【集体智慧编程】提供推荐

一、协作型过滤(Collaborative Filtering)


二、寻找相近用户

数据集

critics = {
'Lisa Rose':
{'Lady in the Water': 2.5, 'Snakes on a Plane': 3.5, 'Just My Luck': 3.0, 'Superman Returns': 3.5, 'You, Me and Dupree': 2.5, 'The Night Listener': 3.0},

'Gene Seymour':
{'Lady in the Water': 3.0, 'Snakes on a Plane': 3.5, 'Just My Luck': 1.5, 'Superman Returns': 5.0, 'The Night Listener': 3.0, 'You, Me and Dupree': 3.5},

'Michael Phillips':
{'Lady in the Water': 2.5, 'Snakes on a Plane': 3.0, 'Superman Returns': 3.5, 'The Night Listener': 4.0},

'Claudia Puig':
{'Snakes on a Plane': 3.5, 'Just My Luck': 3.0, 'The Night Listener': 4.5, 'Superman Returns': 4.0, 'You, Me and Dupree': 2.5},

'Mick LaSalle':
{'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0, 'Just My Luck': 2.0, 'Superman Returns': 3.0, 'The Night Listener': 3.0, 'You, Me and Dupree': 2.0},

'Jack Matthews':
{'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0, 'The Night Listener': 3.0, 'Superman Returns': 5.0, 'You, Me and Dupree': 3.5},

'Toby':
{'Snakes on a Plane':4.5, 'You, Me and Dupree':1.0, 'Superman Returns':4.0}
}

	Lady in the Water	Snakes on a plane	Just My Luck	Superman Returns	You, Me and Dupree	The Night Listener
Rose	2.5	3.5	3.0	3.5	2.5	3.0
Seymour	3.0	3.5	5.0	5.0	2.5	3.0
Phillips	2.5	3.0	3.5	3.5		4.0
Puig	3.5		3.0	4.0	2.5	4.5
LaSalle	3.0	4.0	2.0	3.0	2.0	3.0
Mattnews	3.0	4.0		5.0	3.5	3.0
Toby	?	4.5	?	4.0	1.0	?

欧几里得距离

>> from math import sqrt

>>sqrt(pow(x1-x2,2) + pow(y1-y2,2))

>> 1 / (1 + sqrt(pow(x1-x2,2) + pow(y1-y2,2))) ==> 归一化 0~1

def sim_distance(prefs, person1, person2):
si = {}
for item in prefs[person1]:        ===> 寻找p1和p2通过评论过的movie
if item in prefs[person2]:
si[item] = 1

if len(si) == 0:
return 0

sum_of_squares = 0.0

for item in prefs[person1]:       ==> 欧几里得距离公式计算相似度
if item in prefs[person2]:
sum_of_squares += pow(prefs[person1][item] - prefs[person2][item], 2)

return 1 / (1 + sum_of_squares)   ==> 归一化

皮尔逊相关度

http://lobert.iteye.com/blog/2024999

def sim_pearson(prefs, p1, p2):
si = {}
for item in prefs[p1]:
if item in prefs[p2]:
si[item] = 1

n = len(si)
if n == 0:
return 1

sum1 = 0.0
sum2 = 0.0
sum1Sq = 0.0
sum2Sq = 0.0
pSum = 0.0
for it in si:
sum1 += prefs[p1][it]
sum2 += prefs[p2][it]
sum1Sq += pow(prefs[p1][it], 2)
sum2Sq += pow(prefs[p2][it], 2)
pSum += prefs[p1][it] * prefs[p2][it]

num = pSum - (sum1 * sum2 / n)
den = sqrt((sum1Sq - pow(sum1, 2) / n) * (sum2Sq - pow(sum2, 2) / n))

if den == 0:
return 0

return num / den

推荐物品

为Toby推荐：

计算所有用户与Toby的相似度(sim_distance,sim_pearson)

def getRecommendations(prefs, person, similarity=sim_pearson):
totals = {}
simSums = {}

for other in prefs:
if other == person:
continue

#
sim = similarity(prefs, person, other)        ==> 计算参数person与其他所有用户的相似度

if sim <= 0:
continue

for item in prefs[other]:
if item not in prefs[person] or prefs[person][item] == 0: ==> 推荐没有看过的movie
totals.setdefault(item, 0)
totals[item] += prefs[other][item] * sim

simSums.setdefault(item, 0)
simSums[item] += sim

# rankings = []
# for item,total in totals.items():
#    rankings[total / simSums[item]] = item

rankings = [(total / simSums[item], item) for item, total in totals.items()]
rankings.sort()                                    ==> 按照相似度降序排序
rankings.reverse()

return rankings

	相似度	Night	sim * Night	Lady	sim * Lady	Luck	sim * Luck
Rose	0.99	3.0	0.99 * 3.0	2.5	0.99 * 2.5	3.0	0.99 * 3.0
Seymour	0.38	3.0	0.38 * 3.0	3.0	0.38 * 3.0	1.5	0.38 * 1.5
Puig	0.89	4.5	0.89 * 4.5			3.0	0.89 * 3.0
LaSalle	0.92	3.0	0.92 * 3.0	3.0	0.92 * 3.0	2.0	0.92 * 2.0
Matthews	0.66	3.0	0.66 * 3.0	3.0	0.66 * 3.0
总计			12.89		8.38		8.07
相似度总计			0.99+0.38+0.89+0.92+0.66=3.84		0.99+0.38+0.92+0.66=2.83		0.99+0.38+0.89+0.92=3.18
总计/相似度总计			3.35		2.83		2.53

三、基于物品的过滤

基于用户的协作型过滤，要求我们使用来自每一位的全部评分构建数据集。这种方法对于数量以千计的用户或是物品规模或是没有问题，但是对于上百万客户的商品的大型网站而言，将一个用户与其他所有用户进行比较，然后再对每位用户评过分的商品进行比较，其速度可能是无法忍受的。同样。一个商品销售量为数百万的网站，也许用户偏好方面彼此间很少见会有重叠，这可能会令用户相似性判断变得十分困难。

在拥有大量数据集的情况下，基于物品的协作型过滤能够更好的得出结论，而且允许我们将大量计算任务预先执行，从而