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 18:51transform
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 usingscript_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