I'm stuck trying to implement and plot in python the intra-cluster of each cluster in k-means to get best number of k. Which is represented using this formula
Which is the sum of the square distances of data points which belong to a certain cluster from the centroid and normalized by the size of the cluster Ck.
Then we can compute the intra cluster variance for all clusters by just adding up the individual cluster or specific variances using this formula:
Can I get help implementing Wk and W?
The custom k-mean implementaion:
def kmeans(X, k):
iterations=0
data = pd.DataFrame(X)
cluster = np.zeros(X.shape[0])
#taking random samples from the datapoints as an initialization of centroids
centroids = data.sample(n=k).values
while True:
# for each observation
for i, row in enumerate(X):
mn_dist = float('inf')
# distance of the point from all centroids
for idx, centroid in enumerate(centroids):
# calculating euclidean distance
d = np.sqrt((centroid[0]-row[0])**2 + (centroid[1]-row[1])**2)
# assign closest centroid
if mn_dist > d:
mn_dist = d
cluster[i] = idx
#updating centroids by taking the mean value of all datapoints of each cluster
new_centroids = pd.DataFrame(X).groupby(by=cluster).mean().values
iterations+=1
# if centroids are same then break.
if np.count_nonzero(centroids-new_centroids) == 0:
break
else: #else update old centroids with new ones
centroids = new_centroids
return centroids, cluster, iterations
Related
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:
init : {‘k-means++’, ‘random’ or an ndarray}
Method for initialization, defaults to ‘k-means++’:
If an ndarray is passed, it should be of shape (n_clusters, n_features) and gives the initial centers.
I would like to pass an ndarray, but I only have 1 reliable centroid, not 2.
Is there a way to maximize the entropy between the K-1st centroids and the Kth? Alternatively, is there a way to manually initialize K-1 centroids and use K++ for the remaining?
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Related questions:
This seeks to define K centroids with n-1 features. (I want to define k-1 centroids with n features).
Here is a description of what I want, but it was interpreted as a bug by one of the developers, and is "easily implement[able]"
I'm reasonably confident this works as intended, but please correct me if you spot an error. (cobbled together from geeks for geeks):
import sys
def distance(p1, p2):
return np.sum((p1 - p2)**2)
def find_remaining_centroid(data, known_centroids, k = 1):
'''
initialized the centroids for K-means++
inputs:
data - Numpy array containing the feature space
known_centroid - Numpy array containing the location of one or multiple known centroids
k - remaining centroids to be found
'''
n_points = data.shape[0]
# Initialize centroids list
if known_centroids.ndim > 1:
centroids = [cent for cent in known_centroids]
else:
centroids = [np.array(known_centroids)]
# Perform casting if necessary
if isinstance(data, pd.DataFrame):
data = np.array(data)
# Add a randomly selected data point to the list
centroids.append(data[np.random.randint(
n_points), :])
# Compute remaining k-1 centroids
for c_id in range(k - 1):
## initialize a list to store distances of data
## points from nearest centroid
dist = np.empty(n_points)
for i in range(n_points):
point = data[i, :]
d = sys.maxsize
## compute distance of 'point' from each of the previously
## selected centroid and store the minimum distance
for j in range(len(centroids)):
temp_dist = distance(point, centroids[j])
d = min(d, temp_dist)
dist[i] = d
## select data point with maximum distance as our next centroid
next_centroid = data[np.argmax(dist), :]
centroids.append(next_centroid)
# Reinitialize distance array for next centroid
dist = np.empty(n_points)
return centroids[-k:]
Its usage:
# For finding a third centroid:
third_centroid = find_remaining_centroid(X_train, np.array([presence_seed, absence_seed]), k = 1)
# For finding the second centroid:
second_centroid = find_remaining_centroid(X_train, presence_seed, k = 1)
Where presence_seed and absence_seed are known centroid locations.
I'm working on an anomaly detection task using KMeans.
