1.闵可夫斯基距离
闵可夫斯基距离可以概括曼哈顿距离与欧几里得距离。
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。这种海量数据的表格,用pandas处理十分方便
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 on their distance to username""" distances = [] for user in users: if user != username: distance = minkefu(users[user], users[username], 2) distances.append((distance, user)) # sort based on distance -- closest first distances.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))