birch | Simple hierarchical configuration for Python packages

by shaypal5 Python Version: 0.0.35 License: Non-SPDX

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | birch Summary

birch is a Python library. birch has no bugs, it has no vulnerabilities, it has build file available and it has low support. However birch has a Non-SPDX License. You can install using 'pip install birch' or download it from GitHub, PyPI.

Simple hierarchical configuration for Python packages.

Support

Quality

Security

License

Reuse

Support

birch has a low active ecosystem.

It has 13 star(s) with 8 fork(s). There are 3 watchers for this library.

It had no major release in the last 12 months.

There are 0 open issues and 1 have been closed. On average issues are closed in 50 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of birch is 0.0.35

Quality

birch has 0 bugs and 0 code smells.

Security

birch has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

birch code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

birch has a Non-SPDX License.

Non-SPDX licenses can be open source with a non SPDX compliant license, or non open source licenses, and you need to review them closely before use.

Reuse

birch releases are available to install and integrate.

Deployable package is available in PyPI.

Build file is available. You can build the component from source.

Top functions reviewed by kandi - BETA

kandi has reviewed birch and discovered the below as its top functions. This is intended to give you an instant insight into birch implemented functionality, and help decide if they suit your requirements.

Return a dict containing the command - line arguments
Get value from configuration
Build a ConfigParser from root
Get a configuration value
Extract version information from VCS
Create a versioneer config file
Install versioneer
Return the dpath dpath for the given namespace
Return the dpath to the cache directory
Return the xpath of the xdg_cfg file for the given namespace
Return the path to the dpath
Extract version information from the VCS
Scans the setup py file and checks for missing values

Get all kandi verified functions for this library.

birch Key Features

No Key Features are available at this moment for birch.

birch Examples and Code Snippets

No Code Snippets are available at this moment for birch.

Community Discussions

Trending Discussions on birch

Encoding to JSON format is not encoding the toggled boolean value in Swift

Left join with count() and show matches separately within table

SQL join multiple tables with one column becoming two

Are there any verification methods how good cluster I got?

The variables are not changing the original variables inside a function

Shorten if-elif-elif...-else

react router v5 to v6 nested route not working

Creating combinations of 3 arrays using for of loops

Unnesting elements of large list into dataframe

Clustering on Python and Bokeh; select widget which allows user to change clustering algorithm

QUESTION

Encoding to JSON format is not encoding the toggled boolean value in Swift

Asked 2022-Apr-11 at 20:19

I am making an app that has information about different woods, herbs and spices, and a few other things. I am including the ability to save their favorite item to a favorites list, so I have a heart button that the user can press to add it to the favorites. Pressing the button toggles the isFavorite property of the item and then leaving the page calls a method that encodes the data to save it to the user's device. The problem that I am running into is that it is not encoding the updated value of the isFavorite property. It is still encoding the value as false, so the favorites list is not persisting after closing and reopening the app.

Here is my Wood.swift code, this file sets up the structure for Wood items. I also included the test data that I was using to make sure that it displayed properly in the Wood extension:

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ANSWER

Answered 2022-Apr-11 at 20:19

Your problem is that structs are value types in Swift. Essentially this means that the instance of Wood that you have in WoodsDetailView is not the same instance that is in your array in your model (WoodData); It is a copy (Technically, the copy is made as soon as you modify the isFavourite property).

In SwiftUI it is important to maintain separation of responsibilities between the view and the model.

Changing the favourite status of a Wood is something the view should ask the model to do.

This is where you have a second issue; In your detail view you are creating a separate instance of your model; You need to refer to a single instance.

You have a good start; you have put your model instance in the environment where views can access it.

