算法4-5.1.1鍵索引計數法
阿新 • • 發佈:2019-03-10
one sel tin jackson har ins 5.1 索引 row
test_case =[('Anderson',2),('Brown',3),('Davis',3),('Garcia',4),('Harris',1),('Jackson',3),('Johnson',4), ('Jones',3),('Martin',1),('Martinez',2),('Miller',2),('Moore',1),('Robinson',2),('Smith',4), ('Taylor',3),('Thomas',4),('Thompson',4),('White',2),('Williams',3),('Wilson',4)] class Student(): def __init__(self,name,key): self.name = name self.key = key a=[] for name,key in test_case: a.append(Student(name,key)) count =[0 for i in range(6)] # 計算出現頻率 for i in range(len(a)): count[a[i].key+1]+=1 # 將頻率轉換位起始索引 for i in range(len(count)-1): count[i+1]+=count[i] # 數據分類 aux = [0 for i in range(len(a))] for i in range(len(a)): aux[count[a[i].key]] = a[i] count[a[i].key] += 1 for st in aux: print(st.name,st.key)
算法4-5.1.1鍵索引計數法