In [1]:
import pandas as pd
from sklearn.cluster import KMeans
typelabel = {'肝气郁结证型系数':'A',
'热毒蕴结证型系数':'B',
'冲任失调证型系数':'C',
'气血两虚证型系数':'D',
'脾胃虚弱证型系数':'E',
'肝肾阴虚证型系数':'F'}
k = 4
data = pd.read_excel('/home/jeff/python_data/chapter8/chapter8/demo/data/data.xls')
keys = list(typelabel.keys())
result = pd.DataFrame()
for i in range(len(keys)):
# 用kmeans算法进行聚类离散化
print('正在进行”%s“的聚类...' % keys[i])
kmodel = KMeans(n_clusters = k, n_jobs = 4)
kmodel.fit(data[[keys[i]]].as_matrix())
r1 = pd.DataFrame(kmodel.cluster_centers_, columns = [typelabel[keys[i]]])
r2 = pd.Series(kmodel.labels_).value_counts() # 分类统计
r2 = pd.DataFrame(r2, columns = [typelabel[keys[i]] + 'n'])
r = pd.concat([r1, r2],axis = 1).sort_values(typelabel[keys[i]])
r.index = [1, 2, 3, 4]
r.loc[:, typelabel[keys[i]]] = r[typelabel[keys[i]]].rolling(window = 2).mean()
r.loc[1, typelabel[keys[i]]] = 0.0
result = result.append(r.T)
result = result.sort_index()
result.round(2)
Out[1]:
In [2]:
#自定义连接函数，用于实现L_{k-1}到C_k的连接
def connect_string(x, ms):
x = list(map(lambda i:sorted(i.split(ms)), x))
l = len(x[0])
r = []
for i in range(len(x)):
for j in range(i,len(x)):
if x[i][:l-1] == x[j][:l-1] and x[i][l-1] != x[j][l-1]:
r.append(x[i][:l-1]+sorted([x[j][l-1],x[i][l-1]]))
return r
#寻找关联规则的函数
def find_rule(d, support, confidence, ms = '--'):
result = pd.DataFrame(index=['support', 'confidence']) #定义输出结果
support_series = 1.0*d.sum()/len(d) #支持度序列
column = list(support_series[support_series > support].index) #初步根据支持度筛选
k = 0
while len(column) > 1:
k = k+1
print(u'\n正在进行第%s次搜索...' %k)
column = connect_string(column, ms)
print(u'数目：%s...' %len(column))
sf = lambda i: d[i].prod(axis=1, numeric_only = True) #新一批支持度的计算函数
#创建连接数据，这一步耗时、耗内存最严重。当数据集较大时，可以考虑并行运算优化。
d_2 = pd.DataFrame(list(map(sf,column)), index = [ms.join(i) for i in column]).T
support_series_2 = 1.0*d_2[[ms.join(i) for i in column]].sum()/len(d) #计算连接后的支持度
column = list(support_series_2[support_series_2 > support].index) #新一轮支持度筛选
support_series = support_series.append(support_series_2)
column2 = []
for i in column: #遍历可能的推理，如{A,B,C}究竟是A+B-->C还是B+C-->A还是C+A-->B？
i = i.split(ms)
for j in range(len(i)):
column2.append(i[:j]+i[j+1:]+i[j:j+1])
cofidence_series = pd.Series(index=[ms.join(i) for i in column2]) #定义置信度序列
for i in column2: #计算置信度序列
cofidence_series[ms.join(i)] = support_series[ms.join(sorted(i))]/support_series[ms.join(i[:len(i)-1])]
for i in cofidence_series[cofidence_series > confidence].index: #置信度筛选
result[i] = 0.0
result[i]['confidence'] = cofidence_series[i]
result[i]['support'] = support_series[ms.join(sorted(i.split(ms)))]
result = result.T.sort_values(['confidence','support'], ascending = False) #结果整理，输出
print(u'\n结果为：')
print(result)
return result
In [3]:
import pandas as pd
import time
data = pd.read_csv('/home/jeff/python_data/chapter8/chapter8/demo/data/apriori.txt', header = None, dtype=object)
start = time.clock()
print('转换原始数据为0-1矩阵')
ct = lambda x : pd.Series(1, index = x[pd.notnull(x)])
b = map(ct, data.as_matrix())
data = pd.DataFrame(list(b)).fillna(0)
end = time.clock()
print('\n转换完毕，用时：%0.2f秒' %(end - start))
del b
support = 0.06
confidence = 0.75
ms = '---'
start = time.clock()
print('开始搜索关联规则...')
find_rule(data, support, confidence, ms)
end = time.clock()
print('\n搜索完成，用时：%0.2f秒' %(end - start))