Mailman 3 November 2023 - Python-list

>>> %matplotlib inline results in SyntaxError: invalid syntax
by Mike S Dec. 25, 2023

Dec. 25, 2023

I have installed Python 3.4.4 on XPSP3 and am trying to work my way through this tutorial. A Complete Tutorial on Ridge and Lasso Regression in Python http://www.analyticsvidhya.com/blog/2016/01/complete-tutorial-ridge-lasso-r… In Step 2 "Why Penalize the Magnitude of Coefficients?" we are shown this code: #Importing libraries. The same will be used throughout the article. import numpy as np import pandas as pd import random import matplotlib.pyplot as plt %matplotlib inline I get an error on the last line. I am running this code in Idle Python 3.4.4 Shell... Python 3.4.4 (v3.4.4:737efcadf5a6, Dec 20 2015, 19:28:18) [MSC v.1600 32 bit (Intel)] on win32 Type "copyright", "credits" or "license()" for more information. >>> import numpy as np >>> import pandas as pd >>> import random >>> import matplotlib.pyplot as plt >>> %matplotlib inline SyntaxError: invalid syntax What am I doing wrong? Suggested reading? TIA, Mike

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on writing a number as 2^s * q, where q is odd
by Julieta Shem Dec. 3, 2023

Dec. 3, 2023

How would you write this procedure? --8<---------------cut here---------------start------------->8--- def powers_of_2_in(n): s = 0 while "I still find factors of 2 in n...": q, r = divmod(n, 2) if r == 0: s = s + 1 n = n // 2 else: return s, n --8<---------------cut here---------------end--------------->8---

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Re: on writing a number as 2^s * q, where q is odd
by jak Nov. 30, 2023

Nov. 30, 2023

Dom Grigonis ha scritto: > def powers_of_2_in(n): > s = 0 > while n % 2 == 0: > s += 1 > n = n // 2 > return s, n > Good solution, unfortunately if the input data is zero, the function never ends. >> On 30 Nov 2023, at 02:44, Julieta Shem via Python-list <python-list(a)python.org> wrote: >> >> How would you write this procedure? >> >> --8<---------------cut here---------------start------------->8--- >> def powers_of_2_in(n): >> s = 0 >> while "I still find factors of 2 in n...": >> q, r = divmod(n, 2) >> if r == 0: >> s = s + 1 >> n = n // 2 >> else: >> return s, n >> --8<---------------cut here---------------end--------------->8--- >> -- >> https://mail.python.org/mailman/listinfo/python-list >

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Accessing configuration across multiple modules
by Loris Bennett Nov. 28, 2023

Nov. 28, 2023

Hi, I am using Typer to create a command-line program with multiple levels of subcommands, so a typical call might look like mytool --config-file=~/test/mytool.conf serviceXYZ list people In the top-level mytool.main, I evaluate the option '--config-file' and read the config file to initialize the logging. This works fine. However, in the module which lists people, namely mytool.serviceXYZ.cli_people I need to set up a connection to an LDAP server in order to actually read the data. If the LDAP connection details are also in the config file, what is the best way of making them accessible at the point where the object wrapping the LDAP server is initialized? I found this a suggestion here which involves creating a separate module for the configuration and then importing it https://codereview.stackexchange.com/questions/269550/python-share-global-v… I think I could probably get that to work, but are there any better alternatives? Cheers, Loris -- This signature is currently under constuction.

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Re: Printing dict value for possibly undefined key
by Loris Bennett Nov. 28, 2023

