Jeff,
thank you. However, if I knew how to do this, I would probably not
have asked :-)-O
I think I have been reasonably comprehensive in describing my issue, but
let me do it now with the real life problem:
My malpractice insurance gives me a discount if I consult up to 22
hours per week in a 3 months period.
I add every patient, date and minutes whenever I see her into a MySQL
database. I want to file the report of my hours worked with them for
the first 3 month period (November to January and not properly quarterly
unfortunately :-)-0), and while I can generate this with LyX/LateX and
knitR producing a (super)tabular table containing the full list, and
tables for time per week and time per month I really can't figure out is
how to average the hours worked per week for each month (even if weeks
don't align with months properly :-)-O)
While I am at it how would I get this to sort properly (year, month) if
I used the proper names of the months, ie '%Y %B' or '%B %Y'?
CONSMINUTES %>%
select(datum, dauer) %>%
group_by(month = format(datum, '%Y %m'),
week = format(datum, '%V')) %>%
summarise_if(is.numeric, sum) %>%
mutate(hm=sprintf("%d Hour%s %d Minutes", dauer %/% 60,
ifelse((dauer %/% 60) == 1, " ", "s"), dauer %% 60)) %>%
select(-dauer)
Any help, or just pointers to where I can read this up, are highly
appreciated.
greetings, el
On 2021-03-25 22:37 , Jeff Newmiller wrote:
> This is a very unclear question. Weeks don't line up with months..
> so you need to clarify how you would do this or at least give an
> explicit example of input data and result data.
>
> On March 25, 2021 11:34:15 AM PDT, Dr Eberhard W Lisse
<nos...@lisse.na> wrote:
>> Thanks, that is helpful.
>>
>> But, how do I group it to produce hours worked per week per month?
>>
>> el
>>
>>
>> On 2021-03-25 19:03 , Greg Snow wrote:
>>> Here is one approach:
>>>
>>> tmp <- data.frame(min=seq(0,150, by=15))
>>>
>>> tmp %>%
>>> mutate(hm=sprintf("%2d Hour%s %2d Minutes",
>>> min %/% 60, ifelse((min %/% 60) == 1, " ", "s"),
>>> min %% 60))
>>>
>>> You could replace `sprintf` with `str_glue` (and update the syntax
>>> as well) if you realy need tidyverse, but you would also loose some
>>> formatting capability.
>>>
>>> I don't know of tidyverse versions of `%/%` or `%%`. If you need
>>> the numeric values instead of a string then just remove the
>>> `sprintf` and use mutate directly with `min %/% 60` and `min %% 60`.
>>>
>>> This of course assumes all of your data is in minutes (by the time
>>> you pipe to this code) and that all hours have 60 minutes (I don't
>>> know of any leap hours.
>>>
>>> On Sun, Mar 21, 2021 at 8:31 AM Dr Eberhard W Lisse <nos...@lisse.na>
>> wrote:
>>>>
>>>> Hi,
>>>>
>>>> I have minutes worked by day (with some more information)
>>>>
>>>> which when using
>>>>
>>>> library(tidyverse)
>>>> library(lubridate)
>>>>
>>>> run through
>>>>
>>>> CONSMINUTES %>%
>>>> select(datum, dauer) %>%
>>>> arrange(desc(datum))
>>>>
>>>> look somewhat like
>>>>
>>>> # A tibble: 142 x 2
>>>> datum dauer
>>>> <date> <int>
>>>> 1 2021-03-18 30
>>>> 2 2021-03-17 30
>>>> 3 2021-03-16 30
>>>> 4 2021-03-16 30
>>>> 5 2021-03-16 30
>>>> 6 2021-03-16 30
>>>> 7 2021-03-11 30
>>>> 8 2021-03-11 30
>>>> 9 2021-03-11 30
>>>> 10 2021-03-11 30
>>>> # … with 132 more rows
>>>>
>>>> I can extract minutes per hour
>>>>
>>>> CONSMINUTES %>%
>>>> select(datum, dauer) %>%
>>>> group_by(week = format(datum, '%Y %V'))%>%
>>>> summarise_if(is.numeric, sum)
>>>>
>>>> and minutes per month
>>>>
>>>> CONSMINUTES %>%
>>>> select(datum, dauer) %>%
>>>> group_by(month = format(datum, '%Y %m'))%>%
>>>> summarise_if(is.numeric, sum)
>>>>
>>>> I need to show the time worked per week per month in the format of
>>>>
>>>> '# hours # minutes'
>>>>
>>>> and would like to also be able to show the average time per week
>>>> per month.
>>>>
>>>> How can I do that (preferably with tidyverse :-)-O)?
>>>>
>>>> greetings, el
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