python-3.x – Row Coding

Django : Can’t import ‘module’. Check that module AppConfig.name is correct

by Tarik

According to the documentation, AppConfig.name is a full python path to the application. AppConfig.name Full Python path to the application, e.g. ‘django.contrib.admin’. This attribute defines which application the configuration applies to. It must be set in all AppConfig subclasses. It must be unique across a Django project. https://docs.djangoproject.com/en/2.2/ref/applications/#django.apps.AppConfig.name Try this: class CoreConfig(AppConfig): name=”compfactu.core”

Name not defined in type annotation [duplicate]

by Tarik

You have a forward declaration; functions (to be bound as methods) are created before the class is, so the name Vector doesn’t yet exist. Only when all of the class body has been executed, can Python create the class object and bind the name Vector to it. Simply use a string with the name instead: …

How to upload new versions of project to PyPI with twine?

by Tarik

PyPI does not allow for the reuse of distribution filenames (project name + version number + distribution type). This ensures that a given distribution for a given release for a given project will always resolve to the same file, and cannot be surreptitiously changed one day by the projects maintainer or a malicious party (it …

Conditionally fill column values based on another columns value in pandas

by Tarik

You probably want to do df[‘Normalized’] = np.where(df[‘Currency’] == ‘$’, df[‘Budget’] * 0.78125, df[‘Budget’])

How to make new anaconda env from yml file

by Tarik

conda env create allows an option –file for an environment file: conda env create –name envname –file=environments.yml

base_name argument not specified, and could not automatically determine the name from the viewset, as it does not have a .queryset attribute

by Tarik

In the latest DRF, you need to explicitly set base_name in your viewset url if you don’t have queryset defined. So, something like this should do good: router.register(r’my-model/’, MyModelView, basename=”MyModel”) See this: docs

During handling of the above exception, another exception occurred

by Tarik

Currently, you are having an issue with raising the ValueError exception inside another caught exception. The reasoning for this solution doesn’t make much sense to me, but if you change it. raise Exception(‘Invalid json: {}’.format(e)) To raise Exception(‘Invalid json: {}’.format(e)) from None Making your end code. with open(json_file) as j: try: json_config = json.load(j) except …

Python3 subprocess output

by Tarik

I suggest that you use subprocess.getoutput() as it does exactly what you want—run a command in a shell and get its string output (as opposed to byte string output). Then you can split on whitespace and grab the first element from the returned list of strings. Try this: import subprocess stdoutdata = subprocess.getoutput(“wc –lines /var/log/syslog”) …

Reading named command arguments

by Tarik

You can use the Optional Arguments like so. With this program: #!/usr/bin/env python3 import argparse, sys parser=argparse.ArgumentParser() parser.add_argument(“–bar”, help=”Do the bar option”) parser.add_argument(“–foo”, help=”Foo the program”) args=parser.parse_args() print(f”Args: {args}\nCommand Line: {sys.argv}\nfoo: {args.foo}”) print(f”Dict format: {vars(args)}”) Make it executable: $ chmod +x prog.py Then if you call it with: $ ./prog.py –bar=bar-val –foo foo-val It prints: …