Python推荐算法讲解

枫铃2年前 (2022-02-05)Python368

1.闵可夫斯基距离：计算用户相似度

闵可夫斯基距离可以概括曼哈顿距离与欧几里得距离。
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其中r越大，单个维度差值大小会对整体产生更大的影响。这个很好理解，假设当r=2时一个正方形对角线长度，永远是r=3时正方体对角线的投影，因此r越大，单个维度差异会有更大影响。（所以这也可能是很多公司的推荐算法并不准确的原因之一）

我们在对一个新用户进行推荐时，可以计算在同等维度下其他用户的闵可夫斯基距离。这种海量数据的二维表格，用pandas处理十分方便
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下面有一个闵可夫距离计算的实例

from math import sqrt

users = {"Angelica": {"Blues Traveler": 3.5, "Broken Bells": 2.0, "Norah Jones": 4.5, "Phoenix": 5.0, "Slightly Stoopid": 1.5, "The Strokes": 2.5, "Vampire Weekend": 2.0},
         "Bill":{"Blues Traveler": 2.0, "Broken Bells": 3.5, "Deadmau5": 4.0, "Phoenix": 2.0, "Slightly Stoopid": 3.5, "Vampire Weekend": 3.0},
         "Chan": {"Blues Traveler": 5.0, "Broken Bells": 1.0, "Deadmau5": 1.0, "Norah Jones": 3.0, "Phoenix": 5, "Slightly Stoopid": 1.0},
         "Dan": {"Blues Traveler": 3.0, "Broken Bells": 4.0, "Deadmau5": 4.5, "Phoenix": 3.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 2.0},
         "Hailey": {"Broken Bells": 4.0, "Deadmau5": 1.0, "Norah Jones": 4.0, "The Strokes": 4.0, "Vampire Weekend": 1.0},
         "Jordyn":  {"Broken Bells": 4.5, "Deadmau5": 4.0, "Norah Jones": 5.0, "Phoenix": 5.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 4.0},
         "Sam": {"Blues Traveler": 5.0, "Broken Bells": 2.0, "Norah Jones": 3.0, "Phoenix": 5.0, "Slightly Stoopid": 4.0, "The Strokes": 5.0},
         "Veronica": {"Blues Traveler": 3.0, "Norah Jones": 5.0, "Phoenix": 4.0, "Slightly Stoopid": 2.5, "The Strokes": 3.0}
        }


def minkefu(rating1, rating2, n):
    """Computes the Manhattan distance. Both rating1 and rating2 are dictionaries
       of the form {'The Strokes': 3.0, 'Slightly Stoopid': 2.5}"""
    distance = 0
    commonRatings = False 
    for key in rating1:
        if key in rating2:
            distance += abs((rating1[key] - rating2[key])**n)
            commonRatings = True
    if commonRatings:
        return distance**1/n
    else:
        return -1 #Indicates no ratings in common


def computeNearestNeighbor(username, users):
    """creates a sorted list of users based>= []
    for user in users:
        if user != username:
            distance = minkefu(users[user], users[username], 2)
            distances.append((distance, user))
    # sort based>.sort()
    return distances

def recommend(username, users):
    """Give list of recommendations"""
    # first find nearest neighbor
    nearest = computeNearestNeighbor(username, users)[0][1]

    recommendations = []
    # now find bands neighbor rated that user didn't
    neighborRatings = users[nearest]
    userRatings = users[username]
    for artist in neighborRatings:
        if not artist in userRatings:
            recommendations.append((artist, neighborRatings[artist]))
    # using the fn sorted for variety - sort is more efficient
    return sorted(recommendations, key=lambda artistTuple: artistTuple[1], reverse = True)

# examples - uncomment to run

print( recommend('Hailey', users))

2.皮尔逊相关系数：如何解决主观评价差异带来的推荐误差

在第一部分提到r越大，单个维度差值大小会对整体产生更大的影响。因此，我们需要有一个解决方案来应对个体的主观评价差异。这个东西就是皮尔逊相关系数。
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上述公式计算复杂度很大，需要进行n!*m!次遍历，后续有一个近似计算公式可以大大降低算法复杂度。

皮尔逊相关系数（-1，1）用于衡量两个向量（用户）的相关性，如两个用户意见基本一致，那皮尔逊相关系数靠近1，如果两个用户意见基本相反，那皮尔逊相关系数结果靠近-1。在这里，我们需要弄明白两个问题：

（1）怎么确定多维向量之间的皮尔逊相关系数

（2）怎么利用闵可夫斯基距离结合起来，优化我们的推荐模型

对第（1）问题，在这里有一个近似计算公式
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用代码来表示则为

def pearson(rating1, rating2):
    sum_xy = 0
    sum_x = 0
    sum_y = 0
    sum_x2 = 0
    sum_y2 = 0
    n = 0
    for key in rating1:
        if key in rating2:
            n += 1
            x = rating1[key]
            y = rating2[key]
            sum_xy += x * y
            sum_x += x
            sum_y += y
            sum_x2 += pow(x, 2)
            sum_y2 += pow(y, 2)
    # now compute denominator
    denominator = sqrt(sum_x2 - pow(sum_x, 2) / n) * sqrt(sum_y2 - pow(sum_y, 2) / n)
    if denominator == 0:
        return 0
    else:
        return (sum_xy - (sum_x * sum_y) / n) / denominator