I am trying to evaluate the clusters generated by K-means with different metrics, but I am not sure about whether the results are good or not.
I have 40 documents to cluster in 6 categories.
I first converted them into tf-idf vectors, then I clustered them by K-means (k = 6). Finally, I tried to evaluate the results by different metrics.
Because I have the real labels of the documents, I tried to calculate the F1 score and accuracy. But I also want to know the performance for the metrics that do not need real labels such as silhouette score.
For F1 score and accuracy, the results are about 0.65 and 0.88 respectively, while for the silhouette score, it is only about 0.05, which means I may have overlapping clusters.
In this case, can I say that the results are acceptable? Or should I handle the overlapping issue by trying other methods instead of tf-idf to represent the documents or other algorithms to cluster?
With such tiny data sets, you really need to use a measure that is adjusted for chance.
Do the following: label each document randomly with an integer 1..6
What F1 score to you get? Now repeat this 100x times, what is the best result you get? A completely random result can score pretty well on such tiny data!
Because of this problem, the standard measure used in clustering it's the adjusted Rand index (ARI). A similar adjustment also exists for NMI: Adjusted Mutual Information or AMI. But AMI is much less common.
Related
I am using the https://www.mathworks.com/matlabcentral/fileexchange/32197-clustering-results-measurement for evaluating my clustering accuracy in MATLAB, it provides accuracy and rand_index, the performance is normal as expect. However, when I try to use NMI as a metric, the clustering performance is extremely low, I am using the source code (https://www.mathworks.com/matlabcentral/fileexchange/29047-normalized-mutual-information).
Actually I have two Nx1 vectors as inputs, one is the actual label while another is the label assignments. I basically check each of every element insides and I found that even I have 82% rand_index, the NMI is only 0.3209. Below is the example for Iris Dataset https://archive.ics.uci.edu/ml/datasets/iris with MATLAB built-in K-Means.
data = iris(:,1:data_dim);
k = 3;
[result_label,centroid] = kmeans(data,k,'MaxIter',10000);
actual_label = iris(:,end);
NMI = nmi(actual_label,result_label);
[Acc,rand_index,match] = AccMeasure(actual_label',result_label');
The result:
Auto ACC: 0.820000
Rand_Index: 0.701818
NMI: 0.320912
The Rand Index will tend towards 1 as the number of data points increases (even when comparing random clusterings) so you never really expect to see small values of Rand when you have a big data set.
At the same time, Accuracy can be high when all of your points fall into the same large cluster.
I have a feeling that the NMI is producing a more reliable comparison. To verify, trying running a dimensionality reduction and plot the data points with color based on the two clusterings. Visual statistics are often the best for developing an intuition about data.
If you want to explore more, a convenient python package for clustering comparisons is CluSim.
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Cluster analysis in R: determine the optimal number of clusters
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I am going to build a K-means clustering model for outlier detection. For that, I need to identify the best number of clusters needs to be selected.
For now, I have tried to do this using Elbow Method. I plotted the sum of squared error vs. the number of clusters(k) but, I got a graph like below which makes confusion to identify the elbow point.
I need to know, why do I get a graph like this and how do I identify the optimal number of clusters.
K-means is not suitable for outlier detection. This keeps popping up here all the time.
K-means is conceptualized for "pure" data, with no false points. All measurements are supposed to come from the data, and only vary by some Gaussian measurement error. Occasionally this may yield some more extreme values, but even these are real measurements, from the real clusters, and should be explained not removed.
K-means itself is known to not work well on noisy data where data points do not belong to the clusters
It tends to split large real clusters in two, and then points right in the middle of the real cluster will have a large distance to the k-means centers
It tends to put outliers into their own clusters (because that reduces SSQ), and then the actual outliers will have a small distance, even 0.
Rather use an actual outlier detection algorithm such as Local Outlier Factor, kNN, LOOP etc. instead that were conceptualized with noisy data in mind.
Remember that the Elbow Method doesn't just 'give' the best value of k, since the best value of k is up to interpretation.
The theory behind the Elbow Method is that we in tandem both want to minimize some error function (i.e. sum of squared errors) while also picking a low value of k.
The Elbow Method thus suggests that a good value of k would lie in a point on the plot that resembles an elbow. That is the error is small, but doesn't decrease drastically when k increases locally.
In your plot you could argue that both k=3 and k=6 resembles elbows. By picking k=3 you'd have picked a small k, and we see that k=4, and k=5 doesn't do much better in minimizing the error. Same goes with k=6.
I have 150 images, 15 each of 10 different people. So basically I know which image should belong together, if clustered.
These images are of 73 dimensions (feature-vector) and I clustered them into 10 clusters using kmeans function in matlab.
Later, I processed these 150 data points and reduced its dimension from 73 to 3 for my work and applied the same kmeans function on them.
I want to compare the results obtained on these data sets (processed and unprocessed) by applying the same k-means function and wish to know if the processing which reduced it to lower dimension improves the kmeans clustering or not.
I thought comparing the variance of each cluster can be one parameter for comparison, however I am not sure if I can directly compare and evaluate my results (within cluster sum of distances etc.) as both the cases are of different dimension. Could anyone please suggest a way where I can compare the kmean results, some way to normalize them or any other comparison that I can make?
I can think of three options. I am unaware of any well developed methodology to do this specifically with K-means clustering.
Look at the confusion matrices between the two approaches.
Compare the mahalanobis distances between the clusters, and between items in clusters to their nearest other clusters.
Look at the Vornoi cells and see how far your points are from the boundaries of the cells.
