Removing row with smallest value, for a given factor
On Apr 23, 2011, at 1:30 PM, Peter Ehlers wrote:
On 2011-04-23 07:02, David Winsemius wrote:
On Apr 23, 2011, at 9:05 AM, - - wrote:
I have a table.
First column is a date, second column is an index and other columns
contains some other values.
I want to remove, for each date, the row with the smallest index (it
is not necessarily 1).
ex: in the following table, I want to remove row 1 (2013-05-12 with
index 2) and row 8 (2013-05-13 with index 1)
day index values
1 2013-05-12 2 xxxx
2 2013-05-12 3 xxxx
3 2013-05-12 4 xxxx
4 2013-05-12 5 xxxx
5 2013-05-12 6 xxxx
6 2013-05-12 7 xxxx
7 2013-05-12 8 xxxx
8 2013-05-13 1 xxxx
9 2013-05-13 3 xxxx
10 2013-05-13 4 xxxx
11 2013-05-13 5 xxxx
12 2013-05-13 6 xxxx
13 2013-05-13 7 xxxx
14 2013-05-13 8 xxxx
15 2013-05-13 9 xxxx
16 2013-05-13 10 xxxx
17 2013-05-13 12 xxxx
Consider using ave and creating a logical vector that you then negate:
> ave(dat$index, list(dat$day), FUN=function(x) x==min(x))
[1] 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 dat[ -ave(dat$index, list(dat$day), FUN=function(x) x==min(x)), ]
ave() is one of those really handy functions, but I think that you meant dat[ !ave(dat$index, list(dat$day), FUN=function(x) x==min(x)), ]
Yes, that is what I should have answered. Somehow I thought that because it was numeric I could use "-" but that would only be correct if it returned line numbers. I'm not sure why it should return numeric rather logical, but the help page does say the value will be numeric, so I shouldn't be surprised, I suppose.
Here's another way, using the plyr package require(plyr) ddply(dat, .(day), .fun = function(x) subset(x, index != min(index))) Peter Ehlers
-- David Winsemius, MD West Hartford, CT
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David Winsemius, MD West Hartford, CT