Perform pairwise comparison of matrix

问题

I have a matrix of n variables and I want to make an new matrix that is a pairwise difference of each vector, but not of itself. Here is an example of the data.

    Transportation.services Recreational.goods.and.vehicles Recreation.services Other.services
         2.958003                     -0.25983789            5.526694           2.8912009
         2.857370                     -0.03425164            5.312857           2.9698044
         2.352275                      0.30536569            4.596742           2.9190123
         2.093233                      0.65920773            4.192716           3.2567390
         1.991406                      0.92246531            3.963058           3.6298314
         2.065791                      1.06120930            3.692287           3.4422340

I tried running a for loop below, but I'm aware that R is very slow with loops.

Difference.Matrix<- function(data){
 n<-2
new.cols="New Columns"
list = list()
for (i in 1:ncol(data)){

    for (j in n:ncol(data)){

        name <- paste("diff",i,j,data[,i],data[,j],sep=".")
        new<- data[,i]-data[,j]
        list[[new.cols]]<-c(name)
        data<-merge(data,new)
        }
    n= n+1
    }
results<-list(data=data)
return(results)
}

As I said before the code is running very slow and has not even finished a single run through yet. Also I apologize for the beginner level coding. Also I am aware this code leaves the original data on the matrix, but I can delete it later.

Is it possible for me to use an apply function or foreach on this data?

回答1:

You can find the pairs with combn and use apply to create the result:

apply(combn(ncol(d), 2), 2, function(x) d[,x[1]] - d[,x[2]])
##          [,1]      [,2]       [,3]      [,4]      [,5]      [,6]
## [1,] 3.217841 -2.568691  0.0668021 -5.786532 -3.151039 2.6354931
## [2,] 2.891622 -2.455487 -0.1124344 -5.347109 -3.004056 2.3430526
## [3,] 2.046909 -2.244467 -0.5667373 -4.291376 -2.613647 1.6777297
## [4,] 1.434025 -2.099483 -1.1635060 -3.533508 -2.597531 0.9359770
## [5,] 1.068941 -1.971652 -1.6384254 -3.040593 -2.707366 0.3332266
## [6,] 1.004582 -1.626496 -1.3764430 -2.631078 -2.381025 0.2500530

You can add appropriate names with another apply. Here the column names are very long, which impairs the formatting, but the labels tell what differences are in each column:

x <- apply(combn(ncol(d), 2), 2, function(x) d[,x[1]] - d[,x[2]])
colnames(x) <- apply(combn(ncol(d), 2), 2, function(x) paste(names(d)[x], collapse=' - '))
> x
     Transportation.services - Recreational.goods.and.vehicles Transportation.services - Recreation.services
[1,]                                                  3.217841                                     -2.568691
[2,]                                                  2.891622                                     -2.455487
[3,]                                                  2.046909                                     -2.244467
[4,]                                                  1.434025                                     -2.099483
[5,]                                                  1.068941                                     -1.971652
[6,]                                                  1.004582                                     -1.626496
     Transportation.services - Other.services Recreational.goods.and.vehicles - Recreation.services
[1,]                                0.0668021                                             -5.786532
[2,]                               -0.1124344                                             -5.347109
[3,]                               -0.5667373                                             -4.291376
[4,]                               -1.1635060                                             -3.533508
[5,]                               -1.6384254                                             -3.040593
[6,]                               -1.3764430                                             -2.631078
     Recreational.goods.and.vehicles - Other.services Recreation.services - Other.services
[1,]                                        -3.151039                            2.6354931
[2,]                                        -3.004056                            2.3430526
[3,]                                        -2.613647                            1.6777297
[4,]                                        -2.597531                            0.9359770
[5,]                                        -2.707366                            0.3332266
[6,]                                        -2.381025                            0.2500530

来源：https://stackoverflow.com/questions/22446825/perform-pairwise-comparison-of-matrix