First, change the detail view to remove the binding, refer to the model from the environment and ask the model to do the work:

Source https://stackoverflow.com/questions/71831369

QUESTION

Left join with count() and show matches separately within table

Asked 2022-Feb-05 at 15:39

I counted birds on different dates and areas. Some birds got a tracking ID. That gives my a table with this header. t1:

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ANSWER

Answered 2022-Feb-05 at 13:26

You should create one more grouping level for the birds without BIRD_TRACKING_ID and use conditional aggregation:

Source https://stackoverflow.com/questions/70998086

QUESTION

SQL join multiple tables with one column becoming two

Asked 2022-Jan-26 at 12:42

I have 3 tables that I want to join. One table has fields that need to be broken out of one column to become two columns. Here are sample tables.

Doors

DoorID DoorType 1 A 2 A 3 B 4 B 5 A 6 A 7 A

Parts

PartID DoorID Description MaterialID 1 1 Hinge 1 2 1 Hinge 1 3 2 Hinge 1 4 2 Hinge 1 5 3 Hinge 1 6 3 Hinge 1 7 1 Plate 3 8 1 Plate 3 9 2 Plate 3 10 2 Plate 3 11 3 Plate 3 12 3 Plate 3 13 4 Plate 3 14 4 Plate 3 15 5 Hinge 2 16 5 Hinge 2 17 Deck 33 18 Unfinished Left End 33 19 Partition 38 20 5 Plate 4 21 5 Plate 4

Materials

MaterialID Name 1 3/8 Hinge 2 5/8 Hinge 3 3/8 Plate 4 5/8 Plate 33 1/2 Birch Ply 38 3/4 Birch Ply

What I'd like to end up with is a count of each door type and what hinge and hinge plate is on each door type like the following. All of the A doors are counted together except the one that has no hinges. All of the B doors are counted together. Any part that's not a hinge or plate is ignored. Any material that's not a hinge or plate is ignored. Any door type that has no hinge or plate leaves those fields blank.

Door Count with Hinge Data:

Qty Hinge Plate 4 3/8 Hinge 3/8 Plate 2 5/8 Hinge 5/8 Plate 1

I've been beating my head against a wall on this for hours and am not getting anywhere. I'm very new to SQL as well. I have discovered that the application I'm using doesn't support CASE statements at all, but it can use IIF functions.

Any help would be much appreciated.

...

ANSWER

Answered 2022-Jan-26 at 12:42

You can try something like this

Source https://stackoverflow.com/questions/70858184

QUESTION

Are there any verification methods how good cluster I got?

Asked 2022-Jan-20 at 09:28

I have a task about clusterization in python. When I did this clusterization I need to check the result with business logic.

I dont see the pattern in solved clusters. Next, I decided to do post analysis with correlation. I take one cluster and calculate a correlation pairwise. In calculation I used whole feature unlike a clusterization when I used only 3. I got a high level of correlation from 0.99 to 1 in whole cluster. For me it means that algorithm watched the logic in cluster.

But, i did this clusterization to solved a problem with banks data (i wont to see the client's pattern like (issued amount > 50.000,age < 22, salary < 80.000 - this client, for instance bad)). And I cant see the business logic, for me it's random data.

With this description I have a question. How can i check the logic in the clusters except a simple self-checking ?

I think there are 2 reasons. First, my clusterization is bad and I need to write a new one. Second, the data is bad and I need to check data and do a post analysis

I did a BIRCH cluster with StandardScaler.

...

ANSWER

Answered 2022-Jan-20 at 09:28

All of verification methods are 'empirical'.

You can compare the different methods of clusterization and choose the best one.
The correlation comparison methods: a) If correlation approximately 1. You need to calculate a row's average and median. Next step you compare this two value and drop bads row. b) If corr are different in whole matrix. Calculate averages for all rows and compare the value and the mean average; choose god one like this 'value > mean(avg)'

Source https://stackoverflow.com/questions/70751515

QUESTION

The variables are not changing the original variables inside a function

Asked 2022-Jan-05 at 01:06

I have variables, say variables A outside the functions. In one function, mine_function, a random integer is added to the separate variables in variables A. But when I call the inv_function() after calling the mine_command(), the variables A stay 0. They were not added. What do I do?

