search for: 19e6l

...t;,header=TRUE) MyDate2 = read.csv("C:\\Date2.txt",header=TRUE) MyDate1 Date1 = MyDate1$Date class(Date1) mode(Date1) Date1 Date1 = as.Date(Date1, "%m/%d/%y") class(Date1) mode(Date1) Date1 MyDate2 Date2 = MyDate2$EnDate class(Date2) mode(Date2) Date2 Date2 = strptime(Date2 + 19e6L, "%Y%m%d") class(Date2) mode(Date2) Date2 Date2 = as.Date(Date2) class(Date2) mode(Date2) Date2 Results follow. Thanks, Mark > MyDate1 = read.csv("C:\\Date1.txt",header=TRUE) > MyDate2 = read.csv("C:\\Date2.txt",header=TRUE) > > MyDate1 Date...

...a.frame Y from X where X$PL_Pos is summed by each unique X$MyDate. Y should end up with two (or more) columns Y$MyDate and Y$PL_Sum with its value being the cumsum of all the values in X for that date. - a 'daily cumsum'. Thanks, Mark TStoDate = function (TSDate) { X = strptime(TSDate + 19e6L, "%Y%m%d") return(as.Date(X)) } X = structure(list(EnDate = c(1090803, 1090804, 1090805, 1090806, 1090806, 1090810, 1090811, 1090812, 1090813, 1090817, 1090819, 1090820, 1090820, 1090824, 1090825, 1090825, 1090826, 1090826, 1090827, 1090827, 1090827), PL_Pos = c(174, -26, 614, 344, -26,...

Hi, Is there a function that will convert this sort of date code which looks like "years from 1900 + month_number + day_number" with no spaces? As an example Jan. 3rd 2008 would be written as 1080103. Thanks, Mark

...ode. What I need to do is run it on all the files in the directory tested and augment the data frame I'm building to have more results columns.How can I do that? Here's the code: library(chron) # .Holidays / is.holiday / is.weekend TStoDate = function (TSDate) { X = strptime(TSDate + 19e6L, "%Y%m%d") return(as.Date(X)) } FirstDate = 1090601 LastDate = 1101101 StartDate = TStoDate(FirstDate) EndDate = TStoDate(LastDate) # Create a sequence of days from start to finish # Then remove weekends and holidays and turn into a data.frame dd <- seq(StartDate, EndDate, by = &q...