CODEWITHSUNDEEP
rmatrix
digestivee
digestivee

Reputation: 740

R - Construct conditional probability matrix

I have a dataframe which consists of four columns, the two first are used for identifying a user and a product, and the last two are conditional probabilities. My final dataframe looks like this:

         id1    id2       p(id2|id1)   p(id1|id2)
1        1      1         0.1111111    4.290376e-04
2        1      2         0.22222222   8.286866e-03
3        1      3         0.22222222   2.639876e-04
4        1      4         0.44444444   2.850284e-03
5        2      1         0.09090909   1.644470e-03
6        2      5         0.2727273    3.286420e-04
7        2      6         0.4545455    1.002740e-03
8        2      3         0.1818182    1.738019e-05

and with many more users coming after. As you can see, we can have more than one different value for id2 belonging to the same id1. I want to find the probability of getting a certain id2, given that a user already has some id2, i.e. I am interested in finding 

p(id2 = x | id2 = y) = sum_id1 ( p(id2 = x | id1 ) * p(id1 | id2 = y) )

and construct it as a matrix for all x and y. In this case we have 6 different id2, so the resulting matrix should look something like this

     1             2              3             4       5      6
1    NA            0.0009207628   3.091197e-05  ....    ....   ....
2    9.534169e-05  NA             ...
3    0.0003943363  ...
4    ...
5    ...
6    ...

We get the element (1,2) as 

p(id2=1 | id2 = 2) = p(id2 = 1 | id1 = 1) * p(id1 = 1 | id2 = 2) 
= 0.1111111*8.286866e-03 = 0.0009207628.

For element (1,3) we get 

p(id2 = 1 | id2 = 3) = p(id2 = 1 | id1 = 1) * p(id1 = 1 | id2 = 3)
+ p(id2 = 1 | id1 = 2) * p(id1 = 2 | id2 = 3) 
= 0.1111111 * 2.639876e-04 + 0.09090909 * 1.738019e-05 = 3.091197e-05

I hope it is clear what I want to accomplish. Does anyone have any idea how I can construct this matrix in R?

Thanks in advance

Upvotes: 2

Views: 388

Answers (1)

Oriol Mirosa
Oriol Mirosa

Reputation: 2826

It's not pretty, but I think this works:

vec1 = c(0.1111111, 0.22222222, 0.22222222, 0.44444444, 0.09090909,
         0.2727273, 0.4545455, 0.1818182)
vec2 = c(4.290376e-04, 8.286866e-03, 2.639876e-04, 2.850284e-03,
         1.644470e-03, 3.286420e-04, 1.002740e-03, 1.738019e-05)

df <- data.frame(id1 = rep(c(1, 2), each = 4),
                 id2 = c(seq.int(4), 1, 5, 6, 3),
                 p.id2.id1 = vec1,
                 p.id1.id2 = vec2)

> df    
##   id1 id2  p.id2.id1    p.id1.id2
## 1   1   1 0.11111110 4.290376e-04
## 2   1   2 0.22222222 8.286866e-03
## 3   1   3 0.22222222 2.639876e-04
## 4   1   4 0.44444444 2.850284e-03
## 5   2   1 0.09090909 1.644470e-03
## 6   2   5 0.27272730 3.286420e-04
## 7   2   6 0.45454550 1.002740e-03
## 8   2   3 0.18181820 1.738019e-05

mat = matrix(NA, nrow = length(unique(df$id2)), ncol = length(unique(df$id2)))

for (i in 1:length(unique(df$id2))) {
  for (j in 1:length(unique(df$id2))) {
    if (i != j) {
      val = 0
      for (k in 1:length(unique(df$id1))) {
        if (length(df$p.id2.id1[df$id1 == k & df$id2 == i]) > 0 &
            length(df$p.id2.id1[df$id1 == k & df$id2 == j]) > 0) {
          val <-val + df$p.id2.id1[df$id1 == k & df$id2 == i] *
            df$p.id1.id2[df$id1 == k & df$id2 == j]
        }
      }
      mat[i, j] <- val
    }
  }  
}

Here's the result:

> mat

##              [,1]         [,2]         [,3]         [,4]         [,5]         [,6]
## [1,]           NA 0.0009207628 3.091197e-05 0.0003166982 2.987655e-05 9.115818e-05
## [2,] 9.534169e-05           NA 5.866391e-05 0.0006333964 0.000000e+00 0.000000e+00
## [3,] 3.943363e-04 0.0018415258           NA 0.0006333964 5.975310e-05 1.823164e-04
## [4,] 1.906834e-04 0.0036830515 1.173278e-04           NA 0.000000e+00 0.000000e+00
## [5,] 4.484919e-04 0.0000000000 4.740052e-06 0.0000000000           NA 2.734746e-04
## [6,] 7.474864e-04 0.0000000000 7.900087e-06 0.0000000000 1.493827e-04           NA

Upvotes: 1

Related Questions