search for: months

...some confusion in applying a function over a column. Here's my function. I just need to shift non-March month-ends to March month-ends. Initially I tried seq.dates, but one cannot give a negative increment (decrement) here. return(as.Date(seq.dates(format(xdate,"%m/%d/%Y"),by="months",len=4)[4]) ) Hence this simple function: > mydate <- as.Date("2006-01-01") > > # Function to shift non-March company-reporting dates to March. > Set2March <- function(xdate){ + # Combines non-March months into March months: + # Dec2006 -> Mar2007 + # Mar2006...

Hi All I have a column/variable called time difference. It has a whole list of numbers from 0 through to the hundreds eg 236. I want to assign a corresponding "name" to each variable from a predefined list: Month or less, 1 -2 months, 2-3 months etc So the result would look something like: Time Difference Month 1 Month or less 365 1-2 years 52 2-3 months Etc I have tried using if...

I have 2 datafiles 'target' and 'observed' as shown below (I will gladly email these 2 small files to whomever). X25. And X75. Indicate the value of 25th and 75th-percentile of the target ('what should be') and the observed ('what is'). The i.value is simply the month. > target X i.value X25. X75. 1 one.month 1 10.845225 17.87237 2

Hello! I have a data set similar to the data set "monthly" in the example below: monthly<-data.frame(month=c(20090301,20090401,20090501,20100301,20100401,20090301,20090401,20090501,20100301,20100401),monthly.value=c(100,200,300,101,201,10,20,30,11,21),market=c("Market A","Market A", "Market A","Market A", "Market A","Market

so I am trying to sum month over month the amount that a user has posted. So for example: User 1: Jan $3000 Feb $4000 March $1500, etc. I can get this to work if I sum totals (aggregate of all users) but just not by user. Here is my code in the controller: def index @users = User.find :all, :order => ''name ASC'' @deal_groups = Deal.find(:all).group_by {|t|

Perhaps it is just that I don't even know the correct term to search for, but I can find nothing that explains how to wrap around from the end to a start of a row in a matrix. For example, you have a matrix of 2 years of data, where rows are years, and columns are months. month.data = matrix(c(3,4,6,8,12,90,5,14,22, 8), nrow = 2, ncol=5) I would like to take the average of months 5:1 for each year (for row 1 =12.5). However, I am passing the start month (5) and the end month (1) as variables. I would like to do something like year.avg = apply(month.data[, start...

..., as i wrote, i have 69 cities and the actuall table runs down very deep.) Column 2 represnts the year for which the data is given (Actuall data for each station is of different length but atleast of 24 years). Column 3 and 4 reprent the month and the day of the data. obviously each year has 12 months and each month as different number of days, but to make table easily understable only 3 months and 3 days of each month are considered. febrary for leap years should also be considered. col5 represents population of that city st. year month day population in million Ta 1966 1 1 2.4 Ta 1966 1 2...

Hi all, I am looking for a function for following calculation. start.month = "July" end.month = "January" months = f(start.month, end.month, by=1) * f is the function that I am looking for. Actually I want to get months = c("July", "August",.............."January") If start.month = 6 and end.month = 1 then I could use (not properly) seq() function and then I would get month as...

Dear all i got a problem in monthly mean temperature. here i am attaching the data set as well as the plot i got with the following command plot(month,type='n') plot(month,X1999) this command gave the plot where the month names are in alphabetic order, i want the plot in monthly sequence could you please suggest me how can i solve my problem? thanking you regard madan

The last line of the example in droplevels' manual page seems to be incorrect to me. I think it should read: "table(droplevels(aq$Month))". Amazingly (I don't understand) both variants seem to produce the same result (R 3.3.3): --- > aq <- transform(airquality, Month = factor(Month, labels = month.abb[5:9])) > aq <- subset(aq, Month != "Jul") >

...>> construct a function that isolates the n'th working day of each month for >> zoo object (time series) to create monthly data from daily observations. >> >> I found out that the code works fine except for the 29 till 31st dates of >> each month as it skips some months (February for example). >> >> If you could help me isolate the problem I would be grateful as I can not >> find a way to explain to R to keep the last working day of month if I >> choose the 29th, 30th or 31st dates... >> >> I enclose a working version of the...

...ave a data.frame with 3 columns. The first column represents an "incident" type The second column represents a "month" The third column represents a "time" Code for a sample data.frame incidents <- rep(c('a','b','d','e'), each =25) months <- rep(c(1,2), each =10) times <-rnorm(100) # make my sample data DF <- data.frame(Incidents=as.factor(incidents),Months=as.factor(months),Time=times) # now calculate a mean for the "by" groups of incident type and month pivot <- aggregate(DF$Time,by=lis...

Hi dear all, the following code is correct. but I want to use non-numeric x-axis, for example if I replace time <- seq(0,72,6) by month <- c("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec","Pag") Ofcourse I use factor(month) instead of

Dear all, I have a code where I subset a data frame to match entries within levels of an factor (actually, the full script uses three difference factors do do that). I'm very happy with the precision with which I can work with R, but since I loop over factor levels, and the data frame is big, the process is slow. So I've been trying to speed up the process using by(), but I got stuck at

In PHP, there was an argument you could pass to the Date function to get the number of days in the current month: echo date("t"); // Outputs "28" for February I don''t see anything like this in Ruby/Rails. Right now, I''m using a very ugly line to pull the last day of the month: @number_of_days = (Date.strptime(Date.today.strftime("%Y-%m-01"))

Hello, I've been using a pre-release version of R v 2.8.0 for Windows for the last couple months. I think that there have been consistent problems with subsetting data sets, but I had usually been able to find work-arounds or was unable to confirm this as a bug. I think now I have, and would love advice on what to do if I've made some error. The data set in question ("c") has...

...;d like code that prints out the name of this month (January) the next month (February) and the one after that (March) I''ve been able to print this month: <% t = Time.now %> <%= t.strftime(" This Month is: %B") %> But I can''t figure out how to get the next months to print out! -- Posted via http://www.ruby-forum.com/.

Within an R dataset, I have a date field called “date_”. (The dates are in the format “YYYY-MM-DD”, e.g. “1995-12-01”.) How can I add or subtract “1 month” from this date, to get “1996-01-01” or “ “1995-11-01”. --------------------------------- [[alternative HTML version deleted]]

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried

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