution
by Jacob Kruger April 4, 2024

April 4, 2024

This started happening this past week, and, while it's worked fine in the past, the moment I try to launch the pyinstaller process at all, to generate compiled output, or even if just launch it with no command line options, I receive the following error message: pkg_resources.DistributionNotFound: The 'altgraph' distribution was not found and is required by the application The full contents of the output string when I even try to just launch pyinstaller with no commands/arguments is the following: Traceback (most recent call last): File "<frozen runpy>", line 198, in _run_module_as_main File "<frozen runpy>", line 88, in _run_code File "C:\pythonScripts\monitoring_nssm\venv\Scripts\pyinstaller.exe\__main__.py", line 7, in <module> File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\__main__.py", line 228, in _console_script_run run() File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\__main__.py", line 170, in run parser = generate_parser() ^^^^^^^^^^^^^^^^^ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\__main__.py", line 136, in generate_parser import PyInstaller.building.build_main File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\building\build_main.py", line 28, in <module> from PyInstaller.building.api import COLLECT, EXE, MERGE, PYZ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\building\api.py", line 32, in <module> from PyInstaller.building.splash import Splash # argument type validation in EXE ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\building\splash.py", line 23, in <module> from PyInstaller.depend import bindepend File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\depend\bindepend.py", line 25, in <module> from PyInstaller.depend import dylib, utils File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\depend\utils.py", line 31, in <module> from PyInstaller.lib.modulegraph import modulegraph File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\PyInstaller\lib\modulegraph\modulegraph.py", line 34, in <module> from altgraph.ObjectGraph import ObjectGraph File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\altgraph\__init__.py", line 144, in <module> __version__ = pkg_resources.require("altgraph")[0].version ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\pkg_resources\__init__.py", line 952, in require needed = self.resolve(parse_requirements(requirements)) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\pkg_resources\__init__.py", line 813, in resolve dist = self._resolve_dist( ^^^^^^^^^^^^^^^^^^^ File "C:\pythonScripts\monitoring_nssm\venv\Lib\site-packages\pkg_resources\__init__.py", line 854, in _resolve_dist raise DistributionNotFound(req, requirers) pkg_resources.DistributionNotFound: The 'altgraph' distribution was not found and is required by the application # ---end of output--- I have tried completely removing python's installation, and, reinstalling it, but, same issue more or less immediately. If I freeze pip's installed list within this specific virtual environment, it lists the following: altgraph==0.17.4 packaging==24.0 pefile==2023.2.7 pyinstaller==6.5.0 pyinstaller-hooks-contrib==2024.3 pywin32-ctypes==0.2.2 # ---end of requirements.txt--- And, if, just for testing, I launch python interpreter, and, ask it to import altgraph, it provides the same last line of error output? If relevant, running with python 3.11.8, under windows 11 64-bit, and, can't think of anything that specifically occurred/changed this past week, besides normal things like windows updates, etc., but, don't really think that's likely to be relevant, unless something to do with pywin32 has caused an issue? Should I try installing python 3.12 version instead and see if it changes? Also, if relevant, while running under latest up-to-date version of windows 11 64 bit, have in any case enabled case-sensitivity on the folder I store all my python code in, just in case. TIA -- Jacob Kruger +2782 413 4791 "Resistance is futile!...Acceptance is versatile..."

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xkcd.com/353 ( Flying with Python )
by HenHanna April 3, 2024

April 3, 2024

https://xkcd.com/353/ ( Flying with Python ) https://xkcd.com/1306/ what does SIGIL mean? Other xkcd that you like?

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A missing iterator on itertools module?
by ast April 3, 2024

April 3, 2024

Hello Suppose I have these 3 strings: s1 = "AZERTY" s2 = "QSDFGH" s3 = "WXCVBN" and I need an itertor who delivers A Q W Z S C E D C ... I didn't found anything in itertools to do the job. So I came up with this solution: list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN"))) ['A', 'Q', 'W', 'Z', 'S', 'X', 'E', 'D', 'C', 'R', 'F', 'V', 'T', 'G', 'B', 'Y', 'H', 'N'] Do you havbe a neat solution ?

