[R] looking for a faster way to compare two columns of a matrix

[Michael Bedward]

Hello Gustavo,

Not sure if I've got all the details of your metric, but what about this...

xx <- x[ , combn(5,2)]
i <- seq(2, ncol(xx), 2)
colSums(xx[,i-1] > xx[,i] & xx[,i] > 0) / colSums(xx[,i] > 0)

Michael


On 24 September 2010 02:53, Gustavo Carvalho <gustavo.bio+R at gmail.com> wrote:
> Please consider this matrix:
>
> x <- structure(c(5, 4, 3, 2, 1, 6, 3, 2, 1, 0, 3, 2, 1, 0, 0, 2, 1,
> 1, 0, 0, 2, 0, 0, 0, 0), .Dim = c(5L, 5L))
>
> For each pair of columns, I want to calculate the proportion of entries
> different than 0 in column j (i > j) that have lower values than the entries
> in the same row in column i:
>
> x[, 1:2]
> sum((x[,1] > x[,2]) & (x[,2] > 0))/sum(x[,2] > 0)
>
> Thus, for this pair, 3 of the 4 entries in the second column are
> lower than the entries in the same row in the first column.
>
> When both columns of a given pair have the same number of cells different than
> 0, the value of the metric is 0.
>
> x[, 3:4]
> colSums(x[, 3:4] > 0)
>
> The same if column j has more valid (> 0) entries.
>
> I've been doing this using this idea:
>
> combinations <- combn(1:ncol(x), 2)
> values <- numeric(ncol(combinations))
>
> for (i in 1:ncol(combinations)) {
>  pair <- combinations[,i]
>  first <- x[, pair[1]]
>  second <- x[, pair[2]]
>  if (sum(first > 0) <= sum(second > 0)) next
>  values[i] <- sum(first - second > 0 & second > 0) / sum(second > 0)
> }
> values
>
> Anyway, I was wondering if there is a faster/better way. I've tried
> putting the code from
> the for loop into a function and passing it to combn but, as expected, it didn't
> help much. Any pointers to functions that I should be looking into will be
> greatly appreciated.
>
> Thank you very much,
>
> Gustavo.
>
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