Pandas dataframe that i'm using has a single feature and it is like the following one:
df = array([[12534.],
[12014.],
[12158.],
[11935.],
...,
[ 5120.],
[ 4828.],
[ 4443.]])
I'm able to fit and to predict values with the following instructions:
km = KMeans(n_clusters=2)
km.fit(df)
km.predict(df)
In order to identify anomalies, I would like to calculate the distance between centroid and each single point, but with a dataframe with a single feature i'm not sure that it is the correct approach.
I found examples which used euclidean distance to calculate the distance. An example is the following one:
def k_mean_distance(data, cx, cy, i_centroid, cluster_labels):
distances = [np.sqrt((x - cx) ** 2 + (y - cy) ** 2) for (x, y) in data[cluster_labels == i_centroid]]
return distances
centroids = self.km.cluster_centers_
distances = []
for i, (cx, cy) in enumerate(centroids):
mean_distance = k_mean_distance(day_df, cx, cy, i, clusters)
distances.append({'x': cx, 'y': cy, 'distance': mean_distance})
This code doesn't work for me because centroids are like the following one in my case, since i have a single feature dataframe:
array([[11899.90692187],
[ 5406.54143126]])
In this case, what is the correct approach to find the distance between centroid and points? Is it possible?
Thank you and sorry for the trivial question, i'm still learning
There's scipy.spatial.distance_matrix you can make use of:
# setup a set of 2d points
np.random.seed(2)
df = np.random.uniform(0,1,(100,2))
# make it a dataframe
df = pd.DataFrame(df)
# clustering with 3 clusters
from sklearn.cluster import KMeans
km = KMeans(n_clusters=3)
km.fit(df)
preds = km.predict(df)
# get centroids
centroids = km.cluster_centers_
# visualize
plt.scatter(df[0], df[1], c=preds)
plt.scatter(centroids[:,0], centroids[:,1], c=range(centroids.shape[0]), s=1000)
gives
Now the distance matrix:
from scipy.spatial import distance_matrix
dist_mat = pd.DataFrame(distance_matrix(df.values, centroids))
You can confirm that this is correct by
dist_mat.idxmin(axis=1) == preds
And finally, the mean distance to centroids:
dist_mat.groupby(preds).mean()
gives:
0 1 2
0 0.243367 0.525194 0.571674
1 0.525350 0.228947 0.575169
2 0.560297 0.573860 0.197556
where the columns denote the centroid number and rows denote the mean distance of the points in a cluster.
You can use scipy.spatial.distance.cdist to create a distance matrix:
from scipy.spatial.distance import cdist
dm = cdist(df, centroids)
This should give you a 2-d array, where each row represents an observation in your original dataset, and each column represents a centroid. The x-th row in the y-th column gives the distance between your x-th observation to your y-th cluster centroid. cdist uses Euclidean distance by default, but you can use other metrics (not that it matters much for a dataset with only one feature).
I'm working on a data science project for my intro to Data Science class, and we've decided to tackle a problem relating to desalination plants in california: "Where should we place k plants to minimize the distance to zip codes?"
The data that we have so far is, zip, city, county, pop, lat, long, amount of water.
The issue is, I can't find any resources on how to force the centroid to be constrained to staying on the coast. What we've thought of so far is:
Use a normal kmeans algorithm, but move the centroid to the coast once clusters have settled (bad)
Use a normal kmeans algorithm with weights, making the coastal zips have infinite weight (I've been told this isn't a great solution)
What do you guys think?
K-means does not minimize distances.
It minimizes squared errors, which is quite different.
The difference is roughly that of the median, and the mean in 1 dimensional data. The error can be massive.
Here is a counter example, assuming we have the coordinates:
-1 0
+1 0
0 -1
0 101
The center chosen by k-means would be 0,25. The optimal location is 0,0.
The sum of distances by k-means is > 152, the optimum location has distance 104. So here, the centroid is almost 50% worse than the optimum! But the centroid (= multivariate mean) is what k-means uses!
k-means does not minimize the Euclidean distance!
This is one variant how "k-means is sensitive to outliers".
