search for: mondays

Hi! I need use rsync for incremental backup, but i need save that incremental data on diferent directory, for example I have a Full backup on directory ./FULL, and carpets for ./monday ... I need to compare the sources with ./FULL and the diferents files save on ../monday .. but not in ./FULL ---------- ----------- SERVER A | | SERVER B |

I have a xts object made of daily closing prices I have acquired using quantmod. Here is my code: library(xts) library(quantmod) library(lubridate) # Gets SPY data getSymbols("SPY") # Subset Prices to just closing price SP500 <- Cl(SPY) # Show day of the week for each date using 2-6 for monday-friday SP500wd <- wday(SP500) # Add Price and days of week together

Hello, I have a table (in a txt file) which look like this: Monday 12 78 89 Tuesday 34 44 67 Wednesday 78 98 2 Thursday 34 55 4 Then the table repeats Monday , Tuesday, ... followed by several numbers My goal is to read values after the table. My problem is a little more complicated, but I just present a simpler case for ease of illustration. Is there any way to ask R to "read

I have a checkbox: <%= check_box(:day, :monday, {:disabled => ''''} , (day.monday == true ? {:checked => ''checked''} : {:checked => false} )) %> It is disabled allright, but it is always uncecked. Is this supposed to happen. It looks like (day.monday == true ? {:checked => ''checked''} : {:checked => false} ) is not

The release of R-1.9.1 is scheduled for Monday, June 21. Automatic generation of daily alpha releases should start Monday, June 7 and switch to beta status on Monday, June 14. It would be good if package maintainers could get any planned changes done as soon as possible, and test their packages carefully against the alpha/beta releases. -- O__ ---- Peter Dalgaard Blegdamsvej 3

I've uploaded clang+llvm-3.5.2-rc1-mips-linux-gnu.tar.xz and clang+llvm-3.5.2-rc1-mips-linux-gnu.tar.xz started testing them. > -----Original Message----- > From: llvmdev-bounces at cs.uiuc.edu [mailto:llvmdev-bounces at cs.uiuc.edu] > On Behalf Of Daniel Sanders > Sent: 16 March 2015 15:36 > To: Tom Stellard; llvmdev at cs.uiuc.edu > Subject: Re: [LLVMdev] Reminder 3.5.2

# File app/models/timesheet.rb, line 27 27: def totals 28: totals = Hash.new 29: totals["Monday"] = totals["Tuesday"] = totals["Wednesday"] = totals["Thursday"] = totals["Friday"] = totals["Saturday"] = totals["Sunday"] = totals["Totals"]=0 #initialise all to zero 31: 32: for item in

dear All, i'm trying to calculate the number of Mondays, Tuesdays, etc that each month within a date range has. I have time series data that spans 60 months and i want to calculate the number of Mondays, Tuesdays, Wed, etc of each month. (I want to control for weekly seasonality but my data is monthly). Is there an easy way to to this in R? or...

OK. What if I have a time series which is collected every Monday, please? What is the proper way to use the start option within the ts command in order to indicate that this is Monday data, please? Thanks again! Sincerely, Erin

Hi there, I am having problem of matching string. what i want is when i type a date such as 2010-11-30, the function will return the day (monday, tuesday, wednesday, thursday, friday or staturday). then i want another function will return true if the return of the day is monday, return false if the return of the day is not monday. I already find the weekdays(as.Date('2010-11-30')) function

Hi, Just a reminder that testing for 3.5.2 will begin on Monday, Mar 16. If you have any patches you want merged please send an email to the relevant commits list and cc me and the code owner. If you have already done this and are waiting for a response, please ping the thread. As always we need testers, so let me know if you want to help with testing. Thanks, Tom

I'm an R noob and have a (maybe) stupid question... I have a table where I have the weekdays and a number for each weekday of entries: Thats what the table looks like... Now I want to have an pie3D plot of this, but obviously the order of the weekdays are not as one would expect... Friday Monday Saturday Sunday Thursday Tuesday Wednesday 119 173 80

Greetings, Normally, we hold the Xen Project Documentation Day on the fourth Monday of the month. However, for the next two months, there are some issues: - Next Monday is the beginning of Thanksgiving week in the US. This is historically a very hard time to accomplish anything (other than shop for the holidays which occur in December), so if we hold Documentation Day on Monday, it is likely

Sorry for the extra email. It send to quickly. procmail: Assigning "SUBJECT= Tornado Monday, 03/27/2017 at 20:27:02. The Point BB.OBSURGRH is" jerry

...le based on a conditional statement on either the same variable or a different variable in the same data set? So if I had the closing prices of the S&P from 01/01/1990-12/31/1990, how could I get the average price of the S&P from 02/01/1990-03/15/1990? Or the average price of the S&P on Mondays (assuming a dummy var is created for 1 = Monday, 0 = else). I understand that you can create subsets and new data sets based on the conditional statements; but is there an easier way to do this by typing a line into the mean() statement? That was extremely easy in SAS where you could say: proc mea...

# The first error rsync: generator.c:1867: check_for_finished_files: Assertion `flist != ((void *)0)' failed. rsync: writefd_unbuffered failed to write 4092 bytes [receiver]: Broken pipe (32) rsync error: error in rsync protocol data stream (code 12) at io.c(1493) [receiver=3.0.0pre2] # Sample command, obfuscated to protect the guilty rsync --archive --hard-links --no-motd \ --numeric-ids

Dear All, I'm afraid this applies to LLVM as well. So, please fill in the OpenProjects pages for GSoC today - 19:00 UTC Monday is morning in California, so Monday will be too late :( ---------- Forwarded message ---------- From: 'Stephanie Taylor' via Google Summer of Code Mentors List <google-summer-of-code-mentors-list at googlegroups.com> Date: Sun, Jan 28, 2018 at 11:05 AM

Hi Can someone help me with this? How can I apply a function to a list of variables. something like this listvar=list("Monday","Tuesday","Wednesday") func=function(x){x[which(x<=10)]=NA} lapply(listvar, func) were Monday=[213,56,345,33,34,678,444] Tuesday=[213,56,345,33,34,678,444] ... in my case I have a neverending list of vectors. Thanks!

Apparently any (?) call of the form cut(date,"weeks") where the date *begins the week*, gives the error Error in 1:(1 + max(which(breaks < maxx))) : result would be too long a vector In addition: Warning message: In max(which(breaks < maxx)) : no non-missing arguments to max; returning -Inf To my surprise, this was first reported as a problem in 2006 (!) (version 2.3.1

hi, I have a and data frame with date-column and some other columns. My first question is what is the fastest way to get the index of an array if I know the value f.e > x = c(4,5,6,7,8) so i know the value is 6.. i.e. the index is 3. What I currently do is loop over the array, I was thinking if there is faster more direct way. The next one...is I have a data frame one of the columns is Date