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Making 'compiled' modules work with multiple python versions on Linux
by Olivier B. April 2, 2024

April 2, 2024

I have a python module that includes some C++ code that links with the Python C API I have now modified the c++ code so that it only uses the Limited API, and linked with python3.lib instead of python311.lib. I can now use that python module with different python versions on Windows But on Linux, it seems that linking to libpython3.so instead of libpython3.11.so.1.0 does not have the same effect, and results in many unresolved python symbols at link time Is this functionality only available on Windows?

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Can you help me with this memoization simple example?
by marc nicole March 31, 2024

March 31, 2024

I am creating a memoization example with a function that adds up / averages the elements of an array and compares it with the cached ones to retrieve them in case they are already stored. In addition, I want to store only if the result of the function differs considerably (passes a threshold e.g. 500000 below). I created an example using a decorator to do so, the results using the decorator is slightly faster than without the memoization which is OK, but is the logic of the decorator correct ? anybody can tell me ? My code is attached below: import time def memoize(f): cache = {} def g(*args): if args[1] == "avg": sum_key_arr = sum(list(args[0])) / len(list(args[0])) elif args[1] == "sum": sum_key_arr = sum(list(args[0])) if sum_key_arr not in cache: for ( key, value, ) in ( cache.items() ): # key in dict cannot be an array so I use the sum of the array as the key if ( abs(sum_key_arr - key) <= 500000 ): # threshold is great here so that all values are approximated! # print('approximated') return cache[key] else: # print('not approximated') cache[sum_key_arr] = f(args[0], args[1]) return cache[sum_key_arr] return g @memoize def aggregate(dict_list_arr, operation): if operation == "avg": return sum(list(dict_list_arr)) / len(list(dict_list_arr)) if operation == "sum": return sum(list(dict_list_arr)) return None t = time.time() for i in range(200, 15000): res = aggregate(list(range(i)), "avg") elapsed = time.time() - t print(res) print(elapsed)

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Re: A missing iterator on itertools module?
by Mark Bourne March 29, 2024

March 29, 2024

Stefan Ram wrote: > ast <none(a)none.fr> wrote or quoted: >> Why did you renamed itertools as _itertools ? > > Assume I have a module A.py: > > import math > def f(): pass > > . Assume I have an additional module B.py: > > import A > > . Now, when I'm editing "B.py" in IDLE and type "A.", IIRC > IDLE will offer me two possible completions: "A.math" and > "A.f". The "A.math" makes no sense to me. `import math` imports the `math` module and binds it to `math` in the global namespace of the `A` module. Since it doesn't have a leading underscore, by default it's considered to be a public attribute of the `A` module, and IDLE is offering all the public attributes of the `A` module for completion. > I want it to go > away. Therefore, I rewrite A.py as: > > import math as _math > def f(): pass > > . Now, Idle will only offer the completion "A.f". > > So, I sometimes use this "import math as _math" style. But then, > it is simpler for me to /always/ use this style; after all: you > can't know whether someone eventually will import your module! You can explicitly declare the public interface of a module by defining `__all__`, listing the names which should be considered part of the module's public interface; see: - https://docs.python.org/3/reference/simple_stmts.html#the-import-statement - https://peps.python.org/pep-0008/#public-and-internal-interfaces Although `from A import *` is generally discouraged, if `A` defines `__all__` then only the names listed in `__all__` are bound in the importing module's namespace. Otherwise, all names from `A` which don't have a leading underscore are considered to be public and bound in the importing module. I don't use IDLE, but it may be that it also uses `__all__` to determine a module's public API. In that case, setting `__all__ = ["f"]` in `A` should prevent it from offering `math` as a completion (nor any other name that's not in the `__all__` list). -- Mark.