It does not get better if you try to constrain it to place "centers" on the coast only...
Also, you may want to at least use Haversine distance, because in California, 1 degree north != 1 degree east, because it's not at the Equator.
Furthermore, you likely should not make the assumption that every location requires its own pipe, but rather they will be connected like a tree. This greatly reduces the cost.
I strongly suggest to treat this as a generic optimization problem, rather than k-means. K-means is an optimization too, but it may optimize the wrong function for your problem...
I would approach this by setting possible points that could be centers, i.e. your coastline.
I think this is close to Nathaniel Saul's first comment.
This way, for each iteration, instead of choosing a mean, a point out of the possible set would be chosen by proximity to the cluster.
I’ve simplified the conditions to only 2 data columns (lon. and lat.) but you should be able to extrapolate the concept. For simplicity, to demonstrate, I based this on code from here.
In this example, the purple dots are places on the coastline. If I understood correctly, the optimal Coastline locations should look something like this:
See code below:
#! /usr/bin/python3.6
# Code based on:
# https://datasciencelab.wordpress.com/2013/12/12/clustering-with-k-means-in-python/
import matplotlib.pyplot as plt
import numpy as np
import random
##### Simulation START #####
# Generate possible points.
def possible_points(n=20):
y=list(np.linspace( -1, 1, n ))
x=[-1.2]
X=[]
for i in list(range(1,n)):
x.append(x[i-1]+random.uniform(-2/n,2/n) )
for a,b in zip(x,y):
X.append(np.array([a,b]))
X = np.array(X)
return X
# Generate sample
def init_board_gauss(N, k):
n = float(N)/k
X = []
for i in range(k):
c = (random.uniform(-1, 1), random.uniform(-1, 1))
s = random.uniform(0.05,0.5)
x = []
while len(x) < n:
a, b = np.array([np.random.normal(c[0], s), np.random.normal(c[1], s)])
# Continue drawing points from the distribution in the range [-1,1]
if abs(a) < 1 and abs(b) < 1:
x.append([a,b])
X.extend(x)
X = np.array(X)[:N]
return X
##### Simulation END #####
# Identify points for each center.
def cluster_points(X, mu):
clusters = {}
for x in X:
bestmukey = min([(i[0], np.linalg.norm(x-mu[i[0]])) \
for i in enumerate(mu)], key=lambda t:t[1])[0]
try:
clusters[bestmukey].append(x)
except KeyError:
clusters[bestmukey] = [x]
return clusters
# Get closest possible point for each cluster.
def closest_point(cluster,possiblePoints):
closestPoints=[]
# Check average distance for each point.
for possible in possiblePoints:
distances=[]
for point in cluster:
distances.append(np.linalg.norm(possible-point))
closestPoints.append(np.sum(distances)) # minimize total distance
# closestPoints.append(np.mean(distances))
return possiblePoints[closestPoints.index(min(closestPoints))]
# Calculate new centers.
# Here the 'coast constraint' goes.
def reevaluate_centers(clusters,possiblePoints):
newmu = []
keys = sorted(clusters.keys())
for k in keys:
newmu.append(closest_point(clusters[k],possiblePoints))
return newmu
# Check whether centers converged.
def has_converged(mu, oldmu):
return (set([tuple(a) for a in mu]) == set([tuple(a) for a in oldmu]))
# Meta function that runs the steps of the process in sequence.
def find_centers(X, K, possiblePoints):
# Initialize to K random centers
oldmu = random.sample(list(possiblePoints), K)
mu = random.sample(list(possiblePoints), K)
while not has_converged(mu, oldmu):
oldmu = mu
# Assign all points in X to clusters
clusters = cluster_points(X, mu)
# Re-evaluate centers
mu = reevaluate_centers(clusters,possiblePoints)
return(mu, clusters)
K=3
X = init_board_gauss(30,K)
possiblePoints=possible_points()
results=find_centers(X,K,possiblePoints)
# Show results
# Show constraints and clusters
# List point types
pointtypes1=["gx","gD","g*"]
plt.plot(
np.matrix(possiblePoints).transpose()[0],np.matrix(possiblePoints).transpose()[1],'m.'
