Hi,
If the rows are already ordered:
x1<- as.data.frame(x)
x[with(x1,ave(seq_along(V2),V2,FUN=function(x) !x%in%min(x)))==1,]
# [,1] [,2] [,3]
#[1,] "1/5/13" "15" "25"
#[2,] "1/9/13" "15" "28"
#[3,] "2/5/13" "18" "35"
#[4,] "2/9/13" "18" "38"
#otherwise
x[with(x1,unlist(tapply(as.Date(V1,"%m/%d/%y"),list(V2),function(x)
x!=min(x)),use.names=FALSE)),]
# [,1] [,2] [,3]
#[1,] "1/5/13" "15" "25"
#[2,] "1/9/13" "15" "28"
#[3,] "2/5/13" "18" "35"
#[4,] "2/9/13" "18" "38"
A.K.
----- Original Message -----
From: Andras Farkas <motyoc...@yahoo.com>
To: "r-help@r-project.org" <r-help@r-project.org>
Cc:
Sent: Tuesday, July 30, 2013 8:13 AM
Subject: [R] selection based on dates
Dear All
please provide your insigths on the following:
I have:
a <-c("1/1/13",15,20)
b <-c("1/5/13",15,25)
c <-c("1/9/13",15,28)
d <-c("2/1/13",18,30)
e <-c("2/5/13",18,35)
f <-c("2/9/13",18,38)
x <-matrix(c(a,b,c,d,e,f),ncol=3,byrow=TRUE)
What I would like to do is to eliminate certain rows of this matrix based on
the date column values. As you can see, in the second column my values (15 and
18) repeat 3 times each, so this column serves as an ID number if you will.
Thus each ID numbers show up with 3 different date values in the first column.
Now I would like to eliminate the rows with the earliest date per ID number. My
result should look like this:
z <-x[-c(1,4),]
as allways, your help is greatly appreciated,
thanks,
Andras
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and provide commented, minimal, self-contained, reproducible code.