Nov. 28, 2023

DL Neil <PythonList(a)DancesWithMice.info> writes: > On 11/25/2023 3:31 AM, Loris Bennett via Python-list wrote: >> Hi, >> I want to print some records from a database table where one of the >> fields contains a JSON string which is read into a dict. I am doing >> something like >> print(f"{id} {d['foo']} {d['bar']}") >> However, the dict does not always have the same keys, so d['foo'] or >> d['bar'] may be undefined. I can obviously do something like >> if not 'foo' in d: >> d['foo']="NULL" >> if not 'bar' in d: >> d['bar']="NULL" >> print(f"{id} {d['foo']} {d['bar']}") >> Is there any more compact way of achieving the same thing? > > > What does "the dict does not always have the same keys" mean? > > a) there are two (or...) keys, but some records don't include both; > > b) there may be keys other than 'foo' and 'bar' which not-known in-advance; > > c) something else. Sorry for being unclear. There is either 'foo' or 'bar' or both, plus some other keys which are always present. > As mentioned, dict.get() solves one of these. > > Otherwise, there are dict methods which collect/reveal all the keys, > all the values, or both - dict.keys(), .values(), .items(), resp. That is a also a good point. I had forgotten about dict.keys() and dict.values(), and hadn't been aware of dict.items(). Cheers, Loris -- This signature is currently under constuction.

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Printing dict value for possibly undefined key
by Loris Bennett Nov. 28, 2023

Nov. 28, 2023

Hi, I want to print some records from a database table where one of the fields contains a JSON string which is read into a dict. I am doing something like print(f"{id} {d['foo']} {d['bar']}") However, the dict does not always have the same keys, so d['foo'] or d['bar'] may be undefined. I can obviously do something like if not 'foo' in d: d['foo']="NULL" if not 'bar' in d: d['bar']="NULL" print(f"{id} {d['foo']} {d['bar']}") Is there any more compact way of achieving the same thing? Cheers, Loris -- This signature is currently under constuction.

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argparse argument post-processing
by Dom Grigonis Nov. 27, 2023

Nov. 27, 2023

Hi all, I have a situation, maybe someone can give some insight. Say I want to have input which is comma separated array (e.g. paths='path1,path2,path3') and convert it to the desired output - list: import argparse parser = argparse.ArgumentParser() parser.add_argument('paths', type=lambda x: list(filter(str.strip, x.split(',')))) So far so good. But this is just an example of what sort of solution I am after. --------- Now the second case. I want input to be space separated array - bash array. And I want space-separated string returned. My current approach is: import argparse parser = argparse.ArgumentParser() parser.add_argument('paths', nargs='+') args = parser.parse_args() paths = ' '.join(args.paths) But what I am looking for is a way to do this, which is intrinsic to `argparse` module. Reason being I have a fair amount of such cases and I don’t want to do post-processing, where post-post-processing happens (after `parser.parse_args()`). I have tried overloading `parse_args` with post-processor arguments, and that seemed fine, but it stopped working when I had sub-parsers, which are defined in different modules and do not call `parse_args` themselves. Any ideas appreciated, Regards, DG

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Re: Context without manager
by Piergiorgio Sartor Nov. 27, 2023

Nov. 27, 2023

On 26/11/2023 18.50, Dieter Maurer wrote: > Piergiorgio Sartor wrote at 2023-11-25 22:15 +0100: >> ... >> Apparently, the "with" context manager is not usable >> in classes, at least not with __init__() & co. > > You can use `with` in classes -- with any context manager. > However, you would usually not use `with` with a file you have opened > in `__init__`. > > If a class defines `__enter__` and `__exit__` (i.e. > the "cntext manager protocol"), then its instances > can be used with the `with` statement. > > The important use case for a context manager is the > situation: > set up a context (--> method `__enter__`) > perform some operations in this context (--> body of `with` statement) > tear down the context (--> method `__exit__`). > If you do not have this case (e.g. usually if you open the file > in a class's `__init__`), you do not use a context manager. Very clear, but what if the class is *not* "open()", but something else _requiring_ using "with"? How to do this in a "__init__()" of a class? In other words, what if "open()" could *only* be used with "with" and not just by assigning "fp = open()"? The problem is I've some SDK of some device which provides context manager *only* classes. I *cannot* do: device = device_open(...) device.do_something() device.close() I *must* do: with device_open() as device: device.do_something() Nevertheless, I _need_ to have a class where the device is opened in the __init__() and used in some methods. Any ideas? bye, -- piergiorgio

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Context without manager
by Richard Damon Nov. 27, 2023

