partition | A fast and flexible framework for data reduction in R | Machine Learning library

 by   USCbiostats HTML Version: v0.1.3 License: Non-SPDX

kandi X-RAY | partition Summary

kandi X-RAY | partition Summary

partition is a HTML library typically used in Artificial Intelligence, Machine Learning applications. partition has no bugs, it has no vulnerabilities and it has low support. However partition has a Non-SPDX License. You can download it from GitHub.

partition is a fast and flexible framework for agglomerative partitioning. partition uses an approach called Direct-Measure-Reduce to create new variables that maintain the user-specified minimum level of information. Each reduced variable is also interpretable: the original variables map to one and only one variable in the reduced data set. partition is flexible, as well: how variables are selected to reduce, how information loss is measured, and the way data is reduced can all be customized.
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            kandi-support Support

              partition has a low active ecosystem.
              It has 16 star(s) with 1 fork(s). There are 4 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 0 open issues and 11 have been closed. On average issues are closed in 29 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of partition is v0.1.3

            kandi-Quality Quality

              partition has no bugs reported.

            kandi-Security Security

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

            kandi-License License

              partition 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.

            kandi-Reuse Reuse

              partition releases are available to install and integrate.
              Installation instructions, examples and code snippets are available.

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            partition Key Features

            No Key Features are available at this moment for partition.

            partition Examples and Code Snippets

            Adds a batched feature partition to a tensor .
            pythondot img1Lines of Code : 80dot img1License : Non-SPDX (Apache License 2.0)
            copy iconCopy
            def _add_batched_ragged_partition(rt, partition, tensor_dict, feature_key,
                                              validate, outer_splits=None):
              """Adds a batched ragged partition tensor to a batched ragged tensor.
            
              Args:
                rt: A RaggedTensor with sh  
            Plot partition boundary .
            pythondot img2Lines of Code : 50dot img2License : Permissive (MIT License)
            copy iconCopy
            def plot_partition_boundary(
                model, train_data, ax, resolution=100, colors=("b", "k", "r")
            ):
                """
                We can not get the optimum w of our kernel svm model which is different from linear
                svm.  For this reason, we generate randomly distribu  
            Load partition graphs from disk .
            pythondot img3Lines of Code : 49dot img3License : Non-SPDX (Apache License 2.0)
            copy iconCopy
            def _load_partition_graphs(self, client_partition_graphs, validate):
                """Load and process partition graphs.
            
                Load the graphs; parse the input and control input structure; obtain the
                device and op type of each node; remove the Copy and debu  

            Community Discussions

            QUESTION

            AWS DynamoDB Partition Key Design
            Asked 2021-Jun-15 at 18:09

            I read this answer, which clarified a lot of things, but I'm still confused about how I should go about designing my primary key.

            First off I want to clarify the idea of WCUs. I get that WCU is the write capacity of max 1kb per second. Does it mean that if writing a piece of data takes 0.25 seconds, I would need 4 of those to be billed 1 WCU? Or each time I write something it consumes 1 WCU, but I could also write X times within 1 second and still be billed 1 WCU?

            Usage

            I want to create a table that stores the form data for a set of gyms (95% will be waivers, the rest will be incidents reports). Most of the time, each forms will be accessed directly via its unique ID. I also want to query the forms by date, form, userId, etc..

            We can assume an average of 50k forms per gym

            Options

            • First option is straight forward: having the formId be the partition key. What I don't like about this option is that scan operations will always filter out 90% of the data (i.e. the forms from other gyms), which isn't good for RCUs.

            • Second option is that I would make the gymId the partition key, and add a sort key for the date, formId, userId. To implement this option I would need to know more about the implications of having 50k records on one partition key.

            • Third option is to have one table per gyms and have the formId as partition key. This seems to be like the best option for now, but I don't really like the idea of having a a large number of tables doing the same thing in my account.

            Is there another option? Which one of the three is better?

            Edit: I'm assuming another option would be SimpleDB?

            ...

            ANSWER

            Answered 2021-May-21 at 20:26

            For your PK design. What data does the app have when a user is going to look for a form? Does it have the GymID, userID, and formID? If so, make a compound key out of that for the PK perhaps? So your PK might look like:

            Source https://stackoverflow.com/questions/67628589

            QUESTION

            SLURM and Python multiprocessing pool on a cluster
            Asked 2021-Jun-15 at 13:42

            I am trying to run a simple parallel program on a SLURM cluster (4x raspberry Pi 3) but I have no success. I have been reading about it, but I just cannot get it to work. The problem is as follows:

            I have a Python program named remove_duplicates_in_scraped_data.py. This program is executed on a single node (node=1xraspberry pi) and inside the program there is a multiprocessing loop section that looks something like:

            ...

            ANSWER

            Answered 2021-Jun-15 at 06:17

            Pythons multiprocessing package is limited to shared memory parallelization. It spawns new processes that all have access to the main memory of a single machine.

            You cannot simply scale out such a software onto multiple nodes. As the different machines do not have a shared memory that they can access.

            To run your program on multiple nodes at once, you should have a look into MPI (Message Passing Interface). There is also a python package for that.

            Depending on your task, it may also be suitable to run the program 4 times (so one job per node) and have it work on a subset of the data. It is often the simpler approach, but not always possible.

            Source https://stackoverflow.com/questions/67975328

            QUESTION

            Is it safe to delete the cleaner-offset-checkpoint file to force the compaction?
            Asked 2021-Jun-15 at 13:24

            I need a way to force the compaction of the __consumer_offsets topic. In a test environment I tried to delete the file cleaner-offset-checkpoint and then kafka deleted many segments as you can see below. Is it safe to delete this file in a production environment?

