Try
a["2016-01-18 15:31:54.079"]
The question though is why R displays the milliseconds as "078", when
it is clearly "079"...
-h
On Mon, Jan 18, 2016 at 8:55 PM, ce wrote:
> Dear all,
>
> I have this code :
>
> library(xts)
> a <- structure(c(1,2), class = c("xts", "zoo"), .indexCLASS = c("P
Well it works when you don't put the milliseconds because it is
matching that part of the date which is provided. Hence, the following
all work:
a["2016-01-18 15:31:54.0"]
a["2016-01-18 15:31:54"]
a["2016-01-18 15:31"]
a["2016-01-18 15"]
However, what I don't get is why it is showing 1453149114.0
On Mon, Jan 18, 2016 at 8:31 PM, ce wrote:
>
> yes it shows , more interesting :
>
>> a["2016-01-18 15:31:54.0"]
> value
> 2016-01-18 15:31:54 2
>
>> a["2016-01-18 15:31:54.07"]
> value
>
>> a["2016-01-18 15:31:54.079"]
> value
> 2016-01-18 15:31:5
yes it shows , more interesting :
> a["2016-01-18 15:31:54.0"]
value
2016-01-18 15:31:54 2
> a["2016-01-18 15:31:54.07"]
value
> a["2016-01-18 15:31:54.079"]
value
2016-01-18 15:31:54 2
why it doesn't work with .07 milisecond but work wit
Dear all,
I have this code :
library(xts)
a <- structure(c(1,2), class = c("xts", "zoo"), .indexCLASS = c("POSIXct",
"POSIXt"), .indexTZ = "", tclass = c("POSIXct", "POSIXt"), tzone = "", index =
structure(c(1453137885.23,
1453149114.079), tzone = "", tclass = c("POSIXct", "POSIXt")), .Dim = c
Thanks David,
That's exactly what I need! Thanks very much for your example.
Best regards,
Julian
On 18 January 2016 at 13:00, David L Carlson wrote:
> Providing us with a reproducible example, makes it much easier to answer your
> question. The object returned by LSD.test() is completel
Hello,
I am new to R support. I am trying to install R on a compute cluster running
Platform MPI and PBS Pro.
I have installed R, Rmpi and snow - at first glance successfully.
When I try to run a simple Rmpi program, however, I get a Segmentation fault on
mpi send.
The error messages reads a
Providing us with a reproducible example, makes it much easier to answer your
question. The object returned by LSD.test() is completely different from the
object described in the "Value" section of its manual page which makes things
confusing. If you look at the first example on the manual page:
Yeah, the philosophy of knitr from the very beginning is that if you
want to draw a plot, simply draw it, and knitr will take care of the
rest of work (http://i.imgur.com/jrwbX.jpg). You rarely need to think
about graphical devices or LaTeX or a specific output format. With
knitr, the example can b
Hello, everyone,
I am using LSD.test() from package "agricolae" to do my anova
analysis. I know I can calculate LSD value by its equation
t*sqrt(MSE*2/n), but I am wondering if there is a code or something
that can give the LSD value more directly in R?
Thanks!
Julian
__
Thanks Peter. That make sense. Nevertheless, what comes at a surprise to me
(and maybe to others) is that one can potentially get different fits by simply
swapping the terms in the model formula.
Best,
Leo.
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: 2016,
On 18 Jan 2016, at 16:49 , Guelman, Leo wrote:
> Is it really the same model?
No, and I didn't say that they would be. I did say that they would be in the
all-factor case, which does seem to be right:
> df$trt <- factor(df$trt)
> fit1 <- glm(y ~ x1:trt + f1:trt, data = df, family = binomial)
Is it really the same model?
Following the example provided by Lars:
set.seed(1)
x1 <- rnorm(100)
f1 <- factor(sample(letters[1:3], 100, replace = TRUE))
trt <- sample(c(-1,1), 100, replace = TRUE)
y <- factor(sample(c(0,1), 100, T))
df <- data.frame(y=y, x1=x1, f1=f1, trt=trt)
fit1 <- glm(y ~ x
> This is "bugged":
> > str(test1)List of 1$ justaname: chr "justaname"- attr(*, "class")= chr
> "eem"
> This is ok:> str(test2)List of 1$ sample: chr "justaname"- attr(*,
> "class")= chr
> "eem2"
> It seems that override of the "<-" is causing problem since in str(test1) the
> variable nam
swizz-john wrote:
> Hi people,
>
> my task is to analyse data that is formatted like this.
>
>
date,bid,name,w1,w2,w3,m1,m2,m3,m4,m5,m6,m7,m8,m9,m10,m11,m12,debt2mkt,cds,equity
> 28jul2009,1,"ABN Amro",56.5,
> 29jul2009,1,"ABN Amro",56.5,
> 30jul2009,1,"ABN Amro",
On 18/01/2016 6:12 AM, Naresh Gurbuxani wrote:
Duncan,
Many thanks for looking at my code and for your suggestion.
Your solution works. But my problem is different. This code gives me a tex
file with a lot of tikz code. If there are several graphs in the document,
then tex file become very
Again: thanks a lot for all your suggestions, the problem is now solved.
This combination of calls finally did the trick:
> n96 <- as.matrix(n96)
> frame2 <- data.frame(gushVM, n96=I(n96))
> str(frame2)
'data.frame': 15 obs. of 2 variables:
$ gushVM: num 2 23.2 14.3 40.9 32.8 29.1 0 79 0 0 .
S Ellison writes:
> Reading ?plsr examples and inspecting the data they use, you need to arrange
> frame1 so that it has the data from n96 included as columns with names of the
> from "n96.xxx" whre xxx can be numbers, names etc.
No, you do not. :) plsr() is happy with a data frame where n96 is
Thank you everyone for the codes and the link. They work well!
Mr. Lemon, thank you for the detailed code and the explanations. I
appreciate it. One thing though, in the last line
sapply(split_strings,fill_strings,list(max_length,element_sets))
should it be unlist instead of list - I get this er
Hi Miluji,
While the other answers are correct in general, I noticed that your request
was for the elements of an incomplete string to be placed in the same
positions as in the complete strings. Perhaps this will help:
strings<-list("pc_m2_45_ssp3_wheat","pc_m2_45_ssp3_wheat",
"ssp3_maize","m2_wh
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