用于聚类算法的编程结构 [英] Which programming structure for clustering algorithm

问题描述

我正在尝试实现以下（分裂）聚类算法（以下简称为算法），完整描述可用 here ）：

I am trying to implement the following (divisive) clustering algorithm (below is presented short form of the algorithm, the full description is available here):

从样本x开始，i = 1，...，n被视为单个n个数据点的簇和为所有对点定义的不相似矩阵D。修正阈值T以决定是否拆分集群。

Start with a sample x, i = 1, ..., n regarded as a single cluster of n data points and a dissimilarity matrix D defined for all pairs of points. Fix a threshold T for deciding whether or not to split a cluster.

1. 首先确定所有数据点对和选择一个最大距离（Dmax）的对。

1. First determine the distance between all pairs of data points and choose a pair with the largest distance (Dmax) between them.

将Dmax与T比较。如果Dmax> T，则使用所选的作为两个新集群中的第一个元素。剩下的n-2个数据点放在两个新的集群之一中。如果D（x_i，x_l）

Compare Dmax to T. If Dmax > T then divide single cluster in two by using the selected pair as the first elements in two new clusters. The remaining n - 2 data points are put into one of the two new clusters. x_l is added to the new cluster containing x_i if D(x_i, x_l) < D(x_j, x_l), otherwise is added to new cluster containing x_i.

在第二阶段，D（x_i，x_j）在群集中找到两个最大距离Dmax的两个新群集之一。如果Dmax < T，集群停止的划分，另一个集群被考虑。然后，该过程将重复从此迭代生成的集群。

At the second stage, the values D(x_i, x_j) are found within one of two new clusters to find the pair in the cluster with the largest distance Dmax between them. If Dmax < T, the division of the cluster stops and the other cluster is considered. Then the procedure repeats on the clusters generated from this iteration.

输出是集群数据记录的层次结构。我请教一个如何实现聚类算法的建议。

Output is a hierarchy of clustered data records. I kindly ask for an advice how to implement the clustering algorithm.

编辑1：我附加了Python函数，它定义了距离（相关系数）和在数据矩阵中找到最大距离的函数。

EDIT 1: I attach Python function which defines distance (correlation coefficient) and function which finds maximal distance in data matrix.

# Read data from GitHub
import pandas as pd
df = pd.read_csv('https://raw.githubusercontent.com/nico/collectiveintelligence-book/master/blogdata.txt', sep = '\t', index_col = 0)
data = df.values.tolist()
data = data[1:10]

# Define correlation coefficient as distance of choice
def pearson(v1, v2):
  # Simple sums
  sum1 = sum(v1)
  sum2 = sum(v2)
  # Sums of the squares
  sum1Sq = sum([pow(v, 2) for v in v1])
  sum2Sq = sum([pow(v, 2) for v in v2]) 
  # Sum of the products
  pSum=sum([v1[i] * v2[i] for i in range(len(v1))])
  # Calculate r (Pearson score)
  num = pSum - (sum1 * sum2 / len(v1))
  den = sqrt((sum1Sq - pow(sum1,2) / len(v1)) * (sum2Sq - pow(sum2, 2) / len(v1)))
  if den == 0: return 0
  return num / den


# Find largest distance
dist={}
max_dist = pearson(data[0], data[0])
# Loop over upper triangle of data matrix
for i in range(len(data)):
  for j in range(i + 1, len(data)):
     # Compute distance for each pair
     dist_curr = pearson(data[i], data[j])
     # Store distance in dict
     dist[(i, j)] = dist_curr
     # Store max distance
     if dist_curr > max_dist:
       max_dist = dist_curr

编辑2：是Dschoni的答案的功能。

EDIT 2: Pasted below are functions from Dschoni's answer.

# Euclidean distance
def euclidean(x,y):
  x = numpy.array(x)
  y = numpy.array(y) 
  return numpy.sqrt(numpy.sum((x-y)**2))

# Create matrix
def dist_mat(data):
  dist = {}
  for i in range(len(data)):
    for j in range(i + 1, len(data)):
      dist[(i, j)] = euclidean(data[i], data[j])
  return dist


# Returns i & k for max distance
def my_max(dict):
    return max(dict)

# Sort function
list1 = []
list2 = []
def sort (rcd, i, k):
  list1.append(i)
  list2.append(k)
  for j in range(len(rcd)):
    if (euclidean(rcd[j], rcd[i]) < euclidean(rcd[j], rcd[k])):
      list1.append(j)
    else:
      list2.append(j)