            Before removing cleaner-offset-checkpoint:

            ...

            ANSWER

            Answered 2021-Jun-15 at 13:24

            cleaner-offset-checkpoint is in kafka logs directory. This file keeps the last cleaned offset of the topic partitions in the broker like below.

            Source https://stackoverflow.com/questions/67982650

            QUESTION

            How to have your own partition key while writing data in Cosmos DB
            Asked 2021-Jun-15 at 10:50

            I am using below code to write my content in Cosmos Db, however in Cosmos Db I see the Partition Key is automatically generated and it is for Id which I have kept as default. My requirement is to have my Own Partition Key . from my below json I would like TypeId to be my partition Key How can I do that in my below code?

            content is of JObject Type and is of below format {{

            ...

            ANSWER

            Answered 2021-Jun-15 at 10:49

            You can form a partition key by concatenating multiple property values into a single artificial partitionKey property.

            Please follow the steps given in below page: https://www.c-sharpcorner.com/article/understanding-partitioning-and-partition-key-in-azure-cosmos-db/

            Let me know if it helps.

            All the best!

            Source https://stackoverflow.com/questions/67984787

            QUESTION

            Get a list of all existing partitions in couchDb
            Asked 2021-Jun-15 at 10:05

            I have a partitioned CouchDB database. Is there any query to get a list of all partitions in a particular database? I have not found anything like that in CouchDb documentation.

            ...

            ANSWER

            Answered 2021-Jun-11 at 21:55

            There is no endpoint that just lists partitioned state for all db's, however the /_dbs_info endpoint is close enough with a little processing.

            Here is a naïve script I spun up using nano and nodejs 10. The script displays database names, prefixed with an asterisk (*) if the database is partitioned.

            Source https://stackoverflow.com/questions/67937832

            QUESTION

            Unable to replace item with chinese word in partition key header in Azure Cosmos DB
            Asked 2021-Jun-15 at 09:35

            I'm trying to use golang to do CURD operation in Azure Cosmos db using github.com/vippsas/go-cosmosdb package.

            Everything works fine except trying to Create、Replace documents with chinese character in the x-ms-documentdb-partitionkey.

            Document sample data, partition key is /method

            ...

            ANSWER

            Answered 2021-Jun-15 at 09:35

            Azure Cosmos db is only supporting Unicode or ASCII in x-ms-documentdb-partitionkey while github.com/vippsas/go-cosmosdb package is using json.Marshal which internally transforms Unicode to Chinese characters automatically.

            The only way to solve it is using English as partition key when creating documents.

            Source https://stackoverflow.com/questions/67845767

            QUESTION

            Is it possible to use this SWITCH PARTITION control option with Azure SQL Server?
            Asked 2021-Jun-15 at 06:44

            I'm doing some ETL, using the standard "Pre-Load" partition pattern: Load the data into a dated partition of a loading table, then SWITCH that partition into the live table.

            I found these options for the SWITCH command:

            ...

            ANSWER

            Answered 2021-Jun-15 at 06:44

            Looks the question was solved by @Larnu's comment, just add it as an answer to close the question.

            If you are using Azure SQL Database, then what the error is telling you is true. Azure SQL Databases are what are known as Partially Contained databases; things like their USER objects have their own Password and the LOGIN objects on the server aren't used for connections. The CONNECTION permission is a server level permission, and thus not supported in Azure SQL Databases.

            Source https://stackoverflow.com/questions/67935455

            QUESTION

            Why does the cte return the error that it does not exist?
            Asked 2021-Jun-14 at 22:04

            Here is my code

            ...

            ANSWER

            Answered 2021-Jun-14 at 21:50

            Create a CTE that returns for each Block_id the step of the first John.
            Then join the table to the CTE:

            Source https://stackoverflow.com/questions/67977325

            QUESTION

            For each product get the second customer who made a purchase
            Asked 2021-Jun-14 at 19:29

            Given a table:

            ...

            ANSWER

            Answered 2021-Jun-14 at 19:29

            You need to pair down the customers first so there is only one record per customer:

            Source https://stackoverflow.com/questions/67975843

            QUESTION

            SQL: How can I count unique instances grouped by client ordered by date?
            Asked 2021-Jun-14 at 15:06

            I have the following table in a Snowflake data warehouse:

            Client_ID Appointment_Date Store_ID Client_1 1/1/2021 Store_1 Client_2 1/1/2021 Store_1 Client_1 2/1/2021 Store_2 Client_2 2/1/2021 Store_1 Client_1 3/1/2021 Store_1 Client_2 3/1/2021 Store_1

            I need to be able to count the number of unique Store_ID for each Client_ID in order of Appointment_Date. Something like following is my desired output:

            Customer_ID Appointment_Date Store_ID Count_Different_Stores Client_1 1/1/2021 Store_1 1 Client_2 1/1/2021 Store_1 1 Client_1 2/1/2021 Store_2 2 Client_2 2/1/2021 Store_1 1 Client_1 3/1/2021 Store_1 2 Client_2 3/1/2021 Store_1 1

            Where I would be actively counting the number of distinct stores a client visits over time. I've tried:

            ...

            ANSWER

            Answered 2021-Jun-14 at 14:26

            If I understand correctly, you want a cumulative count(distinct) as a window function. Snowflake does not support that directly, but you can easily calculate it using row_number() and a cumulative sum:

            Source https://stackoverflow.com/questions/67971987

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

            Vulnerabilities

            No vulnerabilities reported

            Install partition

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