)
for i in list(range(0,len(results[0]))) :
plt.plot(
np.matrix(results[0][i]).transpose()[0], np.matrix(results[0][i]).transpose()[1],pointtypes1[i]
)
pointtypes=["bx","yD","c*"]
# Show all cluster points
for i in list(range(0,len(results[1]))) :
plt.plot(
np.matrix(results[1][i]).transpose()[0],np.matrix(results[1][i]).transpose()[1],pointtypes[i]
)
plt.show()
Edited to minimize total distance.
I'm trying to cluster a matrix of longitude and latitude values with agglomerating clustering of sklearn library.
I want to calculate in how many clusters I have to divide my points to have a maximum distance of fifty meters. I calculate the maximum distance between the points of each clusters until, in all the clusters, there is a maximum distance of fifty meters. If not, I do a clustering with n_clusters + = 1
I calculate the tree and the clusters until I find the proper separation. I would like to not have to calculate the tree in each iteration because I lose a lot of computational time.
I would like to calculate the whole tree once using the function "compute_full_tree" and then just ask the groups indicating the number of clusters I want. It's possible?
cl = 1
while(max(maxDist) > 50):
cl += 1
labels = aglomerativeClust(X, cl)
for l in range(len(X)):
X[l].cluster = str(labels[l])
clusters = [[] for o in range(max(labels)+1)]
for x in X:
clusters[int(x.cluster)].append(x)
maxDist = [0. for o in range(max(labels)+1)]
for m in range(len(clusters)):
for x in clusters[m]:
for y in clusters[m]:
dist=compute_vincenty_distance(x,y)
if(dist > maxDist[m]):
maxDist[m] = dist
def aglomerativeClust(Y, cl):
X = []
for y in Y:
X.append([y.visit.latitude,y.visit.longitude])
model = AgglomerativeClustering(linkage='complete',n_clusters=cl)
model.fit(X)
labels = model.labels_
return labels
I am using SciPy's hierarchical agglomerative clustering methods to cluster a m x n matrix of features, but after the clustering is complete, I can't seem to figure out how to get the centroid from the resulting clusters. Below follows my code:
Y = distance.pdist(features)
Z = hierarchy.linkage(Y, method = "average", metric = "euclidean")
T = hierarchy.fcluster(Z, 100, criterion = "maxclust")
I am taking my matrix of features, computing the euclidean distance between them, and then passing them onto the hierarchical clustering method. From there, I am creating flat clusters, with a maximum of 100 clusters
Now, based on the flat clusters T, how do I get the 1 x n centroid that represents each flat cluster?
A possible solution is a function, which returns a codebook with the centroids like kmeans in scipy.cluster.vq does. Only thing you need is the partition as vector with flat clusters part and the original observations X
def to_codebook(X, part):
"""
Calculates centroids according to flat cluster assignment
Parameters
----------
X : array, (n, d)
The n original observations with d features
part : array, (n)
Partition vector. p[n]=c is the cluster assigned to observation n
Returns
-------
codebook : array, (k, d)
Returns a k x d codebook with k centroids
"""
codebook = []
for i in range(part.min(), part.max()+1):
codebook.append(X[part == i].mean(0))
return np.vstack(codebook)
You can do something like this (D=number of dimensions):
# Sum the vectors in each cluster
lens = {} # will contain the lengths for each cluster
centroids = {} # will contain the centroids of each cluster
for idx,clno in enumerate(T):
centroids.setdefault(clno,np.zeros(D))
centroids[clno] += features[idx,:]
lens.setdefault(clno,0)
lens[clno] += 1
# Divide by number of observations in each cluster to get the centroid
for clno in centroids:
centroids[clno] /= float(lens[clno])
This will give you a dictionary with cluster number as the key and the centroid of the specific cluster as the value.