Eric: I believe you're doing something different than I did. I broke up each month into biweekly periods, 2+ per month. You seem to be grouping the overall entire period into biweekly intervals -- apologies if I'm wrong, but if I understood correctly, that's not the same thing. I do not know which of us -- if either -- has interpreted her query correctly.
Cheers, Bert On Wed, Jun 19, 2019 at 2:35 AM Eric Berger <ericjber...@gmail.com> wrote: > Hi Siti, > I didn't test Bert's code but I assume it's fine. :-) > I would take a different approach than Bert. I prefer to use a package > such as lubridate to handle the date wrangling, and a package such as dplyr > to handle the grouping and max extraction. > It may be overkill for this problem, but these are great packages to > become familiar with. > If one can take the actual week of the year as an acceptable definition of > week, then here's my approach. > > library(lubridate) > library(dplyr) > > # Step 1: start with Bert's code to create sample data > ## create some example data for 3 months in 2000 > d<- 2e7 +c(113:131,201:228, 301:330) ## dates > conc <- runif(length(d)) # concentrations > > # Step 2: collect the data into a data frame > df <- data.frame( dt=d, conc=conc) > > # Step 3: use lubridate's ymd() function to parse the dates, its week() > function to identify the week of the year, and define the new column > 'wkpair' that groups the weeks 2-at-a-time > df2 <- dplyr::mutate( df, > wkpair=as.integer(floor(lubridate::week(lubridate::ymd(dt) )/2)) ) > > # Step 4: group by the wkpair and use dplyr's summarise to get the info > you wanted > df3 <- dplyr::group_by(df2,wkpair) %>% > dplyr::summarise( from=min(dt), to=max(dt), maxconc=max(conc)) > %>% > dplyr::select(from,to,maxconc) > > df3 > > # A tibble: 6 x 3 > from to maxconc > <dbl> <dbl> <dbl> > 1 20000113 20000121 0.963 > 2 20000122 20000204 0.988 > 3 20000205 20000218 0.939 > 4 20000219 20000303 0.883 > 5 20000304 20000317 0.863 > 6 20000318 20000330 0.765 > > HTH, > Eric > > > > On Tue, Jun 18, 2019 at 9:39 PM Bert Gunter <bgunter.4...@gmail.com> > wrote: > >> My apologies. I negected to cc r-help. -- Bert >> >> >> >> On Tue, Jun 18, 2019 at 11:21 AM Bert Gunter <bgunter.4...@gmail.com> >> wrote: >> >> > >> > I assume that 20000215 means year 2000, month 2, day 15. >> > I also assume that you want maxes for the first 2 weeks of a month, the >> > second 2 weeks, and any remaining days. >> > I also assume that this might be desired for arbitrary numbers of years, >> > months, and days. >> > >> > The following is one way to do this. As it's a slight pain to cut and >> > paste email data as text into R (use ?dput or R code to run to provide >> > example data instead), I just made up my own. You'll have to do the >> > following within a data frame through extraction or by using with() of >> > course. >> > >> > ## create some example data for 3 months in 2000 >> > d<- 2e7 +c(113:131,201:228, 301:330) ## dates >> > conc <- runif(length(d)) # concentrations >> > >> > ## convert the date to character to extract year, month, and day >> > cdate <- as.character(d) >> > ## use substr to to the extraction >> > year <- substr(cdate,1,4) >> > mon <- substr(cdate,5,6) >> > day <- substr(cdate, 7,8) >> > >> > ## convert day to numeric and use cut() to group into the biweekly >> periods. >> > d14 <- cut(as.numeric(day), c(0,14.5,28.5, 32)) >> > >> > ## Use tapply() to create your desired table of results. >> > tapply(conc, list(year, d14, mon), max, na.rm = TRUE) >> > >> > ## Results >> > >> > , , 01 >> > >> > (0,14.5] (14.5,28.5] (28.5,32] >> > 2000 0.7357389 0.9655391 0.7962965 >> > >> > , , 02 >> > >> > (0,14.5] (14.5,28.5] (28.5,32] >> > 2000 0.8193979 0.9487207 NA >> > >> > , , 03 >> > >> > (0,14.5] (14.5,28.5] (28.5,32] >> > 2000 0.9718919 0.9997093 0.168659 >> > >> > >> > Cheers, >> > Bert >> > >> > Bert Gunter >> > >> > >> > >> > >> > On Tue, Jun 18, 2019 at 8:53 AM SITI AISYAH BINTI ZAKARIA < >> > aisyahzaka...