search for: keyby

Displaying 6 results from an estimated 6 matches for "keyby".

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2014 Mar 15
0
allocation error and high CPU usage from kworker and migration: memory fragmentation?
...e, parent, by=c('store', 'year') # adds parent column, an integer setkeyv(move, by) # this reduces move to a data.table with at most 6000 rows, but always 4 columns move <- move[, list(revenue=sum(price*units), revenue_PL=sum(price*units*is_PL)), keyby=by] move[, category := gsub(search, replace, filename)] return(move) } -- James Sams sams.james at gmail.com
2012 Oct 26
3
how to make simulation faster
Dear R users, I need to run 1000 simulations to find maximum likelihood estimates. I print my output as a vector. However, it is taking too long. I am running 50 simulations at a time and it is taking me 30 minutes. Once I tried to run 200 simulations at once, after 2 hours I stopped it and saw that only about 40 of them are simulated in those 2 hours. Is there any way to make my simulations
2013 Aug 30
0
ddply for comparing simulation results
...>> Loading required package: data.table >> data.table 1.8.8 For help type: help("data.table") >> > system.time({ >> + df_leads <- data.table(df_leads) >> + df_leads_sum <- df_leads[ >> + , list(count = .N) >> + , keyby = c('isSimulated', 'userId') >> + ] >> + }) >> user system elapsed >> 0.75 0.01 0.76 >> > >> > head(df_leads_sum) >> isSimulated userId count >> 1: 0 1 5 >> 2: 0 2...
2018 Nov 29
3
Unexpected argument-matching when some are missing
On Thu, Nov 29, 2018 at 10:51 AM S Ellison <S.Ellison at lgcgroup.com> wrote: > > > When trying out some variations with `[.data.frame` I noticed some (to me) > > odd behaviour, > > Not just in 'myfun' ... > > plot(x=1:10, y=) > plot(x=1:10, y=, 10:1) > > In both cases, 'y=' is ignored. In the first, the plot is for y=NULL (so not
2013 Apr 29
3
Counting number of consecutive occurrences per rows
Hi, I would appreciate if somebody could help me with following calculation. I have a dataframe, by 10 minutes time, for mostly one year data. This is small example: > dput(test) structure(list(jul = structure(c(14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655, 14655), origin = structure(0, class = "Date")), time =
2013 Jul 12
4
simplify a dataframe
Hello I have the following problem : group the lines of a dataframe when no information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays) and when the value of Debut of lines i = value Fin of lines i-1 I can obtain it with a do loop. Is it possible to avoid the loop ? The dataframe initial is df1 dput(df1) structure(list(Matricule = c(1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L, 6L, 8L, 8L, 8L,