# [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
>> >>       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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>> >>
>> >
>>
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>>
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