P.S. this is from one file. The main file imports these functions to the main file to run.

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ANSWER

Answered 2022-Jan-05 at 01:06

This is due to ints being an immutable data type. When you modify them using the += operator, it replaces the entire variable, rather than modifying the existing one. This means that if you look at the variables inside the inventory tuple, they won't be changed, since they weren't updated.

You might want to consider using a dictionary data type to store your inventory, rather than a bunch of variables, since dictionaries are mutable.

Source https://stackoverflow.com/questions/70517817

QUESTION

Shorten if-elif-elif...-else

Asked 2021-Dec-31 at 08:40

I have a program, and one function is to sell an item that the user owns. It prompts the user to input the name (id) and amount, and it sells. But there are a lot of items the user can own, so there are lots of if else elif statements. How do I shorten this? (P.S. I am using Replit, and Replit currently has Python 3.8) Here is the sell function, for reference.

...

ANSWER

Answered 2021-Dec-31 at 08:23

You can reduce the branching a lot using dictionaries and objects for what you are doing. And it is the power of OOPs. Here is an example how you will remodel your code.

Source https://stackoverflow.com/questions/70540234

QUESTION

react router v5 to v6 nested route not working

Asked 2021-Dec-09 at 18:01

I've been trying to solve the following problem : I try to upgrade this Frontend Mentor project https://haydee75.github.io/galleria/ from React Router v5 to v6. I tried to replace the code between with :

...

ANSWER

Answered 2021-Dec-09 at 18:01

If I'm understanding your question/issue correctly, you want to render the Gallery and Paint components each on their own routes independently, and fix the slideshow linking from painting to painting. For this use the first routing snippet so they are independent routes and not nested.

Source https://stackoverflow.com/questions/70278436

QUESTION

Creating combinations of 3 arrays using for of loops

Asked 2021-Nov-06 at 21:45

I'm trying to list all combinations of strings from 3 arrays, and assign all of those combinations to separate variables. I'm able to do it with only two arrays, but not with three for a reason I don't know. This is the code:

...

ANSWER

Answered 2021-Nov-06 at 21:45

I rewrited a little bit your code. I think it will work right now

Source https://stackoverflow.com/questions/69868091

QUESTION

Unnesting elements of large list into dataframe

Asked 2021-Oct-26 at 14:44

I am currently working with a large dataset I retrieved from the crossref API in which I retrieved information on scientific papers based on a DOI search.

Currently the large list contains of ~3500 elements. Each of these elements is a list of their own consisting of the metadata 'meta', the actual relevant data 'data' and an irrelevant list 'facets'.

This is an example of two of the lists based on two DOI's:

...

ANSWER

Answered 2021-Oct-25 at 16:55

Like this? Note - it is better to include a Minimal reprex that includes a toy data set, rather than a snapshot of what you have. This way the question will likely get answers faster.

Source https://stackoverflow.com/questions/69711158

QUESTION

Clustering on Python and Bokeh; select widget which allows user to change clustering algorithm

Asked 2021-Aug-11 at 12:48

I am trying to build a feature in a Bokeh dashboard which allows the user to cluster data. I am using the following example as a template, here is the link:- Clustering in Bokeh example

Here is the code from this example:-

...

ANSWER

Answered 2021-Aug-11 at 12:48

I don't know sklearn but comparing both your examples I can see the following:

the Select is a Bokeh model which has value attribute of type string. So select.value is a string
the dbscan is an algorithm function

So when you do algorithm = dbscan you assign an algorithm function to your algorithm variable and when you do algorithm = select.value in your second example you assign just a string to it so it won't work because string doesn't have the fit() function. You should do something like this:

Source https://stackoverflow.com/questions/68726083

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install birch

You can install using 'pip install birch' or download it from GitHub, PyPI.
You can use birch like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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