[R] Sum data according to date in sequence

Rui Barradas ru|pb@rr@d@@ @end|ng |rom @@po@pt
Sat Nov 4 17:56:20 CET 2023


Às 01:49 de 03/11/2023, roslinazairimah zakaria escreveu:
> Hi all,
> 
> This is the data:
> 
>> dput(head(dt1,20))structure(list(StationName = c("PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1",
> "PALO ALTO CA / CAMBRIDGE #1"), date = c("1/14/2016", "1/14/2016",
> "1/14/2016", "1/15/2016", "1/15/2016", "1/15/2016", "1/15/2016",
> "1/16/2016", "1/16/2016", "1/16/2016", "1/16/2016", "1/16/2016",
> "1/16/2016", "1/16/2016", "1/17/2016", "1/17/2016", "1/17/2016",
> "1/17/2016", "1/17/2016", "1/18/2016"), time = c("12:09", "19:50",
> "20:22", "8:25", "14:23", "18:17", "21:46", "10:19", "12:12",
> "14:12", "16:22", "19:16", "19:19", "20:24", "9:54", "12:16",
> "13:53", "19:03", "22:00", "8:58"), EnergykWh = c(4.680496, 6.272414,
> 1.032782, 11.004884, 10.096824, 6.658797, 4.808874, 1.469384,
> 2.996239, 0.303222, 4.988339, 8.131804, 0.117156, 3.285669, 1.175608,
> 3.677487, 1.068393, 8.820755, 8.138583, 9.0575)), row.names = c(NA,
> 20L), class = "data.frame")
> 
> 
> I would like to sum EnergykW data by the date. E.g. all values for
> EnergykWh on 1/14/2016
> 
> 
> On Fri, Nov 3, 2023 at 8:10 AM jim holtman <jholtman using gmail.com> wrote:
> 
>> How about send a 'dput' of some sample data.  My guess is that your date
>> is 'character' and not 'Date'.
>>
>> Thanks
>>
>> 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 Thu, Nov 2, 2023 at 4:24 PM roslinazairimah zakaria <
>> roslinaump using gmail.com> wrote:
>>
>>> Dear all,
>>>
>>> I have this set of data. I would like to sum the EnergykWh according date
>>> sequences.
>>>
>>>> head(dt1,20)                   StationName      date  time EnergykWh
>>> 1  PALO ALTO CA / CAMBRIDGE #1 1/14/2016 12:09  4.680496
>>> 2  PALO ALTO CA / CAMBRIDGE #1 1/14/2016 19:50  6.272414
>>> 3  PALO ALTO CA / CAMBRIDGE #1 1/14/2016 20:22  1.032782
>>> 4  PALO ALTO CA / CAMBRIDGE #1 1/15/2016  8:25 11.004884
>>> 5  PALO ALTO CA / CAMBRIDGE #1 1/15/2016 14:23 10.096824
>>> 6  PALO ALTO CA / CAMBRIDGE #1 1/15/2016 18:17  6.658797
>>> 7  PALO ALTO CA / CAMBRIDGE #1 1/15/2016 21:46  4.808874
>>> 8  PALO ALTO CA / CAMBRIDGE #1 1/16/2016 10:19  1.469384
>>> 9  PALO ALTO CA / CAMBRIDGE #1 1/16/2016 12:12  2.996239
>>> 10 PALO ALTO CA / CAMBRIDGE #1 1/16/2016 14:12  0.303222
>>> 11 PALO ALTO CA / CAMBRIDGE #1 1/16/2016 16:22  4.988339
>>> 12 PALO ALTO CA / CAMBRIDGE #1 1/16/2016 19:16  8.131804
>>> 13 PALO ALTO CA / CAMBRIDGE #1 1/16/2016 19:19  0.117156
>>> 14 PALO ALTO CA / CAMBRIDGE #1 1/16/2016 20:24  3.285669
>>> 15 PALO ALTO CA / CAMBRIDGE #1 1/17/2016  9:54  1.175608
>>> 16 PALO ALTO CA / CAMBRIDGE #1 1/17/2016 12:16  3.677487
>>> 17 PALO ALTO CA / CAMBRIDGE #1 1/17/2016 13:53  1.068393
>>> 18 PALO ALTO CA / CAMBRIDGE #1 1/17/2016 19:03  8.820755
>>> 19 PALO ALTO CA / CAMBRIDGE #1 1/17/2016 22:00  8.138583
>>> 20 PALO ALTO CA / CAMBRIDGE #1 1/18/2016  8:58  9.057500
>>>
>>> I have tried this:
>>> library(dplyr)
>>> sums <- dt1 %>%
>>>    group_by(date) %>%
>>>    summarise(EnergykWh = sum(EnergykWh))
>>>
>>> head(sums,20)
>>>
>>> The date is not by daily sequence but by year sequence.
>>>
>>>> head(sums,20)# A tibble: 20 × 2
>>>     date      EnergykWh
>>>     <chr>         <dbl> 1 1/1/2017     25.3   2 1/1/2018     61.0   3
>>> 1/1/2019      0.627 4 1/1/2020     10.7   5 1/10/2017    69.4   6
>>> 1/10/2018    54.5   7 1/10/2019    49.1   8 1/10/2020    45.9   9
>>> 1/11/2017    73.9  10 1/11/2018    53.3  11 1/11/2019    93.5  12
>>> 1/11/2020    66.7  13 1/12/2017    78.6  14 1/12/2018    42.2  15
>>> 1/12/2019    22.7  16 1/12/2020    80.9  17 1/13/2017    85.6  18
>>> 1/13/2018    46.4  19 1/13/2019    40.0  20 1/13/2020   121.
>>>
>>>
>>>
>>> Thank you very much for any help given.
>>>
>>>
>>> --
>>> *Roslinazairimah Zakaria*
>>> *Tel: +609-5492370; Fax. No.+609-5492766*
>>>
>>> *Email: roslinazairimah using ump.edu.my <roslinazairimah using ump.edu.my>;
>>> roslinaump using gmail.com <roslinaump using gmail.com>*
>>> Faculty of Industrial Sciences & Technology
>>> University Malaysia Pahang
>>> Lebuhraya Tun Razak, 26300 Gambang, Pahang, Malaysia
>>>
>>>          [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>>> PLEASE do read the posting guide
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>>>
>>
> 
Hello,

Here are two solutions.

1. Base R

Though I don't coerce the date column to class "Date", it seems to work.


aggregate(EnergykWh ~ date, dt1, sum)
#>        date EnergykWh
#> 1 1/14/2016  11.98569
#> 2 1/15/2016  32.56938
#> 3 1/16/2016  21.29181
#> 4 1/17/2016  22.88083
#> 5 1/18/2016   9.05750


2. Package dplyr.
First column date is coerced from class "character" to class "Date".
Then the grouped sums are computed.


suppressPackageStartupMessages(
   library(dplyr)
)

dt1 %>%
   mutate(date = as.Date(date, "%m/%d/%Y")) %>%
   summarise(EnergykWh = sum(EnergykWh), .by = date)
#>         date EnergykWh
#> 1 2016-01-14  11.98569
#> 2 2016-01-15  32.56938
#> 3 2016-01-16  21.29181
#> 4 2016-01-17  22.88083
#> 5 2016-01-18   9.05750


As you can see, the results are the same.

Also, this exact problem is one of the most asked on StackOverflow. 
Maybe you could try searching there for a solution. My code above is 
also exactly the code in [1], though I had already this answer written. 
I only checked after :(.


[1] 
https://stackoverflow.com/questions/61548758/r-how-sum-values-by-group-by-date


Hope this helps,

Rui Barradas



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