在Python中优化用于创建一起评级的项目列表的算法
发布时间：2020-12-20 11:14:18 所属栏目：Python 来源：网络整理
导读：给出购买事件列表(customer_id,item) 1-hammer1-screwdriver1-nails2-hammer2-nails3-screws3-screwdriver4-nails4-screws 我正在尝试构建一个数据结构,告诉我用另一个项目购买商品的次数.不是同时买的,而是因为我开始保存数据而买的.结果看起来像 { hammer
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给出购买事件列表(customer_id,item)
1-hammer 1-screwdriver 1-nails 2-hammer 2-nails 3-screws 3-screwdriver 4-nails 4-screws 我正在尝试构建一个数据结构,告诉我用另一个项目购买商品的次数.不是同时买的,而是因为我开始保存数据而买的.结果看起来像 {
hammer : {screwdriver : 1,nails : 2},screwdriver : {hammer : 1,screws : 1,nails : 1},screws : {screwdriver : 1,nails : {hammer : 1,screwdriver : 1}
}
表示用钉子两次(人1,3)和一把螺丝刀(人1)买了一把锤子,用螺丝刀买了一次螺钉(人3),依此类推…… 我目前的做法是 users = dict其中userid是键,而购买的项目列表是值 pseudo:
for each event(customer,item)(sorted by item):
add user to users dict if not exists,and add the items
add item to items dict if not exists,and add the user
----------
for item,user in rows:
# add the user to the users dict if they don't already exist.
users[user]=users.get(user,[])
# append the current item_id to the list of items rated by the current user
users[user].append(item)
if item != last_item:
# we just started a new item which means we just finished processing an item
# write the userlist for the last item to the usersForItem dictionary.
if last_item != None:
usersForItem[last_item]=userlist
userlist=[user]
last_item = item
items.append(item)
else:
userlist.append(user)
usersForItem[last_item]=userlist
所以,在这一点上,我有两个决定 – 谁买了什么,以及谁买了什么.这是它变得棘手的地方.现在填充了usersForItem,我遍历它,遍历购买该项目的每个用户,并查看用户的其他购买.我承认这不是最狡猾的做事方式 – 我试图确保在得到Python之前得到正确的结果(我是). relatedItems = {}
for key,listOfUsers in usersForItem.iteritems():
relatedItems[key]={}
related=[]
for ux in listOfReaders:
for itemRead in users[ux]:
if itemRead != key:
if itemRead not in related:
related.append(itemRead)
relatedItems[key][itemRead]= relatedItems[key].get(itemRead,0) + 1
calc jaccard/tanimoto similarity between relatedItems[key] and its values
有没有更有效的方法可以做到这一点？此外,如果这种类型的操作有适当的学术名称,我很乐意听到它. 编辑：澄清包括这样一个事实,即我不会限制购买同时购买的物品.物品可以随时购买. 解决方法events = """
1-hammer
1-screwdriver
1-nails
2-hammer
2-nails
3-screws
3-screwdriver
4-nails
4-screws""".splitlines()
events = sorted(map(str.strip,e.split('-')) for e in events)
from collections import defaultdict
from itertools import groupby
# tally each occurrence of each pair of items
summary = defaultdict(int)
for val,items in groupby(events,key=lambda x:x[0]):
items = sorted(it[1] for it in items)
for i,item1 in enumerate(items):
for item2 in items[i+1:]:
summary[(item1,item2)] += 1
summary[(item2,item1)] += 1
# now convert raw pair counts into friendlier lookup table
pairmap = defaultdict(dict)
for k,v in summary.items():
item1,item2 = k
pairmap[item1][item2] = v
# print the results
for k,v in sorted(pairmap.items()):
print k,':',v
得到： hammer : {'nails': 2,'screwdriver': 1}
nails : {'screws': 1,'hammer': 2,'screwdriver': 1}
screwdriver : {'screws': 1,'nails': 1,'hammer': 1}
screws : {'nails': 1,'screwdriver': 1}
(这通过购买事件解决您的初始请求分组项目.要按用户分组,只需将事件列表的第一个键从事件编号更改为用户ID.) （编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |