# R version 2.3.1 (2006-06-01) Debian Linux "testing"
# Is the following behaviour a bug, feature or just a lack of
# understanding on my part? I see that this was discussed here
# last March with no apparent resolution.
d <-
as.factor(c("1970-04-04","1970-08-11","1970-10-18"))
x <- c(9,10,11)
ch <- data.frame(Date=d,X=x)
d <-
as.factor(c("1970-06-04","1970-08-11","1970-08-18"))
y <- c(109,110,111)
sp <- data.frame(Date=d,Y=y)
df <- merge(ch,sp,all=TRUE,by="Date")
# the rows with dates missing all ch vars are tacked on the end.
# the rows with dates missing all sp vars are sorted in with
# the row with a date with vars from both ch and sp
# is.ordered(df$Date) returns FALSE
# The rows of df are not sorted as they should be as sort=TRUE
# is the default. Adding sort=TRUE does nothing.
# So try this:
# dd <- df[order(df$Date),]
# But that doesn't work.
# Nor does sort(df$Date)
# But sort(as.vector(df$Date)) does work.
# As does order(as.vector(df$Date)), so this works:
dd <- df[order(as.vector(df$Date)),]
# ?????
If you want it to act like a date store it as a Date:
dx <-
as.Date(c("1970-04-04","1970-08-11","1970-10-18"))
###
x <- c(9,10,11)
ch <- data.frame(Date=dx,X=x)
dy <-
as.Date(c("1970-06-04","1970-08-11","1970-08-18"))
###
y <- c(109,110,111)
sp <- data.frame(Date=dy,Y=y)
merge(ch, sp, all = TRUE)
By the way you might consider using zoo objects here:
library(zoo)
chz <- zoo(x, dx)
spz <- zoo(y, dy)
merge(chz, spz)
See:
vignette("zoo")
On 9/25/06, Bruce LaZerte <bdl at fwr.on.ca>
wrote:> # R version 2.3.1 (2006-06-01) Debian Linux "testing"
>
> # Is the following behaviour a bug, feature or just a lack of
> # understanding on my part? I see that this was discussed here
> # last March with no apparent resolution.
>
> d <-
as.factor(c("1970-04-04","1970-08-11","1970-10-18"))
> x <- c(9,10,11)
> ch <- data.frame(Date=d,X=x)
>
> d <-
as.factor(c("1970-06-04","1970-08-11","1970-08-18"))
> y <- c(109,110,111)
> sp <- data.frame(Date=d,Y=y)
>
> df <- merge(ch,sp,all=TRUE,by="Date")
> # the rows with dates missing all ch vars are tacked on the end.
> # the rows with dates missing all sp vars are sorted in with
> # the row with a date with vars from both ch and sp
> # is.ordered(df$Date) returns FALSE
>
> # The rows of df are not sorted as they should be as sort=TRUE
> # is the default. Adding sort=TRUE does nothing.
> # So try this:
> # dd <- df[order(df$Date),]
> # But that doesn't work.
> # Nor does sort(df$Date)
> # But sort(as.vector(df$Date)) does work.
> # As does order(as.vector(df$Date)), so this works:
> dd <- df[order(as.vector(df$Date)),]
> # ?????
>
> ______________________________________________
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
On Mon, 25 Sep 2006, Bruce LaZerte wrote:> # R version 2.3.1 (2006-06-01) Debian Linux "testing" > > # Is the following behaviour a bug, feature or just a lack of > # understanding on my part? I see that this was discussed here > # last March with no apparent resolution.Reference? It is the third alternative. A factor is sorted by its codes: consider> x <- factor(1:3, levels=as.character(3:1)) > x[1] 1 2 3 Levels: 3 2 1> sort(x)[1] 3 2 1 Levels: 3 2 1 and that is what is happening here: for your example the levels of df$Date are> levels(df$Date)[1] "1970-04-04" "1970-08-11" "1970-10-18" "1970-06-04" "1970-08-18" so the result is sorted correctly. If you want to sort a character column in lexicographic order, don't make it into a factor. Similarly for a date column: use class "Date".> d <- as.factor(c("1970-04-04","1970-08-11","1970-10-18")) > x <- c(9,10,11) > ch <- data.frame(Date=d,X=x) > > d <- as.factor(c("1970-06-04","1970-08-11","1970-08-18")) > y <- c(109,110,111) > sp <- data.frame(Date=d,Y=y) > > df <- merge(ch,sp,all=TRUE,by="Date") > # the rows with dates missing all ch vars are tacked on the end. > # the rows with dates missing all sp vars are sorted in with > # the row with a date with vars from both ch and sp > # is.ordered(df$Date) returns FALSE > > # The rows of df are not sorted as they should be as sort=TRUE > # is the default. Adding sort=TRUE does nothing. > # So try this: > # dd <- df[order(df$Date),] > # But that doesn't work. > # Nor does sort(df$Date) > # But sort(as.vector(df$Date)) does work. > # As does order(as.vector(df$Date)), so this works: > dd <- df[order(as.vector(df$Date)),] > # ?????-- Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595