数学SVD推荐系统，内环路

Question

我翻译以下SVD recommendation system，用Ruby编写的，分配问题数学：

require 'linalg' 

users = { 1 => "Ben", 2 => "Tom", 3 => "John", 4 => "Fred" } 
m = Linalg::DMatrix[ 
      #Ben, Tom, John, Fred 
      [5,5,0,5], # season 1 
      [5,0,3,4], # season 2 
      [3,4,0,3], # season 3 
      [0,0,5,3], # season 4 
      [5,4,4,5], # season 5 
      [5,4,5,5] # season 6 
      ] 

# Compute the SVD Decomposition 
u, s, vt = m.singular_value_decomposition 
vt = vt.transpose 

# Take the 2-rank approximation of the Matrix 
# - Take first and second columns of u (6x2) 
# - Take first and second columns of vt (4x2) 
# - Take the first two eigen-values (2x2) 
u2 = Linalg::DMatrix.join_columns [u.column(0), u.column(1)] 
v2 = Linalg::DMatrix.join_columns [vt.column(0), vt.column(1)] 
eig2 = Linalg::DMatrix.columns [s.column(0).to_a.flatten[0,2], s.column(1).to_a.flatten[0,2]] 

# Here comes Bob, our new user 
bob = Linalg::DMatrix[[5,5,0,0,0,5]] 
bobEmbed = bob * u2 * eig2.inverse 

# Compute the cosine similarity between Bob and every other User in our 2-D space 
user_sim, count = {}, 1 
v2.rows.each { |x| 
    user_sim[count] = (bobEmbed.transpose.dot(x.transpose))/(x.norm * bobEmbed.norm) 
    count += 1 
    } 

# Remove all users who fall below the 0.90 cosine similarity cutoff and sort by similarity 
similar_users = user_sim.delete_if {|k,sim| sim < 0.9 }.sort {|a,b| b[1] <=> a[1] } 
similar_users.each { |u| printf "%s (ID: %d, Similarity: %0.3f) \\n", users[u[0]], u[0], u[1] } 

# We'll use a simple strategy in this case: 
# 1) Select the most similar user 
# 2) Compare all items rated by this user against your own and select items that you have not yet rated 
# 3) Return the ratings for items I have not yet seen, but the most similar user has rated 
similarUsersItems = m.column(similar_users[0][0]-1).transpose.to_a.flatten 
myItems = bob.transpose.to_a.flatten 

not_seen_yet = {} 
myItems.each_index { |i| 
    not_seen_yet[i+1] = similarUsersItems[i] if myItems[i] == 0 and similarUsersItems[i] != 0 
} 

printf "\\n %s recommends: \\n", users[similar_users[0][0]] 
not_seen_yet.sort {|a,b| b[1] <=> a[1] }.each { |item| 
    printf "\\tSeason %d .. I gave it a rating of %d \\n", item[0], item[1] 
} 

print "We've seen all the same seasons, bugger!" if not_seen_yet.size == 0 

以下是相应的Mathematica代码：

Clear[s, u, v, s2, u2, v2, m, n, testdata, trainingdata, user, user2d]; 
find1nn[trainingdata_, user_] := { 
    {u , s, v} = SingularValueDecomposition[Transpose[trainingdata]]; 
    (* Reducr to 2 dimensions. *) 
    u2 = u[[All, {1, 2}]]; 
    s2 = s[[{1, 2}, {1, 2}]]; 
    v2 = v[[All, {1, 2}]]; 
    user2d = user.u2.Inverse[s2]; 
    {m, n} = Dimensions[v2]; 
    closest = -1; 
    index = -1; 
    For[a = 1, a < m, a++, 
    {distance = 1 - CosineDistance[v2[[a, {1, 2}]], user2d];, 
     If[distance > closest, {closest = distance, index = a}];}]; 
    closestuserratings = trainingdata[[index]]; 
    closestuserratings 
    } 
rec[closest_, userx_] := { 
    d = Dimensions[closest]; 
    For[b = 1, b <= d[[2]], b++, 
    If[userx[[b]] == 0., userx[[b]] = closest[[1, b]]] 
    ] 
    userx 
    } 
finalrec[td_, user_] := rec[find1nn[td, user], user] 
(*Clear[s,u,v,s2,u2,v2,m,n,testdata,trainingdata,user,user2d]*) 
testdata = {{5., 5., 3., 0., 5., 5.}, {5., 0., 4., 1., 4., 4.}, {0., 
    3., 0., 5., 4., 5.}, {5., 4., 3., 3., 5., 5.}}; 
bob = {5., 0., 4., 0., 4., 5.}; 
(*recommend[testdata,bob]*) 
find1nn[testdata, bob] 
finalrec[testdata, bob] 

