Dear Jeff and Jim,
Thank you for sorting out this for me.
Replacing text(as.Date("2005-03-13"),5900,"b") with:
text(as.POSIXct(("2005-03-13 09:00:00"),5900,"b") resolved the problem.
Warmest regards
Ogbos
On Sun, Feb 17, 2019 at 3:05 AM Jim Lemon wrote:
>
> Hi Ogbos,
> Remember that there is an un
Hi Ogbos,
Remember that there is an underlying numeric representation of dates.
as.Date assigns dates a number of days from an agreed origin
(1970-01-01) while POSIX dates assign a number of seconds. If you plot
in one system and then try to add points in another, it's not going to
work.
Jim
On S
Jim has pointed out the strptime function, but you can use as.POSIXct function
also... so
as.POSIXct( "2005-03-13 09:00:00" )
as.POSIXct( "03/13/2005 9:00", format="%m/%d/%Y %H:%M" )
or other variations should work.
On February 16, 2019 5:41:29 PM PST, Ogbos Okike
wrote:
>Dear Jeff,
>My erro
Dear Jeff,
My error please and sorry about that.
Not "1960-05-04 09:00:00".
I meant to write as.Date("2005-03-13"),-9,"b") and "2005-03-13 09:00:00".
My problem is the additional time factor.
I can text anywhere on my plot when dealing with /mm/dd but I can't
handle the new /mm/dd/hh/mm/
Hi Ogbos,
It may be easier to use strptime:
dta<-data.frame(year=rep(2005,5),month=rep("05",5),
day=c("01","06","11","16","21"),
hour=c(2,4,6,8,10),minute=rep(0,5),second=rep(0,5),value=1:5)
dta$Ptime<-strptime(paste(paste(dta$year,dta$month,dta$day,sep="-"),
paste(dta$hour,dta$minute,dta$secon
This is another approach using match():
df1 = data.frame(x1 = letters[1:26],x2 = NA, stringsAsFactors=FALSE)
df2 = data.frame(x1 = letters[10:15],x2 = c("1a","2a","3a","4a","5a","6a"),
stringsAsFactors =FALSE)
df1$x2[match(df2$x1, df1$x1)] <- df2$x2
-
I have no idea how text(as.Date("2005-03-13"),-9,"b") would mark your plot
anywhere near 1960-05-04 09:00:00. Perhaps someone else does. Or perhaps you
can provide an actual minimal working example of what you had working before
you changed to POSIXct.
On February 16, 2019 1:08:38 PM PST, Ogbos
Dear Jeff,
One more problem please.
When I used as.Date(ISOdate(dta$year, dta$month, dta$day,dta$hour)) to
handle date, I could use text(as.Date("2005-03-13"),-9,"b") to label
my plot.
Now that I am using as.POSIXct(ISOdatetime(year,
month,day,hour,0,0))), can you please tell me how to text "b" o
Hello,
An alternative, same dataset.
df[apply(df, 1, function(x) all(is.finite(x))), ]
Hope this helps,
Rui Barradas
Às 16:14 de 16/02/2019, Martin Møller Skarbiniks Pedersen escreveu:
On Sat, 16 Feb 2019 at 16:07, AbouEl-Makarim Aboueissa <
abouelmakarim1...@gmail.com> wrote:
I have a lo
Sorry, that's
function(x)all(is.finite(x) | is.na(x) )
of course.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Feb 16, 2019 at 8:25 AM Bert Gunter w
Many ways. I assume you know that Inf and -Inf are (special) numeric values
that can be treated like other numerics. i.e.
> 1 == - Inf
[1] FALSE
So straightforward indexing (selection) would do it.
But there is also ?is.infinite and ?is.finite, so
apply(yourdat, 1, function(x)all(is.finite(x)))
On Sat, 16 Feb 2019 at 16:07, AbouEl-Makarim Aboueissa <
abouelmakarim1...@gmail.com> wrote:
>
> I have a log-transformed data frame with some *-Inf* data values.
>
> *my question: *how to remove all rows with *-Inf* data value from that
data
> frame?
Hi,
Here is a solution which uses apply.
F
Dear Abou
Depends on exact details of your variables but
?is.finite
Gives you the basic tool.
On 16/02/2019 15:05, AbouEl-Makarim Aboueissa wrote:
Dear All: good morning
I have a log-transformed data frame with some *-Inf* data values.
*my question: *how to remove all rows with *-Inf* data
Dear All: good morning
I have a log-transformed data frame with some *-Inf* data values.
*my question: *how to remove all rows with *-Inf* data value from that data
frame?
with many thanks
abou
__
*AbouEl-Makarim Aboueissa, PhD*
*Professor, Statistics and Data Science*
*
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