Question

我有一个（对称的）邻接矩阵，它是根据报纸文章中的名称（例如：Greg，Mary，Sam，Tom）的共同出现而创建的（例如：a，b，c，d）。见下文。

如何计算非零矩阵元素（http://en.wikipedia.org/wiki/Lift_(data_mining)）的升力值？

我对一个有效的实现感兴趣，它也可以用于非常大的矩阵（例如一百万个非零元素）。

我感谢任何帮助。

# Load package
library(Matrix)

# Data
A <- new("dgTMatrix"
    , i = c(2L, 2L, 2L, 0L, 3L, 3L, 3L, 1L, 1L)
    , j = c(0L, 1L, 2L, 0L, 1L, 2L, 3L, 1L, 3L)
    , Dim = c(4L, 4L)
    , Dimnames = list(c("Greg", "Mary", "Sam", "Tom"), c("a", "b", "c", "d"))
    , x = c(1, 1, 1, 1, 1, 1, 1, 1, 1)
    , factors = list()
)

# > A
# 4 x 4 sparse Matrix of class "dgTMatrix"
#      a b c d
# Greg 1 . . .
# Mary . 1 . 1
# Sam  1 1 1 .
# Tom  . 1 1 1

# One mode projection of the data 
# (i.e. final adjacency matrix, which is the basis for the lift value calculation)
A.final <- tcrossprod(A)

# > A.final
# 4 x 4 sparse Matrix of class "dsCMatrix"
#      Greg Mary Sam Tom
# Greg    1    .   1   .
# Mary    .    2   1   2
# Sam     1    1   3   2
# Tom     .    2   2   3

Answer 1

这可能会对您有所帮助，但肯定不是最有效的实施方案。

ComputeLift <- function(data, projection){
# Initialize a matrix to store the results.
lift <- matrix(NA, nrow=nrow(projection), ncol=ncol(projection))
# Select all pairs in the projection matrix
for(i in 1:nrow(projection)){
    for(j in 1:ncol(projection)){
        # The probability to observe both names in the same article is the
        # number of articles where the names appear together divided by the
        # total number of articles
        pAB <- projection[i,j]/ncol(data)
        # The probability for a name to appear in an article is the number of
        # articles where the name appears divided by the total number of articles
        pA <- sum(A[i,])/ncol(data)
        pB <- sum(A[j,])/ncol(data)
        # The lift is computed as the probability to observe both names in an
        # article divided by the product of the probabilities to observe each name.
        lift[i,j] <-  pAB/(pA*pB)
    }
 }
lift
}

ComputeLift(data=A, projection=A.final)

提升值计算

1 个答案: