Get clusters from dendrogram python
I have a list of words on which I performed TF-IDF Algorithm to get the list of top 100 words. After which I am supposed to perform clustering. For now I am able to do both of the tasks (I am sharing relevant part of the code and input file, screenshot of output). Show My Query is that I wanted the list of the clusters that are formed in the output Dendrogram, How can I do that? The Dendrogram function returns a Tuple The following is extract from the input file.
The following is the code that I am using
The below link contains the output of the Dendrogram formed https://www.screencast.com/t/2MEc3ohBe Plot the hierarchical clustering as a dendrogram. The dendrogram illustrates how each cluster is composed by drawing a U-shaped link between a non-singleton cluster and its children. The top of the U-link indicates a cluster merge. The two legs of the U-link indicate which clusters were merged. The length of the two legs of the U-link represents the distance between the child clusters. It is also the cophenetic distance between original observations in the two children clusters. Parameters ZndarrayThe linkage matrix encoding the hierarchical clustering to render as a dendrogram. See the The The dendrogram can be hard to read when the original observation matrix from which the linkage is derived is large. Truncation is used to condense the dendrogram. There are several modes: None No truncation is performed (default). Note: 'lastp' The last 'level' No more than Note: For brevity, let \(t\) be the Includes a
list The direction to plot the dendrogram, which can be any of the following strings: 'top' Plots the root at the top, and plot descendent links going downwards. (default). 'bottom' Plots the root at the bottom, and plot descendent links going upwards. 'left' Plots the root at the left, and plot descendent links going right. 'right' Plots the root at the right, and plot descendent links going left. labelsndarray, optionalBy default, For each node n, the order (visually, from left-to-right) n’s two descendent links are plotted is determined by this parameter, which can be any of the following values: False Nothing is done. 'ascending' or True The child with the minimum number of original objects in its cluster is plotted first. 'descending' The child with the maximum number of original objects in its cluster is plotted first. Note, For each node n, the order (visually, from left-to-right) n’s two descendent links are plotted is determined by this parameter, which can be any of the following values: False Nothing is done. 'ascending' or True The child with the minimum distance between its direct descendents is plotted first. 'descending' The child with the maximum distance between its direct descendents is plotted first. Note When True, leaf nodes representing \(k>1\) original observation are labeled with the number of observations they contain in parentheses. no_plotbool, optionalWhen True, the final rendering is not performed. This is useful if only the data structures computed for the rendering are needed or if matplotlib is not available. no_labelsbool, optionalWhen True, no labels appear next to the leaf nodes in the rendering of the dendrogram. leaf_rotationdouble, optionalSpecifies the angle (in degrees) to rotate the leaf labels. When unspecified, the rotation is based on the number of nodes in the dendrogram (default is 0). leaf_font_sizeint, optionalSpecifies the font size (in points) of the leaf labels. When unspecified, the size based on the number of nodes in the dendrogram. leaf_label_funclambda or function, optionalWhen Indices \(k < n\) correspond to original observations while indices \(k \geq n\) correspond to non-singleton clusters. For example, to label singletons with their node id and non-singletons with their id, count, and inconsistency coefficient, simply do: # First define the leaf label function. def llf(id): if id < n: return str(id) else: return '[%d %d %1.2f]' % (id, count, R[n-id,3]) # The text for the leaf nodes is going to be big so force # a rotation of 90 degrees. dendrogram(Z, leaf_label_func=llf, leaf_rotation=90) # leaf_label_func can also be used together with ``truncate_mode`` parameter, # in which case you will get your leaves labeled after truncation: dendrogram(Z, leaf_label_func=llf, leaf_rotation=90, truncate_mode='level', p=2)show_contractedbool, optional When True the heights of non-singleton nodes contracted into a leaf node are plotted as crosses along the link connecting that leaf node. This really is only useful when truncation is used (see If given, link_color_function is called with each non-singleton id corresponding to each U-shaped link it will paint. The function is expected to return the color to paint the link, encoded as a matplotlib color string code. For example: dendrogram(Z, link_color_func=lambda k: colors[k]) colors the direct links below each untruncated non-singleton node If None and
no_plot is not True, the dendrogram will be plotted on the current axes. Otherwise if no_plot is not True the dendrogram will be plotted on the given This matplotlib color string sets the color of the links above the color_threshold. The default is A dictionary of data structures computed to render the dendrogram. Its has the following keys: 'color_list' A list of color names. The k’th element represents the color of the k’th link. 'icoord' and 'dcoord' Each of them is a list of lists. Let 'ivl' A list of labels corresponding to the leaf nodes. 'leaves' For each i, 'leaves_color_list' A list of color names. The k’th element represents the color of the k’th leaf. Notes It is expected that the distances in Examples >>> from scipy.cluster import hierarchy >>> import matplotlib.pyplot as plt A very basic example: >>> ytdist = np.array([662., 877., 255., 412., 996., 295., 468., 268., ... 400., 754., 564., 138., 219., 869., 669.]) >>> Z = hierarchy.linkage(ytdist, 'single') >>> plt.figure() >>> dn = hierarchy.dendrogram(Z) Now, plot in given axes, improve the color scheme and use both vertical and horizontal orientations: >>> hierarchy.set_link_color_palette(['m', 'c', 'y', 'k']) >>> fig, axes = plt.subplots(1, 2, figsize=(8, 3)) >>> dn1 = hierarchy.dendrogram(Z, ax=axes[0], above_threshold_color='y', ... orientation='top') >>> dn2 = hierarchy.dendrogram(Z, ax=axes[1], ... above_threshold_color='#bcbddc', ... orientation='right') >>> hierarchy.set_link_color_palette(None) # reset to default after use >>> plt.show() How do you find the number of clusters in a dendrogram Python?In the dendrogram locate the largest vertical difference between nodes, and in the middle pass an horizontal line. The number of vertical lines intersecting it is the optimal number of clusters (when affinity is calculated using the method set in linkage).
How many clusters do we get from dendrogram?Looking at this dendrogram, you can see the three clusters as three branches that occur at about the same horizontal distance. The two outliers, 6 and 13, are fused in rather arbitrarily at much higher distances. This is the interpretation.
How do I display a dendrogram in Python?Dendrograms in Python. Basic Dendrogram. A dendrogram is a diagram representing a tree. The figure factory called create_dendrogram performs hierarchical clustering on data and represents the resulting tree. ... . Set Color Threshold.. Set Orientation and Add Labels.. Plot a Dendrogram with a Heatmap. See also the Dash Bio demo.. |