对于某些原因，它没有分配用户内部的索引，但在外面。什么可能导致这种情况发生？

Answer 1

请查阅数学文档中变量的本地化教程。问题在于你的rec功能。问题在于你无法正常修改Mathematica中的输入变量（如果你的函数有一个保持属性，那么你可以做到这一点，以便将所讨论的参数传递给它未评估，但事实并非如此这里）：

rec[closest_, userxi_] := 
Block[{d, b, userx = userxi}, {d = Dimensions[closest]; 
    For[b = 1, b <= d[[2]], b++, 
    If[userx[[b]] == 0., userx[[b]] = closest[[1, b]]]]; 
    userx} 

Answer 2

没有试图了解你想要达到的，在这里你有一个更Mathematca十岁上下，但相当于（我希望）工作的代码。

显式循环都没有了，而且很多不必要的瓦尔消除。所有变量现在都是本地的，所以不需要使用Clear []。

find1nn[trainingdata_, user_] := 
    Module[{u, s, v, v2, user2d, m, distances}, 
    {u, s, v} = SingularValueDecomposition[Transpose[trainingdata]]; 
    v2 = v[[All, {1, 2}]]; 
    user2d = user.u[[All, {1, 2}]].Inverse[s[[{1, 2}, {1, 2}]]]; 
    m = First@Dimensions[v2]; 
    distances = (1 - CosineDistance[v2[[#, {1, 2}]], user2d]) & /@ Range[m - 1]; 
    {trainingdata[[Ordering[distances][[-1]]]]}]; 

rec[closest_, userxi_] := userxi[[#]] /. {0. -> closest[[1, #]]} & /@ 
          Range[Dimensions[closest][[2]]]; 

finalrec[td_, user_] := rec[find1nn[td, user], user]; 

我相信它仍然可以优化了不少。

Answer 3

这是我在这个镜头基于贝利萨留的代码，并与Sjoerd的改进。

find1nn[trainingdata_, user_] := 
    Module[{u, s, v, user2d, distances}, 
    {u, s, v} = SingularValueDecomposition[trainingdata\[Transpose], 2]; 
    user2d = user . u . Inverse@s; 
    distances = # ~CosineDistance~ user2d & /@ Most@v; 
    trainingdata[[ distances ~Ordering~ 1 ]] 
    ] 

rec[closest_, userxi_] := If[# == 0, #2, #] & ~MapThread~ {userxi, closest[[1]]} 

Answer 4

Clear[s, u, v, s2, u2, v2, m, n, testdata, trainingdata, user, user2d]; 
recommend[trainingdata_, user_] := { 
    {u , s, v} = SingularValueDecomposition[Transpose[trainingdata]]; 
    (* Reducera till 2 dimensioner. *) 
    u2 = u[[All, {1, 2}]]; 
    s2 = s[[{1, 2}, {1, 2}]]; 
    v2 = v[[All, {1, 2}]]; 
    user2d = user.u2.Inverse[s2]; 
    {m, n} = Dimensions[v2]; 
    closest = -1; 
    index = -1; 
    For[a = 1, a < m, a++, 
    {distance = 1 - CosineDistance[v2[[a, {1, 2}]], user2d];, 
     If[distance > closest, {closest = distance, index = a}];}]; 
    closestuserratings = trainingdata[[index]]; 
    d = Dimensions[closestuserratings]; 
    updateduser = Table[0, {i, 1, d[[1]]}]; 
    For[b = 1, b <= d[[1]], b++, 
    If[user[[b]] == 0., updateduser[[b]] = closestuserratings[[b]], 
    updateduser[[b]] = user[[b]]] 
    ] 
    updateduser 
    } 
testdata = {{5., 5., 3., 0., 5., 5.}, {5., 0., 4., 1., 4., 4.}, {0., 
    3., 0., 5., 4., 5.}, {5., 4., 3., 3., 5., 5.}}; 
bob = {5., 0., 4., 0., 4., 5.}; 
recommend[testdata, bob] 

{{5。空，空为0，4空，1空，空4，5，空}}

现在的作品，但为什么空值？