The problem with 3, is the distance metrics get skewed, 3D distance vs. 73D distances are not commensurate, so I'm not a fan of that approach. I'd recommend reading some books on K-means if you are adamant of that path, rank speculation is fun, but standing on the shoulders of giants is better.
I'm currently trying to preprocess my training data ready for a multi-layered perceptron. The data I downloaded consists of 20,000 instances and 16 attributes, all of which are coordinate values of pixels as part of letter recognition. The data itself has already been scaled from its original form into values between 0 - 15 before being published.
However since it's already been scaled, is it still necessary to perform normalization on it? I've tried to read around and look at previous examples but have come up with conflicting points. In some papers, it has stated that scaling is a form of normalization, where as others have said that normalization would be bringing that values to a range of 0-1.
Since I'm using WEKA I've attempted their normalize filter during a pre-processing stage and it caused the accuracy to decrease by around 2% which makes me think it could be unnecessary. But again, I've read that it may only have a positive effect later in training.
So my question is:
What is the difference between scaling to a range such as 0 - 15 and normalizing it? Should I still normalize it on top of this scaling thats already done?
In your case you do not need to. Normalizing data is done so that an attribute with a different scale will not decide outcome of distance operations, ultimately decide clustering or classification results.
An example you have two attributes weight and income. Weight will be 10 and 200kg at most. While income can be 10,000$ and 20,000,000$. But most of the people's income will be 10,000 and 120,000, while above this values will be outliers. If you do not normalize your data before using Multi Layer Perceptron, outcome of your neural network will be decided by these outliers.
In your case this situation is already mitigated due to your scaling therefore you do not need normalizing.
Can I use k-means algorithm for a single attribute?
Is there any relationship between the attributes and the number of clusters?
I have one attribute's performance, and I want to classify the data into 3 clusters: poor, medium, and good.
Is it possible to create 3 clusters with one attribute?
K-Means is useful when you have an idea of how many clusters actually exists in your space. Its main benefit is its speed. There is a relationship between attributes and the number of observations in your dataset.
Sometimes a dataset can suffer from The Curse of Dimensionality where your number of variables/attributes is much greater than your number of observations. Basically, in high dimensional spaces with few observations, it becomes difficult to separate observations in hyper dimensions.
You can certainly have three clusters with one attribute. Consider the quantitative attribute in which you have 7 observations
1
2
100
101
500
499
501
Notice there are three clusters in this sample centered: 1.5, 100.5, and 500.
If you have one dimensional data, search stackoverflow for better approaches than k-means.
K-means and other clustering algorithms shine when you have multivariate data. They will "work" with 1-dimensional data, but they are not very smart anymore.
One-dimensional data is ordered. If you sort your data (or it even is already sorted), it can be processed much more efficiently than with k-means. Complexity of k-means is "just" O(n*k*i), but if your data is sorted and 1-dimensional you can actually improve k-means to O(k*i). Sorting comes at a cost, but there are very good sort implementations everywhere...
Plus, for 1-dimensional data there is a lot of statistics you can use that are not very well researched or tractable on higher dimensions. One statistic you really should try is kernel density estimation. Maybe also try Jenks Natural Breaks Optimization.
However, if you want to just split your data into poor/medium/high, why don't you just use two thresholds?
As others have answered already, k-means requires prior information about the count of clusters. This may appear to be not very helpful at the start. But, I will cite the following scenario which I worked with and found to be very helpful.
Color segmentation
Think of a picture with 3 channels of information. (Red, Green Blue) You want to quantize the colors into 20 different bands for the purpose of dimensional reduction. We call this as vector quantization.
Every pixel is a 3 dimensional vector with Red, Green and Blue components. If the image is 100 pixels by 100 pixels then you have 10,000 vectors.
R,G,B
128,100,20
120,9,30
255,255,255
128,100,20
120,9,30
.
.
.
Depending on the type of analysis you intend to perform, you may not need all the R,G,B values. It might be simpler to deal with an ordinal representation.
In the above example, the RGB values might be assigned a flat integral representation
R,G,B
128,100,20 => 1
120,9,30 => 2
255,255,255=> 3
128,100,20 => 1
120,9,30 => 2
You run the k-Means algorithm on these 10,000 vectors and specify 20 clusters. Result - you have reduced your image colors to 20 broad buckets. Obviously some information is lost. However, the intuition for this loss being acceptable is that when the human eyes is gazing out over a patch of green meadow, we are unlikely to register all the 16 million RGB colours.
YouTube video
https://www.youtube.com/watch?v=yR7k19YBqiw
I have embedded key pictures from this video for your understanding. Attention! I am not the author of this video.
Original image
After segmentation using K means
Yes it is possible to use clustering with single attribute.
No there is no known relation between number of cluster and the attributes. However there have been some study that suggest taking number of clusters (k)=n\sqrt{2}, where n is the total number of items. This is just one study, different study have suggested different cluster numbers. The best way to determine cluster number is to select that cluster number that minimizes intra-cluster distance and maximizes inter-cluster distance. Also having background knowledge is important.
The problem you are looking with performance attribute is more a classification problem than a clustering problem
Difference between classification and clustering in data mining?
With only one attribute, you don't need to do k-means. First, I would like to know if your attribute is numerical or categorical.
If it's numerical, it would be easier to set up two thresholds. And if it's categorical, things are getting much easier. Just specify which classes belong to poor, medium or good. Then simple data frame operations would be working.
Feel free to send me comments if you are still confused.
Rowen