『科学计算』层次聚类实现

层次聚类理论自行百度，这里是一个按照我的理解的简单实现，

我们先看看数据，

啤酒名 热量 钠含量 酒精 价格Budweiser 144.00 19.00 4.70 .43Schlitz 181.00 19.00 4.90 .43Ionenbrau 157.00 15.00 4.90 .48Kronensourc 170.00 7.00 5.20 .73Heineken 152.00 11.00 5.00 .77Old-milnaukee 145.00 23.00 4.60 .26Aucsberger 175.00 24.00 5.50 .40Strchs-bohemi 149.00 27.00 4.70 .42Miller-lite 99.00 10.00 4.30 .43Sudeiser-lich 113.00 6.00 3.70 .44Coors 140.00 16.00 4.60 .44Coorslicht 102.00 15.00 4.10 .46Michelos-lich 135.00 11.00 4.20 .50Secrs 150.00 19.00 4.70 .76Kkirin 149.00 6.00 5.00 .79Pabst-extra-l 68.00 15.00 2.30 .36Hamms 136.00 19.00 4.40 .43Heilemans-old 144.00 24.00 4.90 .43Olympia-gold- 72.00 6.00 2.90 .46Schlite-light 97.00 7.00 4.20 .47 

程序如下，

import numpy as np
import pandas as pd

data = pd.read_csv(‘./bear.txt‘, delim_whitespace=True)
X = np.array(data.ix[:,1:])
names = [[name] for name in data.ix[:,0]]

def cluster_step(X,names):
    dis = np.empty([len(X),len(X)])
    for i in range(len(X)):
        for j in range(len(X)):
            dis[i][j] = np.sqrt(np.sum(np.square(X[i] - X[j])))
            if i == j:
                dis[i][j] = 999
    x, y = [(np.argmin(dis))//len(X), np.mod(np.argmin(dis),len(X))]
    X[x] = (X[x] + X[y])/2
    X = np.delete(X, y, axis=0)
    names[x].extend(names[y])
    names.remove(names[y])
    return x, y, X, names, dis

def cluster(X, num, names):
    classes = len(X)
    while classes != num:
        _x, _y, X, names, _dis = cluster_step(X, names)
        with open(‘./result.txt‘, ‘a‘) as f:
            f.write(‘\r‘+str(_x))
            f.write(‘\r‘+str(_y))
            f.write(‘\r‘ + str(_dis[_x,_y]))
            f.write(‘\r‘+str(_dis))
            f.write(‘\r‘+str(names))
            f.flush()
        classes -= 1
    return names

if __name__==‘__main__‘:
    names = cluster(X, 4, names)

规则是每次合并后去中心点（每一步会合并两个位置，取均值做新位置）作为类簇位置，距离使用的是欧式距离。

实际上由于每次合并后下一次的节点会减少，和最初的20个点就对不上了，头疼了好一会，后来想到在每一次迭代中把每一个种类名按照类去合并，这样就不需要在最后利用索引去复原啤酒种类了，感觉挺机智。由于这样直接说不直观，我下面给出中间输出，

[[‘Budweiser‘], [‘Schlitz‘], [‘Ionenbrau‘], [‘Kronensourc‘], [‘Heineken‘], [‘Old-milnaukee‘, ‘Heilemans-old‘], [‘Aucsberger‘], [‘Strchs-bohemi‘], [‘Miller-lite‘], [‘Sudeiser-lich‘], [‘Coors‘], [‘Coorslicht‘], [‘Michelos-lich‘], [‘Secrs‘], [‘Kkirin‘], [‘Pabst-extra-l‘], [‘Hamms‘], [‘Olympia-gold-‘], [‘Schlite-light‘]]

[[‘Budweiser‘], [‘Schlitz‘], [‘Ionenbrau‘], [‘Kronensourc‘], [‘Heineken‘], [‘Old-milnaukee‘, ‘Heilemans-old‘], [‘Aucsberger‘], [‘Strchs-bohemi‘], [‘Miller-lite‘, ‘Schlite-light‘], [‘Sudeiser-lich‘], [‘Coors‘], [‘Coorslicht‘], [‘Michelos-lich‘], [‘Secrs‘], [‘Kkirin‘], [‘Pabst-extra-l‘], [‘Hamms‘], [‘Olympia-gold-‘]]

[[‘Budweiser‘, ‘Old-milnaukee‘, ‘Heilemans-old‘], [‘Schlitz‘], [‘Ionenbrau‘], [‘Kronensourc‘], [‘Heineken‘], [‘Aucsberger‘], [‘Strchs-bohemi‘], [‘Miller-lite‘, ‘Schlite-light‘], [‘Sudeiser-lich‘], [‘Coors‘], [‘Coorslicht‘], [‘Michelos-lich‘], [‘Secrs‘], [‘Kkirin‘], [‘Pabst-extra-l‘], [‘Hamms‘], [‘Olympia-gold-‘]]

[[‘Budweiser‘, ‘Old-milnaukee‘, ‘Heilemans-old‘], [‘Schlitz‘], [‘Ionenbrau‘], [‘Kronensourc‘], [‘Heineken‘], [‘Aucsberger‘], [‘Strchs-bohemi‘], [‘Miller-lite‘, ‘Schlite-light‘], [‘Sudeiser-lich‘], [‘Coors‘, ‘Hamms‘], [‘Coorslicht‘], [‘Michelos-lich‘], [‘Secrs‘], [‘Kkirin‘], [‘Pabst-extra-l‘], [‘Olympia-gold-‘]]

... ... ...

每次list长度减少1，某个子list长度加一这样

查看一下输出，

names
Out[1]:
[[‘Budweiser‘,
‘Old-milnaukee‘,
‘Heilemans-old‘,
‘Secrs‘,
‘Strchs-bohemi‘,
‘Ionenbrau‘,
‘Heineken‘,
‘Kkirin‘,
‘Coors‘,
‘Hamms‘,
‘Michelos-lich‘],
[‘Schlitz‘, ‘Aucsberger‘, ‘Kronensourc‘],
[‘Miller-lite‘, ‘Schlite-light‘, ‘Coorslicht‘, ‘Sudeiser-lich‘],
[‘Pabst-extra-l‘, ‘Olympia-gold-‘]]