similar to: extract date

Hello! I am new to this list, so I am sory if this has already been answered. I am looking for a way to enable my notebook to have access to my home network regardles where I am. This is do-able if I have i fixed IP on some location, but if I have a dynamic IP, how can shorewall tell whic user am I so it can let me thru the firewall? I ve seen the MAC parameter in rules, but it doesnt work if

Dear all I have a character vector of dates something like: timevec<-c("15.5.2003 00:00", "15.5.2003 00:01", "15.5.2003 00:02", "15.5.2003 00:03","15.5.2003 00:04") and I would like to transform it to some more convenient date class. Is there a way how to do it directly without previous reformating to ISO like structure and adding a

I am just diving into Rails and I am a bit puzzled by the link_to function. am writing an app to display my photographs in categories. My application layout has a sidebar to show a menu with categories from the categories table in the database (like in taken from the tutorials), to be shown on every page. I managed to get a global @menu_categories object that keeps track of all the categories

I am trying to coerce a data frame column from character to datetime using strptime but keep getting an error because the length of the coerced object is always 9. What am I doing wrong here: ................................................................. > ds <- cbind(1:2, c("02/27/92 23:03:20", "02/27/92 22:29:56")); ds [,1] [,2] [1,]

On Sat, Oct 1, 2016 at 4:40 PM, Teresa Johnson <tejohnson at google.com> wrote: > I completely forgot to add -Wl,--gc-sections. The *-sections options enables > gc-sections to work effectively. (edit: oh wait, it looks like you already have > that one on your link command, so you should be set there!) > > Unfortunately, I just can't reproduce the DSO link error you are

Using strptime() and other functions for dates has been very helpful with the kind of data I often work with. However, I haven't found out how time as such should be specified. All my attempts result in time *and* date: >treatment_time<-c("01:02:03","02:03:04") # hours:minutes:seconds >time.2<-strptime(treatment_time,format="%H:%M:%S") >time.2 [1]

R V2.0.1 on Windows XP. I have read the help pages on strptime() over and over, but can't understand why strptime() is producing the following results. > v <- format("2002-11-31", format="%Y-%m-%d") > v [1] "2002-11-31" > factor(v, levels=v) [1] 2002-11-31 Levels: 2002-11-31 > x <- strptime("2002-11-31",

Thank you Pikal and Bert. My apology for posting parts of my previous email in HTML. Bert's suggestion will work but i am wondering if there is an alternative especially in the case where the data frames are big; that is the difference in lengths among them is large. Below is a list of sample date frames and desired result. EK

Dear All, I wonder how to compute the age from the date of birth and the date of examination. Suppose that have the following data: df <- as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"), c(1,"10/08/1950","20/07/1998"), c(1,"10/08/1950","23/10/1998"))) names(df) <- c("ID", "date_birth",

Dear useRs, How come the first attempt to sort a POSIXt vector fails (Error: non-atomic type in greater), while the second succeeds? (Code inserted below.) The documentation says that POSIXt is used to allow operations such as subtraction, so I'd expect sorting to work. Is this perhaps an OS issue? (I run R 2.0.1 on Win xp.) Thank you, b. #------------code test <- c("2005-02-08

Hi, I'm having a weird result with the length() function: >a [... omited ...] [9994] NA "2003-12-03 16:37:00" "2002-06-26 18:43:00" [9997] "2005-07-04 04:00:00" "2007-02-16 22:09:00" "2007-02-24 15:49:00" [10000] NA > length(LastModified) [1] 9 > length(c(LastModified)) [1] 9 I was expecting to get

At one time (version 1.7), the code below used to work for converting and extracting based on the Date Time. In version 1.8.1, something changed I know, but I cannot for the life of me figure out what... Data: UserName,RequestDate,PO,OrderDate,ExpDelivDate,Vendor,Total "Woody, Jim",12/19/2002,AP15063,1/7/2003,2/10/2003,Ames ,8570 "Harrold,

I read an article today about keyboard interactive auth allowing bruteforcing. I'm afraid I have minimal understanding of what keyboard-interactive really does. What does it do, and should I have my clients set it to off in sshd_config? --- Scott Neugroschl | XYPRO Technology Corporation 4100 Guardian Street | Suite 100 |Simi Valley, CA 93063 | Phone 805 583-2874|Fax 805 583-0124 |

On Thu, 2020-02-13 at 15:07 -0800, Johan Hattne via samba wrote: >> Dear all; >> >> I'm trying to use winbindd to resolve names in an AD setup. I can >> authenticate just fine, but I've noticed that for some users "getent >> passwd" returns a GECOS field populated with displayName from the LDAP >> servers and for others is does not. For

Hi, I want to display dates on my x-axis of the plot. I was trying to use plot() command for the same and passing the values in following manner: The variable "dat" is a data frame. The first column has numeric values and second column has date. e.g. dat [,1] dat[,2] [1,] 300 20060101 [2,] 257

Hello, I do have two different time series A and B, they are different in length and starting point. A starts in Jan, 2012 and ends in Dec, 2012 and B starts in March, 2012 and ends in Nov, 2012. How can I plot those two series A and B in the same plot? I.E., from Jan. 2012 - Feb, 2012, it would have one data point from A and from Mar, 2012-Nov, 2012, it would have two data points from A and B,

Background: OS: Linux Mandrake 10.1 release: R 2.0.0 editor: GNU Emacs 21.3.2 front-end: ESS 5.2.3 --------------------------------- Colleagues I have a wind speed time series with a normal frequency distribution and a spike in the 5 metres/second bin. The most likely explanation is that the instrument was returning duplicate values at this speed. To check this, I want to extract all the unique

... Well, this works in this simple case, but is too clumsy for a general formulation of this problem: given a "dictionary" consisting of two character vectors of unique "names" (or two columns in a data frame), x and y, how does one convert a factor z with levels in x into the corresponding equivalent with levels in y? There are likely a zillion ways to do this with various

On Sun, Oct 2, 2016 at 6:41 AM, Teresa Johnson <tejohnson at google.com> wrote: > > On Sat, Oct 1, 2016 at 11:43 AM, Carsten Mattner <carstenmattner at gmail.com> wrote: > > > > On Sat, Oct 1, 2016 at 4:40 PM, Teresa Johnson <tejohnson at google.com> wrote: > > > > > I completely forgot to add -Wl,--gc-sections. The *-sections > > >

Dear UseRs, I have a very simple question. I have a big matrix (say x) including probabilities (values in (0,1)). I have to store in a list the names of the row and the column where max(x) is located. How can I proceed? Thanks in advance for your assistance! mirko