# [R] Convert daily data to weekly data

jim holtman jho|tm@n @end|ng |rom gm@||@com
Tue May 29 23:50:06 CEST 2018

```Forgot the year if you also want to summarise by that.

> x <- structure(list(X1986.01.01.10.30.00 = c(16.8181762695312,
16.8181762695312,
+                                              18.8294372558594, 16 ....
[TRUNCATED]

> library(tidyverse)

> library(lubridate)

> # convert to long form
> x_long <- gather(x, key = 'date', value = "value", -ID)

> # change the date to POSIXct
> x_long\$date <- ymd_hms(substring(x_long\$date, 2, 19))

> # add the week of the year
> x_long\$week <- week(x_long\$date)

> x_long\$year <- year(x_long\$date)

> # average by ID/week
> avg <- x_long %>%
+   group_by(ID, year, week) %>%
+   summarise(avg = mean(value))
> avg
# A tibble: 6 x 4
# Groups:   ID, year [?]
ID  year  week   avg
<int> <dbl> <dbl> <dbl>
1     1 1986.    1.  16.0
2     2 1986.    1.  16.0
3     3 1986.    1.  17.9
4     4 1986.    1.  16.0
5     5 1986.    1.  17.9
6     6 1986.    1.  16.0

Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

On Tue, May 29, 2018 at 7:02 AM, Miluji Sb <milujisb using gmail.com> wrote:

> Dear Petr,
>
> Thanks for your reply and the solution. The example dataset contains data
> for the first six days of the year 1986. "X1986.01.01.10.30.00" is 01
> January 1986 and the rest of the variable is redundant information. The
> last date is given as "X2016.12.31.10.30.00".
>
> I realized that I missed one information in my previous email, I would like
> to compute the weekly average by the variable ID. Thanks again!
>
> Sincerely,
>
> Shouro
>
> On Tue, May 29, 2018 at 3:24 PM, PIKAL Petr <petr.pikal using precheza.cz>
> wrote:
>
> > Hi
> >
> > Based on your explanation I would advice to use
> >
> > ?cut.POSIXt
> >
> > with breaks "week". However your data are rather strange, you have data
> > frame with names which looks like dates
> >
> > names(test)
> > [1] "X1986.01.01.10.30.00" "X1986.01.02.10.30.00" "X1986.01.03.10.30.00"
> > [4] "X1986.01.04.10.30.00" "X1986.01.05.10.30.00" "X1986.01.06.10.30.00"
> > [7] "ID"
> >
> > and under each name you have 6 numeric values
> > test[,1]
> > [1] 16.81818 16.81818 18.82944 16.81818 18.82944 16.83569
> >
> > You (probably) can get dates by
> > as.Date(substring(names(test),2,11), format="%Y.%m.%d")
> > [1] "1986-01-01" "1986-01-02" "1986-01-03" "1986-01-04" "1986-01-05"
> > [6] "1986-01-06" NA
> >
> > but if you want just average those 6 values below each date you could do
> >
> > colMeans(test)
> >
> > and/or bind it together.
> >
> > > ddd<-as.Date(substring(names(test),2,11), format="%Y.%m.%d")
> > > data.frame(ddd, aver=colMeans(test))
> >                             ddd     aver
> > X1986.01.01.10.30.00 1986-01-01 17.49152
> > X1986.01.02.10.30.00 1986-01-02 16.84200
> > X1986.01.03.10.30.00 1986-01-03 16.51526
> > X1986.01.04.10.30.00 1986-01-04 16.90191
> > X1986.01.05.10.30.00 1986-01-05 16.00480
> > X1986.01.06.10.30.00 1986-01-06 16.04405
> > ID                         <NA>  3.50000
> >
> > Cheers
> > Petr
> >
> > Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a
> > podléhají tomuto právně závaznému prohlášení o vyloučení odpovědnosti:
> > https://www.precheza.cz/01-dovetek/ | This email and any documents
> > attached to it may be confidential and are subject to the legally binding
> > disclaimer: https://www.precheza.cz/en/01-disclaimer/
> >
> > > -----Original Message-----
> > > From: R-help [mailto:r-help-bounces using r-project.org] On Behalf Of Miluji
> > Sb
> > > Sent: Tuesday, May 29, 2018 2:59 PM
> > > To: r-help mailing list <r-help using r-project.org>
> > > Subject: [R] Convert daily data to weekly data
> > >
> > > Dear all,
> > >
> > > I have daily data in wide-format from 01/01/1986 to 31/12/2016 by ID. I
> > would
> > > like to convert this to weekly average data. The data has been
> generated
> > by an
> > > algorithm.
> > >
> > > I know that I can use the lubridate package but that would require me
> to
> > first
> > > convert the data to long-form (which is what I want). I am at a bit of
> > loss of
> > > how to extract the date from the variable names and then converting the
> > data
> > > to weekly average. Any help will be high appreciated.
> > >
> > > ### data
> > > structure(list(X1986.01.01.10.30.00 = c(16.8181762695312,
> > > 16.8181762695312, 18.8294372558594, 16.8181762695312,
> > > 18.8294372558594, 16.835693359375 ), X1986.01.02.10.30.00 =
> > > c(16.2272033691406, 16.2272033691406, 18.0772094726562,
> > > 16.2272033691406, 18.0772094726562, 16.2159423828125 ),
> > > X1986.01.03.10.30.00 = c(15.8944396972656, 15.8944396972656,
> > > 17.7444152832031, 15.8944396972656, 17.7444152832031, 15.91943359375
> > > ), X1986.01.04.10.30.00 = c(16.2752380371094, 16.2752380371094,
> > > 18.125244140625, 16.2752380371094, 18.125244140625, 16.3352355957031
> > > ), X1986.01.05.10.30.00 = c(15.3706359863281, 15.3706359863281,
> > > 17.2806396484375, 15.3706359863281, 17.2806396484375,
> > > 15.3556213378906 ), X1986.01.06.10.30.00 = c(15.3798828125,
> > > 15.3798828125, 17.3136291503906, 15.3798828125, 17.3136291503906,
> > > 15.4974060058594), ID = 1:6), .Names = c("X1986.01.01.10.30.00",
> > > "X1986.01.02.10.30.00", "X1986.01.03.10.30.00", "X1986.01.04.10.30.00",
> > > "X1986.01.05.10.30.00", "X1986.01.06.10.30.00", "ID"), row.names =
> c(NA,
> > 6L),
> > > class = "data.frame")
> > >
> > > Sincerely,
> > >
> > > Milu
> > >
> > > [[alternative HTML version deleted]]
> > >
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