K-Means-Clustering | creates clusters using K-Means Clustering Algorithm | Machine Learning library
kandi X-RAY | K-Means-Clustering Summary
kandi X-RAY | K-Means-Clustering Summary
Script which creates clusters using K-Means Clustering Algorithm with different similarity metrics.
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Top functions reviewed by kandi - BETA
- Generate k - means clustering
- Return the distance between two instances
- Assigns the given clusters to each centroids
- Assign an instance to the centroids
- Load a csv file
- Converts a list of strings to numbers
- Checks if a string is valid
- Convert a line into a tuple
- Perform k - means clustering
- Run k - means clustering
- Displays cluster clusters
- Merge multiple clusters
- Print the table of instances
K-Means-Clustering Key Features
K-Means-Clustering Examples and Code Snippets
const kMeans = (data, k = 1) => {
const centroids = data.slice(0, k);
const distances = Array.from({ length: data.length }, () =>
Array.from({ length: k }, () => 0)
);
const classes = Array.from({ length: data.length }, () =>
def _kmeans_plus_plus(self):
# Points from only the first shard are used for initializing centers.
# TODO(ands): Use all points.
inp = self._inputs[0]
if self._distance_metric == COSINE_DISTANCE:
inp = nn_impl.l2_normalize(inp,
Community Discussions
Trending Discussions on K-Means-Clustering
QUESTION
I've been trying to run RAPIDS on Google Colab pro, and have successfully installed the cuml and cudf packages, however I am unable to run even the example scripts.
TLDR;Anytime I try to run the fit function for cuml on Google Colab I get the following error. I get this when using the demo examples both for installation and then for cuml. This happens for a range of cuml examples (I first hit this trying to run UMAP).
...ANSWER
Answered 2021-May-06 at 17:13Colab retains cupy==7.4.0
despite conda installing cupy==8.6.0
during the RAPIDS install. It is a custom install. I just had success pip installing cupy-cuda110==8.6.0
BEFORE installing RAPIDS, with
!pip install cupy-cuda110==8.6.0
:
I'll be updating the script soon so that you won't have to do it manually, but want to test a few more things out. Thanks again for letting us know!
EDIT: script updated.
QUESTION
I am performing a binary classification of a partially labeled dataset. I have a reliable estimate of its 1's, but not of its 0's.
From sklearn KMeans documentation:
...ANSWER
Answered 2020-Nov-20 at 20:14I'm reasonably confident this works as intended, but please correct me if you spot an error. (cobbled together from geeks for geeks):
QUESTION
I am using a k-modes model (mymodel
) which is created by a data frame mydf1
. I am looking to assign the nearest cluster of mymodel
for each row of a new data frame mydf2
.
Similar to this question - just with k-modes instead of k-means. The predict
function of the flexclust
package only works with numeric data, not categorial.
A short example:
...ANSWER
Answered 2020-Sep-29 at 09:08We can use the distance measure that is used in the kmodes algorithm to assign each new row to its nearest cluster.
QUESTION
I am new to machine learning and i am using
...ANSWER
Answered 2020-Aug-24 at 17:35Iris dataset contains 4 features describing the three different types of flowers (i.e. 3 classes). Therefore, each point in the dataset is located in a 4-dimensional space and the same applies to the centroids, so to describe their position you need the 4 coordinates.
In examples, it's easier to use 2-dimensional data (sometimes 3-dimensional) as it is easier to plot it out and display for teaching purposes, but the centroids will have as many coordinates as your data has dimensions (i.e. features), so with the Iris dataset, you would expect the 4 coordinates.
QUESTION
I'm trying to learn sklearn. As I understand from step 5 of the following example, the predicted clusters can be mislabelled and it would be up to me to relabel them properly. This is also done in an example on sci-kit. Labels must be re-assigned so that the results of the clustering and the ground truth match by color.
How would I know if the labels of the predicted clusters match the initial data labels and how to readjust the indices of the labels to properly match the two sets?
...ANSWER
Answered 2020-Mar-30 at 07:00With clustering, there's no meaningful order or comparison between clusters, we're just finding groups of observations that have something in common. There's no reason to refer to one cluster as 'the blue cluster' vs 'the red cluster' (unless you have some extra knowledge about the domain). For that reason, sklearn will arbitrarily assign numbers to each cluster.
QUESTION
I am testing this code.
...ANSWER
Answered 2020-Jan-03 at 01:33The problem may be with the format of your data. Most models will expect a data frame
QUESTION
I have trained my corpus on w2v and k-means following the instructions given this link.
https://ai.intelligentonlinetools.com/ml/k-means-clustering-example-word2vec/
What I am want to do this a. find the cluster ID for a given word b. get the top 20 nearest words from the cluster for the given word.
I have figured out how to the words in a given cluster. What I want is to find out the words that are closer to my given word in the given cluster.
Any help is appreciated.
...ANSWER
Answered 2019-Dec-25 at 20:58Your linked guide is, with its given data, a bit of misguided. You can't get meaningful 100-dimensional word-vectors (the gensim Word2Vec
class default) from a mere 30-word corpus. The results from such a model will be nonsense, useless for clustering or other downstream steps – so any tutorial purporting to show this process, with true results, should be using far more data.