编辑3：
当我运行@Dschoni提供的代码时，算法按预期工作。然后我修改了 create_distance_list 函数，所以我们可以计算多变量数据点之间的距离。我使用欧几里得距离。对于玩具示例，我加载 iris 数据。我只聚集数据集的前50个实例。

EDIT 3: When I run the code provided by @Dschoni the algorithm works as expected. Then I modified the create_distance_list function so we can compute distance between multivariate data points. I use euclidean distance. For toy example I load iris data. I cluster only the first 50 instances of the dataset.

import pandas as pd
df = pd.read_csv('https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data', header = None, sep = ',')
df = df.drop(4, 1)
df = df[1:50]
data = df.values.tolist()

idl=range(len(data))
dist = create_distance_list(data)
print sort(dist, idl)

结果如下： / p>

The result is as follows:

[[24]，[17]，[4]，[7]，[40]，[13]，[14] [15]，[26,27,38]，[3,16，
39]，[25]，[42]，[18，20，45]，[43]，[1，2，11 ，46]，[12,37,41]，
[5]，[21]，[22]，[10,23,28,29]，[6,34,48]，[0,8 ，33,36,44]，
[31]，[32]，[19]，[30]，[35]，[9，47]]

[[24], [17], [4], [7], [40], [13], [14], [15], [26, 27, 38], [3, 16, 39], [25], [42], [18, 20, 45], [43], [1, 2, 11, 46], [12, 37, 41], [5], [21], [22], [10, 23, 28, 29], [6, 34, 48], [0, 8, 33, 36, 44], [31], [32], [19], [30], [35], [9, 47]]

某些数据点仍然聚集在一起。我通过在 sort 函数中的实际字典中添加少量的数据噪音来解决这个问题：

Some data points are still clustered together. I solve this problem by adding small amount of data noise to actual dictionary in the sort function:

# Add small random noise
for key in actual:    
  actual[key] +=  np.random.normal(0, 0.005)

任何想法如何正确解决这个问题？

Any idea how to solve this problem properly?

推荐答案

欧几里德距离的正确工作示例：

A proper working example for the euclidean distance:

import numpy as np
#For random number generation


def create_distance_list(l):
'''Create a distance list for every
unique tuple of pairs'''
    dist={}
    for i in range(len(l)):
        for k in range(i+1,len(l)):
            dist[(i,k)]=abs(l[i]-l[k])
    return dist

def maximum(distance_dict):
'''Returns the key of the maximum value if unique
or a random key with the maximum value.'''
    maximum = max(distance_dict.values())
    max_key = [key for key, value in distance_dict.items() if value == maximum]
    if len(max_key)>1:
        random_key = np.random.random_integers(0,len(max_key)-1)
        return (max_key[random_key],)
    else:
        return max_key

def construct_new_dict(distance_dict,index_list):
'''Helper function to create a distance map for a subset
of data points.'''
    new={}
    for i in range(len(index_list)):
        for k in range(i+1,len(index_list)):
            m = index_list[i]
            n = index_list[k]
            new[(m,n)]=distance_dict[(m,n)]
    return new

def sort(distance_dict,idl,threshold=4):
    result=[idl]
    i=0
    try:
        while True:
            if len(result[i])>=2:
            actual=construct_new_dict(dist,result[i]) 
                act_max=maximum(actual)
                if distance_dict[act_max[0]]>threshold:
                    j = act_max[0][0]
                    k = act_max[0][1]
                    result[i].remove(j)
                    result[i].remove(k)
                    l1=[j]
                    l2=[k]
                    for iterr in range(len(result[i])):
                        s = result[i][iterr]
                        if s>j:
                            c1=(j,s)
                        else:
                            c1=(s,j)
                        if s>k:
                            c2=(k,s)
                        else: 
                            c2=(s,k)
                        if actual[c1]<actual[c2]:
                            l1.append(s)
                        else:
                            l2.append(s)
                    result.remove(result[i])
    #What to do if distance is equal?
                    l1.sort()
                    l2.sort()
                    result.append(l1)
                    result.append(l2)
                else:
                    i+=1
            else:
                i+=1
    except:
        return result


#This is the dataset
a = [1,2,2.5,5]
#Giving each entry a unique ID
idl=range(len(a))
dist = create_distance_list(a)
print sort(dist,idl)  

我编写的代码是为了可读性，有很多东西可以做得更快，更多可靠和漂亮。这只是给你一个如何做的想法。

I wrote the code for readability, there is a lot of stuff that can made faster, more reliable and prettier. This is just to give you an idea of how it can be done.

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