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I have written my own clustering routine and would like to produce a dendrogram. The easiest way to do this would be to use scipy dendrogram function. However, this requires the input to be in the same format that the scipy linkage function produces. I cannot find an example of how the output of this is formatted. I was wondering whether someone out there can enlighten me.
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This version tracks the distance (size) of each cluster (node_id), and confirms the number of members. This uses the scipy linkage() function which is the same foundation of the Aggregator clusterer.
A linkage matrix is valid if it is a 2-D array (type double) with rows and 4 columns. The first two columns must contain indices between 0 and 2 n − 1 . For a given row i , the following two expressions have to hold: 0 ≤ Z [ i , 0 ] ≤ i + n − 1 0 ≤ Z [ i , 1 ] ≤ i + n − 1.
A linkage function is an essential prerequisite for hierarchical cluster analysis . Its value is a measure of the "distance" between two groups of objects (i.e. between two clusters). Algorithms for hierarchical clustering normally differ by the linkage function used.
cluster. hierarchy ) These functions cut hierarchical clusterings into flat clusterings or find the roots of the forest formed by a cut by providing the flat cluster ids of each observation.
From the scipy docs for the linkage matrix: A (n-1) by 4 matrix Z is returned. At the i-th iteration, clusters with indices Z [i, 0] and Z [i, 1] are combined to form cluster n+i. A cluster with an index less than n corresponds to one of the n original observations.
This is from the scipy.cluster.hierarchy.linkage () function documentation, I think it's a pretty clear description for the output format: A ( n -1) by 4 matrix Z is returned. At the i -th iteration, clusters with indices Z [i, 0] and Z [i, 1] are combined to form cluster n + i.
The benefit of using SciPy library in Python while making ML models is that it also makes a strong programming language available for use in developing less complex programs and applications. LU decomposition is a method that reduce matrix into constituent parts that helps in easier calculation of complex matrix operations.
SciPy is an interactive Python session used as a data-processing library that is made to compete with its rivalries such as MATLAB, Octave, R-Lab,etc. It has many user-friendly, efficient and easy-to-use functions that helps to solve problems like numerical integration, interpolation, optimization, linear algebra and statistics.
I agree with https://stackoverflow.com/users/1167475/mortonjt that the documentation does not fully explain the indexing of intermediate clusters, while I do agree with the https://stackoverflow.com/users/1354844/dkar that the format is otherwise precisely explained.
Using the example data from this question: Tutorial for scipy.cluster.hierarchy
A = np.array([[0.1, 2.5], [1.5, .4 ], [0.3, 1 ], [1 , .8 ], [0.5, 0 ], [0 , 0.5], [0.5, 0.5], [2.7, 2 ], [2.2, 3.1], [3 , 2 ], [3.2, 1.3]]) A linkage matrix can be built using the single (i.e, the closest matching points):
z = hac.linkage(a, method="single") array([[ 7. , 9. , 0.3 , 2. ], [ 4. , 6. , 0.5 , 2. ], [ 5. , 12. , 0.5 , 3. ], [ 2. , 13. , 0.53851648, 4. ], [ 3. , 14. , 0.58309519, 5. ], [ 1. , 15. , 0.64031242, 6. ], [ 10. , 11. , 0.72801099, 3. ], [ 8. , 17. , 1.2083046 , 4. ], [ 0. , 16. , 1.5132746 , 7. ], [ 18. , 19. , 1.92353841, 11. ]]) As the documentation explains the clusters below n (here: 11) are simply the data points in the original matrix A. The intermediate clusters going forward, are indexed successively.
Thus, clusters 7 and 9 (the first merge) are merged into cluster 11, clusters 4 and 6 into 12. Then observe line three, merging clusters 5 (from A) and 12 (from the not-shown intermediate cluster 12) resulting with a Within-Cluster Distance (WCD) of 0.5. The single method entails that the new WCS is 0.5, which is the distance between A[5] and the closest point in cluster 12, A[4] and A[6]. Let's check:
In [198]: norm([a[5]-a[4]]) Out[198]: 0.70710678118654757 In [199]: norm([a[5]-a[6]]) Out[199]: 0.5 This cluster should now be intermediate cluster 13, which subsequently is merged with A[2]. Thus, the new distance should be the closest between the points A[2] and A[4,5,6].
In [200]: norm([a[2]-a[4]]) Out[200]: 1.019803902718557 In [201]: norm([a[2]-a[5]]) Out[201]: 0.58309518948452999 In [202]: norm([a[2]-a[6]]) Out[202]: 0.53851648071345048 Which, as can be seen also checks out, and explains the intermediate format of new clusters.
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