# random-sampling | Java 8 for the problem of random | Machine Learning library

## kandi X-RAY | random-sampling Summary

## kandi X-RAY | random-sampling Summary

A collection of algorithms in Java 8 for the problem of random sampling with a reservoir. Reservoir sampling is a family of randomized algorithms for randomly choosing a sample of k items from a list S containing n items, where n is either a very large or unknown number. Typically n is large enough that the list doesn't fit into main memory. [1] In this context, the sample of k items will be referred to as sample and the list S as stream. This package distinguishes these algorithms into two main categories: the ones that assign a weight in each item of the source stream and the ones that don't. These will be referred to as weighted and unweighted random sampling algorithms respectively. In unweighted algorithms, each item in the stream has probability k/n in appearing in the sample. In weighted algorithms this probability depends on the extra parameter weight. Each algorithm may interpret this parameter in a different way, for example in [2] two possible interpretations are mentioned.

### Support

### Quality

### Security

### License

### Reuse

### Top functions reviewed by kandi - BETA

- Feeds an item into the stream
- Get the sample size
- Overrides the superclass method
- Get the sample size
- Feeds an item
- Feed the elements of an iterable
- Calls the superclass method
- Feed the elements of an iterable
- Synchronized
- Feed the elements of an iterable
- The main implementation of this algorithm
- Get the sample size
- Performs the final transformation
- Returns true if two Iterators are equal
- Returns the number of items in the algorithm
- Returns a set of all the characteristics of this Collector
- Returns the hashCode value for the object
- Returns a string representation of this object
- Folds a function to fold a value into one
- Run VitterZ sampling
- Run a test sample
- Command - line test
- Main method
- Generates a key for the given weight
- Generate a key for the given weight
- Checks if two collections contain the same elements
- Runs a sample
- Main method for testing
- Returns an unweighted selection of the population
- Returns a set of all the characteristics of this collector

## random-sampling Key Features

## random-sampling Examples and Code Snippets

## Community Discussions

Trending Discussions on random-sampling

QUESTION

I have a dataframe with several columns, and I create a new column which randomly samples a single value from either of the other columns. How can I trace back to tell which column the value came from?

I've seen the exact same question and solution here, but it's in python, and couldn't find an R equivalent.

Data 1 :: each row has different values across columns ...ANSWER

Answered 2020-Nov-10 at 11:18One option could be:

QUESTION

I am trying to randomly sample 50% of the data for each of the group following Stratified random sampling from data frame. A reproducible example using mtcars dataset in R looks like below. What I dont understand is, the sample index clearly shows a group of gear labeled as '5', but when the index is applied to the mtcars dataset, the sampled data mtcars2 does not contain any record from gear='5'. What went wrong? Thank you very much.

...ANSWER

Answered 2020-Jun-24 at 14:36I think the approach you've done creates a number `1:length(mtcars$gear)`

for each `gear`

group so you will have repeat row numbers for each group. Then, when you subset it isn't working, see in your output above you have row number `7`

in both `gear`

group `3`

and `4`

.

**Base R**

I would use `split`

first to split by gear:

QUESTION

This should be a simple extension of this question, but my result is not correct and I can't figure it out. I'd like the proportions in the table I'm drawing from to match the proportions of another table. I'd also like to have it stratified by two categories. I think it should be something like:

...ANSWER

Answered 2020-Feb-10 at 22:55I think your `rand()`

comparison is off:

QUESTION

Say I have a dataframe like so:

...ANSWER

Answered 2020-Jan-30 at 18:51`transform`

with `choice`

I forgo efficiency for readability. Note that I generate a random choice for each row but only pick the number I need to fill in the nulls. Theoretically, I can make it such that I only pick random numbers for those missing values.

QUESTION

I have been running a Rails site for a couple of years and some articles are being pulled from the DB based on a weight field. The data structure is:

...ANSWER

Answered 2019-Jul-18 at 16:36Your elasticsearch query is correct and you don't need scripts to perform what you want. It is just a problem with probabilities. For a short answer, replace the multiplier (i.e., `field_value_factor`

) for the weight of 50 by 40 and the multiplier for the weight of 25 by 30 and you will get the expected result.

Basically, the problem is that multiplying a random value by a weight is not producing a weighted distribution where the weight is the multiplier. The multiplier can be derived from the weight, but **there are not the same**.

I can give you an example with your case. For the weight 50, if the random value is above 0.5, it will necessarily have the highest score (0.5 * 50 >= 1 * 25). Since a value of 0.5 as a probability of 50%, you now for sure that the item with weight 50 will be returned at least half of the time.

But even if the random value for weight 50 is below 0.5, it can still be selected. In fact its probability to be selected in this case is 1/3.

I'm just a bit surprised by your result because its probability should be more like 66% (i.e., 50% + 50%/3) and the other probabilities should be around 16.5%. Maybe try to increase the number of runs to be sure.

Solution for any weight using`script_score`

You do not need to compute the multiplier with this solution but you must provide a range, e.g., `min_value`

and `max_value`

for each document. `max_value`

is the sum of `min_value`

and the document wight and `min_value`

is the cumulative sum of the weight of the previous documents.

If you have for example 4 documents with weights 5, 15, 30, 50, then the ranges could be :

- Documents with weight 5 : min_value = 0, max_value = 5
- Documents with weight 15 : min_value = 5, max_value = 5+15 = 20
- Documents with weight 30 : min_value = 20, max_value = 20+30 = 50
- Documents with weight 30 : min_value = 50, max_value = 50+50 = 100

The corresponding elasticsearch query is

QUESTION

I would like to do some Monte Carlo analysis in Haskell. I would like to be able to write code like this:

...ANSWER

Answered 2019-May-10 at 21:41Well, if you want to be able to write code like this:

QUESTION

Consider the dataframe containing N columns as shown below. Each entry is an 8-bit integer.

...ANSWER

Answered 2019-Apr-11 at 17:13Same logic like before when you do the sample , but here I convert between the binary and decimal twice, with unnesting , then join back the result

QUESTION

I have a **data frame**:

ANSWER

Answered 2018-Mar-04 at 19:04You can subset the data first by case `ID == 1`

. To ensure occurrence of 1s and 0s, we use `rep`

function and set `replace`

to False in `sample`

function.

Here's a solution.

QUESTION

I have a large dataset of size N, and want to get a (uniformly) random sample of size n. This question offers two possible solutions:

...ANSWER

Answered 2018-Feb-23 at 15:03I compared the two queries execution times using BigQuery standard SQL with the `natality`

sample dataset (137,826,763 rows) and getting a sample for `source_year`

column of size *n*. The queries are executed without using cached results.

Query1:

QUESTION

From a 10*10 raster I want to unselect for example 90 percent, that is, 10 percent remain visible. To do this I adapted this code, see below. But there is some variation in the resulting pixels (more then 10 or less then 10 pixels remain). Is there a possibility to set precision of random selection?

...ANSWER

Answered 2017-May-31 at 09:33Instead of your `runif`

line, use

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

## Vulnerabilities

No vulnerabilities reported

## Install random-sampling

You can use random-sampling like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the random-sampling component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

## Support

## Reuse Trending Solutions

Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items

Find more librariesStay Updated

Subscribe to our newsletter for trending solutions and developer bootcamps

Share this Page