[R] How to select max data according to week?
Bert Gunter
bgunter@4567 @end|ng |rom gm@||@com
Wed Jun 19 16:04:13 CEST 2019
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 <ericjberger using 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.4567 using gmail.com>
> wrote:
>
>> My apologies. I negected to cc r-help. -- Bert
>>
>>
>>
>> On Tue, Jun 18, 2019 at 11:21 AM Bert Gunter <bgunter.4567 using 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 <
>> > aisyahzakaria using 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. Project and is believed to be
>> >> clean.
>> >>
>> >> ______________________________________________
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>> >> and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >
>>
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>>
>> ______________________________________________
>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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