R help - Sep 2010 - looking for a faster way to compare two columns of a matrix

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.

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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