If you are in fact using far more data, and have succeeded in clustering words, the Word2Vec
model's most_similar()
function will give you the top-N (default 10) nearest-words for any given input word. (Specifically, they will be returned as (word, cosine_similarity)
tuples, ranked by highest cosine_similarity
.)
The Word2Vec
model is of course oblivious to the results of clustering, so you would have to filter those results to discard words outside the cluster of interest.
I'll assume that you have some lookup object cluster
, that for cluster[word]
gives you the cluster ID for a specific word. (This might be a dict, or something that does a KMeans-model predict()
on the supplied vector, whatever.) And, that total_words
is the total number of words in your model. (For example: total_words = len(w2v_model.wv)
. Then your logic should be roughly like
QUESTION
I am using scikit-learn to cluster some data, and I want to compare the results of different clustering techniques. I am immediately faced with the issue that the labels for the clusters are different for different runs, so even if they are clustered exactly the same the similarity of the lists is still very low.
Say I have
...ANSWER
Answered 2019-Mar-21 at 15:07You could try calculating the adjusted Rand index between two results. This gives a score between -1 and 1, where 1 is a perfect match.
Or by taking argmax of confusion matrix:
QUESTION
OK this is going to be a long post. So i am fairly new with R (i am currently using the MR free 3.5, with no checkpoint) but i am trying to work with the tidyverse, which i find very elegant in writing code and a lot of times a lot more simple.
I decided to replicate an exercise from guru99 here. It is a simple k-means exercise. However because i always want to write "generalizeble" code i was trying to automatically rename the variables in mutate with new names. So i searched SO and found this solution here which is very nice.
First what works fine.
...ANSWER
Answered 2019-Nov-16 at 17:41scale()
returns a matrix, and dplyr/tibble isn't automatically coercing it to a vector. By changing your mutate_all()
call to the below, we can have it return a vector. I identified this is what was happening by calling class(df1$speed_scaled)
and seeing the result of "matrix".
QUESTION
We've been using Kmeans for clustering our logs. A typical dataset has 10 mill. samples with 100k+ features.
To find the optimal k - we run multiple Kmeans in parallel and pick the one with the best silhouette score. In 90% of the cases we end up with k between 2 and 100. Currently, we are using scikit-learn Kmeans. For such a dataset, clustering takes around 24h on ec2 instance with 32 cores and 244 RAM.
I've been currently researching for a faster solution.
What I have already tested:
Kmeans + Mean Shift Combination - a little better (for k=1024 --> ~13h) but still slow.
Kmcuda library - doesn't have support for sparse matrix representation. It would require ~3TB RAM to represent that dataset as a dense matrix in memory.
Tensorflow (tf.contrib.factorization.python.ops.KmeansClustering()) - only started investigation today, but either I am doing something wrong, or I do not know how to cook it. On my first test with 20k samples and 500 features, clustering on a single GPU is slower than on CPU in 1 thread.
Facebook FAISS - no support for sparse representation.
There is PySpark MlLib Kmeans next on my list. But would it make sense on 1 node?
Would it be training for my use-case faster on multiple GPUs? e.g., TensorFlow with 8 Tesla V-100?
Is there any magical library that I haven't heard of?
Or just simply scale vertically?
...ANSWER
Answered 2019-Oct-11 at 22:06Choose the algorithm wisely. There are clever algorithms, and there are stupid algorithms for kmeans. Lloyd's is stupid, but the only one you will find in GPUs so far. It wastes a lot of resources with unnecessary computations. Because GPU and "big data" people do not care about resource efficiency... Good algorithms include Elkan's, Hamerly's, Ying-Yang, Exponion, Annulus, etc. - these are much faster than Lloyd's.
Sklearn is one of the better tools here, because it at least includes Elkan's algorithm. But if I am not mistaken, it may be making a dense copy of your data repeatedly. Maybe in chunks so you don't notice it. When I compared k-means from sklearn with my own spherical k-means in Python, my implementation was many times faster. I can only explain this with me using sparse optimizations while the sklearn version performed dense operations. But maybe this has been improved since.
Implementation quality is important. There was an interesting paper about benchmarking k-means. Let me Google it:
Kriegel, H. P., Schubert, E., & Zimek, A. (2017). The (black) art of runtime evaluation: Are we comparing algorithms or implementations?. Knowledge and Information Systems, 52(2), 341-378.
They show how supposedly the same algorithm can have orders f magnitude runtime differences, depending on implementation differences. Spark does not fare very well there... It has too high overheads, too slow algorithms.
You don't need all the data.
K-means works with averages. The quality of the mean very slowly improves as you add more data. So there is little use in using all the data you have. Just use a large enough sample, and the results should be of almost the same quality. You can exploit this also for seeding. Run on a smaller set first, then add more data for refinement.
Because your data is sparse, there is a high chance that k-means is not the right tools anyway. Have you tested the quality of your results? How do you ensure attributes to be appropriately scaled? How much is the result determined simply by where the vectors are 0, and not by the actual non-zero values? Do results actually improve with rerunning k-means so often? What if you d not rerun k-means ever again? What if you just run it on a sample as discussed in 3)? What if you just pick k random centers and do 0 iterations of k-means? What is your best Silhouette? Chances are that you cannot measure the difference and are just wasting time and resources for nothing! So what do you do to ensure reliability of your results?
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Install K-Means-Clustering
You can use K-Means-Clustering 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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