Hello,
I'm afraid you're wrong, this has nothing to do with leading zeros. Just
see:
x <- c("1/2/2011", "1/4/2011", "1/4/2011", "1/4/2011", "1/6/2011",
"1/7/2011",
"1/8/2011", "1/9/2011", "1/10/2011")
as.Date(x, "%m/%d/%Y")
y <- factor(x)
str(y)
as.Date(as.character(y), "%m/%d/%Y")
Note that the correct way of transforming factors into their levels is
to index the levels by the factor's values (which are always positive
integers):
levels(y)[y]
[1] "1/2/2011" "1/4/2011" "1/4/2011" "1/4/2011" "1/6/2011" "1/7/2011"
[7] "1/8/2011" "1/9/2011" "1/10/2011"
Also, if those dates come from a file, maybe you want to read.table or
read.csv with the option
stringsAsFactors = FALSE
and avoid the problem altogether. This has the side effect that all
character variables will be read in as character variables and some
might be categorical. What an ugly sentence! So this option must be used
with caution.
Hope this helps,
Rui Barradas
Em 27-06-2012 04:54, Erin Hodgess escreveu:
Dear R People:
I have dates as factors in the following:
poudel.df$DATE
[1] 1/2/2011 1/4/2011 1/4/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011
[8] 1/9/2011 1/10/2011
Levels: 1/10/2011 1/2/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011 1/9/2011
I want them to be "regular" dates which can be sorted, etc.
But when I did this:
as.character(poudel.df$DATE)
[1] "1/2/2011" "1/4/2011" "1/4/2011" "1/4/2011" "1/6/2011" "1/7/2011"
[7] "1/8/2011" "1/9/2011" "1/10/2011"
and
as.Date(as.character(poudel.df$DATE),"%m/%d/$Y")
[1] NA NA NA NA NA NA NA NA NA
because the dates do not have leading zeros.
There are approximately 30 years of nearly daily data in the entire set.
Any suggestions would be much appreciated.
Sincerely,
Erin
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