@unimap.edu.my> wrote: >> > >> >> Hi, >> >> >> >> I'm Aisyah..I have a problem to run my R coding. I want to select >> maximum >> >> value according to week. >> >> >> >> here is my data >> >> >> >> Date O3_Conc >> >> 20000101 0.033 >> >> 20000102 0.023 >> >> 20000103 0.025 >> >> 20000104 0.041 >> >> 20000105 0.063 >> >> 20000106 0.028 >> >> 20000107 0.068 >> >> 20000108 0.048 >> >> 20000109 0.037 >> >> 20000110 0.042 >> >> 20000111 0.027 >> >> 20000112 0.035 >> >> 20000113 0.063 >> >> 20000114 0.035 >> >> 20000115 0.042 >> >> 20000116 0.028 >> >> >> >> I want to find the max value from column O3_Conc for only 14 days that >> >> refer to biweekly in month. And the next 14 days for the max value. >> >> >> >> I hope that I can get the result like this: >> >> >> >> Date Max O3_Conc >> >> 20000101 - 20000114 0.068 >> >> 20000115 - 20000129 0.061 >> >> >> >> I try many coding but still unavailable. >> >> >> >> this example my coding >> >> >> >> library(plyr) >> >> data.frame(CA0003) >> >> >> >> # format weeks as per requirement (replace "00" with "52" and >> >> adjust corresponding year) >> >> tmp <- list() >> >> tmp$y <- format(df$Date, format="%Y") >> >> tmp$w <- format(df$Date, format="%U") >> >> tmp$y[tmp$w=="00"] <- >> as.character(as.numeric(tmp$y[tmp$w=="00"]) - >> >> 14) >> >> tmp$w[tmp$w=="00"] <- "884" >> >> df$week <- paste(tmp$y, tmp$w, sep = "-") >> >> >> >> # get summary >> >> df2 <- ddply(df, .(week),transform, O3_Conc=max(O3_Conc)) >> >> >> >> # include week ending date >> >> tmp$week.ending <- lapply(df2$week, function(x) rev(df[df$week >> ==x, >> >> "O3_Conc"])[[1]]) >> >> df2$week.ending <- sapply(tmp$week.ending, max(O3_Conc, TRUE) >> >> >> >> output >> >> Site_Id Site_Location >> Date >> >> Year O3_Conc Month Day week >> >> 1 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000101 >> >> 2000 0.033 1 1 NULL-NULL >> >> 2 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000102 >> >> 2000 0.023 1 2 NULL-NULL >> >> 3 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000103 >> >> 2000 0.025 1 3 NULL-NULL >> >> 4 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000104 >> >> 2000 0.041 1 4 NULL-NULL >> >> 5 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000105 >> >> 2000 0.063 1 5 NULL-NULL >> >> 6 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000106 >> >> 2000 0.028 1 6 NULL-NULL >> >> 7 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000107 >> >> 2000 0.068 1 7 NULL-NULL >> >> 8 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000108 >> >> 2000 0.048 1 8 NULL-NULL >> >> 9 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000109 >> >> 2000 0.037 1 9 NULL-NULL >> >> 10 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000110 >> >> 2000 0.042 1 10 NULL-NULL >> >> 11 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >> 20000111 >> >> 2000 0.027 1 11 NULL-NULL >> >> >> >> >> >> >> >> >> >> -- >> >> This message has been scanned by E.F.A. 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