Nov. 27, 2023

Read the Fine context manager documentation. What “with with_expression as var” does is effectively: ob = with_expression var = ob.__enter__() And then at the end of the with, does a ob.__exit__() (With some parameters to __exit__, that could just be None, None, None for the simplest case). Note, YOUR program must now make sure that the __exit__ function is called, and handle any exceptions that got thrown, and that ob and var are put somewhere you can access them at that later time. > On Nov 27, 2023, at 12:24 PM, Piergiorgio Sartor via Python-list <python-list(a)python.org> wrote: > > On 26/11/2023 18.50, Dieter Maurer wrote: >> Piergiorgio Sartor wrote at 2023-11-25 22:15 +0100: >>> ... >>> Apparently, the "with" context manager is not usable >>> in classes, at least not with __init__() & co. >> You can use `with` in classes -- with any context manager. >> However, you would usually not use `with` with a file you have opened >> in `__init__`. >> If a class defines `__enter__` and `__exit__` (i.e. >> the "cntext manager protocol"), then its instances >> can be used with the `with` statement. >> The important use case for a context manager is the >> situation: >> set up a context (--> method `__enter__`) >> perform some operations in this context (--> body of `with` statement) >> tear down the context (--> method `__exit__`). >> If you do not have this case (e.g. usually if you open the file >> in a class's `__init__`), you do not use a context manager. > > Very clear, but what if the class is *not* "open()", > but something else _requiring_ using "with"? > How to do this in a "__init__()" of a class? > > In other words, what if "open()" could *only* be used > with "with" and not just by assigning "fp = open()"? > > The problem is I've some SDK of some device which > provides context manager *only* classes. > > I *cannot* do: > > device = device_open(...) > device.do_something() > device.close() > > I *must* do: > > with device_open() as device: > device.do_something() > > Nevertheless, I _need_ to have a class > where the device is opened in the __init__() > and used in some methods. > > Any ideas? > > bye, > > -- > > piergiorgio > > -- > https://mail.python.org/mailman/listinfo/python-list

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Re: Context without manager
by Richard Damon Nov. 27, 2023

Nov. 27, 2023

Read the Fine context manager documentation. What “with with_expression as var” does is effectively: ob = with_expression var = ob.__enter__() And then at the end of the with, does a ob.__exit__() (With some parameters to __exit__, that could just be None, None, None for the simplest case). Note, YOUR program must now make sure that the __exit__ function is called, and handle any exceptions that got thrown, and that ob and var are put somewhere you can access them at that later time. > On Nov 27, 2023, at 12:24 PM, Piergiorgio Sartor via Python-list <python-list(a)python.org> wrote: > > On 26/11/2023 18.50, Dieter Maurer wrote: >> Piergiorgio Sartor wrote at 2023-11-25 22:15 +0100: >>> ... >>> Apparently, the "with" context manager is not usable >>> in classes, at least not with __init__() & co. >> You can use `with` in classes -- with any context manager. >> However, you would usually not use `with` with a file you have opened >> in `__init__`. >> If a class defines `__enter__` and `__exit__` (i.e. >> the "cntext manager protocol"), then its instances >> can be used with the `with` statement. >> The important use case for a context manager is the >> situation: >> set up a context (--> method `__enter__`) >> perform some operations in this context (--> body of `with` statement) >> tear down the context (--> method `__exit__`). >> If you do not have this case (e.g. usually if you open the file >> in a class's `__init__`), you do not use a context manager. > > Very clear, but what if the class is *not* "open()", > but something else _requiring_ using "with"? > How to do this in a "__init__()" of a class? > > In other words, what if "open()" could *only* be used > with "with" and not just by assigning "fp = open()"? > > The problem is I've some SDK of some device which > provides context manager *only* classes. > > I *cannot* do: > > device = device_open(...) > device.do_something() > device.close() > > I *must* do: > > with device_open() as device: > device.do_something() > > Nevertheless, I _need_ to have a class > where the device is opened in the __init__() > and used in some methods. > > Any ideas? > > bye, > > -- > > piergiorgio > > -- > https://mail.python.org/mailman/listinfo/python-list

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