alcove | PyTorch implementation of ALCOVE | Machine Learning library
kandi X-RAY | alcove Summary
kandi X-RAY | alcove Summary
Kruschke, J. K. (1992). ALCOVE: an exemplar-based connectionist model of category learning. Psychological Review, 99(1), 22.
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Top functions reviewed by kandi - BETA
- Load SHJ images from images
- Extracts the features from the given image
- Load SHJ data from text files
- Visualize training image
- Process SHJ images
- Return the coding of the given loss_type
- Train a model on a SHJ problem
- Evaluate function
- Update weights using batch SGD
- Compute the similarity gradient of the similarity function
- Loads SHJ data from SHJ
- Return the coding for a label
alcove Key Features
alcove Examples and Code Snippets
Community Discussions
Trending Discussions on alcove
QUESTION
ANSWER
Answered 2022-Feb-16 at 13:10Actually, when Filter update data, is just the state, so CarCard don't know that something change. You need put the dataState and the setfunction on Home component and provide the setfucntion callback and dataState to your Filter component and provide the dataState to CarCard.
QUESTION
I have data which looks like:
...ANSWER
Answered 2021-Jul-16 at 13:16You can achieve this by using replace
and fill
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QUESTION
I'm using MariaDB and PHPMyAdmin, but my code is all neutral so happy to switch to Postgres or whatever, but this should be straight forward enough. I haven't designed anything yet, just after best approaches.
I have 3 database tables that describes a premises. Let's say a hotel.
This theoretical hotel has multiple venues - 2 restaurants and a bar. Each of those has a few different rooms/areas. In these rooms are tables that customers can sit at.
In SQL, I imagine the tables would look like this
Venues Venue ID Venue Name 1 Restaurant 1 2 Restaurant 2 3 Bar Rooms Room ID Room Name Parent Venue (foreign key) 1 Patio 1 2 Function Room 1 3 Alcove 3 4 Private Dining 2 Tables Table ID Table Name Parent Room (foreign key) 1 Table 1 1 2 Table 2 1 3 Table 3 1 4 Table 4 2 5 Table 1 3 6 Table 2 3 7 Table 3 3 8 Table 4 3 9 Table 1 4 10 Table 2 4 11 Table 3 4I hope that data is correct :p
What I want to do is define a relationship whereas it's impossible to add a Table Name if it already exists in that venue. It doesn't matter what room the table is in.
E.g if I was to add another "Table 4", it would succeed in being entered if it was entered into Room 4, as Room 4 belongs to Restaurant 2, which does not already have a "Table 4". However if it was entered into any other room, it would fail as Restaurant 1 and Bar already have a "Table 4" in one of their rooms.
Now in the server side code this is fairly easy to check as I can do multiple queries or joins or a myriad of other ways, however I was wondering how to do this in SQL/PhpMyAdmin directly. I'm having a bit of trouble finding my way around MyAdmin.
Cheers
...ANSWER
Answered 2020-Dec-14 at 14:59My recommendation is to redundantly include the parent venue in the tables table. So tables
would have the additional column:
QUESTION
I have a dataframe like the following:
...ANSWER
Answered 2020-Sep-04 at 11:49Wouldn't it just be:
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Install alcove
You can use alcove 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.
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