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Re: A missing iterator on itertools module?
by ast March 28, 2024

March 28, 2024

Le 28/03/2024 à 18:07, Stefan Ram a écrit : > ast <none(a)none.fr> wrote or quoted: >> s1 = "AZERTY" >> s2 = "QSDFGH" >> s3 = "WXCVBN" >> and I need an itertor who delivers >> A Q W Z S C E D C ... >> I didn't found anything in itertools to do the job. >> So I came up with this solution: >> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN"))) > > Maybe you meant "zip(s1,s2,s3)" as the definition of s1, s2, > and s3 otherwise would not be required. Also the "list" is not > necessary because "chain.from_iterable" already is an iterable. > You could also use "*" instead of "list" to print it. So, > > import itertools as _itertools > s =[ "AZERTY", "QSDFGH", "WXCVBN" ] > print( *_itertools.chain.from_iterable( zip( *s ))) > > . But these are only minor nitpicks; you have found a nice solution! Why did you renamed itertools as _itertools ?

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RE: Popping key causes dict derived from object to revert to object (3/25/2024)
by Dr. F. M. (Mike) Covington March 28, 2024

March 28, 2024

Hello, all. I guess I'm just not seeing this dictionary pop() problem you're having. What version of Python are you using? Is there something I'm missing? If I've made an obvious oversight, please forgive me and explain. Thanks! PyCharm (2023.1.1) REPL running Python 3.12: Mike ---------------------- F. M. (Mike) Covington, PhD, PCPP, MCT Automation Consultant, Developer, Instructor

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Re: Popping key causes dict derived from object to revert to object
by Loris Bennett March 25, 2024

March 25, 2024

Grant Edwards <grant.b.edwards(a)gmail.com> writes: > On 2024-03-22, Loris Bennett via Python-list <python-list(a)python.org> wrote: > >> Yes, I was mistakenly thinking that the popping the element would >> leave me with the dict minus the popped key-value pair. > > It does. Indeed, but I was thinking in the context of dict_list = [d.pop('a') for d in dict_list] and incorrectly expecting to get a list of 'd' without key 'a', instead of a list of the 'd['a]'. >> Seem like there is no such function. > > Yes, there is. You can do that with either pop or del: > > >>> d = {'a':1, 'b':2, 'c':3} > >>> d > {'a': 1, 'b': 2, 'c': 3} > >>> d.pop('b') > 2 > >>> d > {'a': 1, 'c': 3} > > > >>> d = {'a':1, 'b':2, 'c':3} > >>> del d['b'] > >>> d > {'a': 1, 'c': 3} > > In both cases, you're left with the dict minus the key/value pair. > > In the first case, the deleted value printed by the REPL because it > was returned by the expression "d.pop('b')" (a method call). > > In the second case is no value shown by the REPL because "del d['b']" > is a statement not an expression. Thanks for pointing out 'del'. My main problem, however, was failing to realise that the list comprehension is populated by the return value of the 'pop', not the popped dict. Cheers, Loris -- This signature is currently under constuction.

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Popping key causes dict derived from object to revert to object
by Loris Bennett March 25, 2024

March 25, 2024

Hi, I am using SQLAlchemy to extract some rows from a table of 'events'. >From the call to the DB I get a list of objects of the type sqlalchemy.orm.state.InstanceState I would like to print these rows to the terminal using the 'tabulate' package, the documentation for which says The module provides just one function, tabulate, which takes a list of lists or another tabular data type as the first argument, and outputs a nicely formatted plain-text table So as I understand it, I need to convert the InstanceState-objects to, say, dicts, in order to print them. However I also want to remove one of the keys from the output and assumed I could just pop it off each event dict, thus: event_dicts = [vars(e) for e in events] print(type(event_dicts[0])) event_dicts = [e.pop('_sa_instance_state', None) for e in event_dicts] print(type(event_dicts[0])) However, this prints <class 'dict'> <class 'sqlalchemy.orm.state.InstanceState'> If I comment out the third line, which pops the unwanted key, I get <class 'dict'> <class 'dict'> Why does popping one of the keys cause the elements of the list to revert back to their original class? Cheers, Loris -- This signature is currently under constuction.

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