Sorting each column of a matrix independently

Hi Desmond,

Perhaps I'm not understanding something, but using sort on separate columns
looks too complicated to me...

According to help(sort), sort is for a vector or factor, and order is for
sorting dataframes (or matrices).

So, you should be able to do:

sorted <- unsorted[order(unsorted[,1], unsorted[,2]),]

Stick a minus sign in front of columns if you want decreasing order:

sorted <- unsorted[order(unsorted[,1], -unsorted[,2]),]

Would sort by the first column in increasing order and the second column by
descreasing order.

Change the order of the columns within the order() statement depending on
which columns have preference during the ordering.

Best wishes,

Todd

Hi,

I have this matrix

unsorted$A with 5 columns (all numeric).

Currently I'm sorting the columns like this:

sorted$A[,1] <- sort(unsorted$A[,1])
sorted$A[,2] <- sort(unsorted$A[,2]) ...

I have tried doing

sorted$A <- unsorted$A[do.call(order, data.frame(unsorted$A)),] but it
sorts the first column then the 2nd column which it ties to the 1st
column, then the 3rd which is tied to the 2nd, etc. I need all the
columns to be independently sorted.

Any help is appreciated.

Desmond

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