Hi Ragia, I can't make out your data or desired result, but it sounds like aggregate() might get you started. If you need more help, please repost your data using dput() and do NOT post in HTML so that we can see what your desired result looks like. Sarah On Fri, Sep 4, 2015 at 10:12 AM, Ragia Ibrahim <ragia11 at hotmail.com> wrote:> Dear Group,kinldy, I have the following data frame df > id value1 1 4 2 1 4 3 1 6 4 1 6 5 2 1.5 6 2 2.5 7 2 2.5 8 2 2.5 > > add rank column regarding id coulmn where rank for the highest value would be 1, others rank would be the (value/ value of heighest)/ number of rows that took the same value > thus the data frame should be > id value Rank1 1 4 0.332 1 4 0.333 1 6 0.54 1 6 0.55 2 1.5 0.6 6 2 2.5 0.337 2 2.5 0.338 2 2.5 0.33 > > how to reach this resultthanks in advanceRagia > [[alternative HTML version deleted]] >-- Sarah Goslee http://www.functionaldiversity.org
thanks for replying, I attached the data frame for source "i" I want to sum the values and get the max value then add a new column called rank . That new column cell value for each source i and for specific id would be (value/max value) * count of rows that have the same criteria "same i and same id" many thanksRagia> Date: Fri, 4 Sep 2015 10:19:35 -0400 > Subject: Re: [R] groups Rank > From: sarah.goslee at gmail.com > To: ragia11 at hotmail.com > CC: r-help at r-project.org > > Hi Ragia, > > I can't make out your data or desired result, but it sounds like > aggregate() might get you started. If you need more help, please > repost your data using dput() and do NOT post in HTML so that we can > see what your desired result looks like. > > Sarah > > On Fri, Sep 4, 2015 at 10:12 AM, Ragia Ibrahim <ragia11 at hotmail.com> wrote: > > Dear Group,kinldy, I have the following data frame df > > id value1 1 4 2 1 4 3 1 6 4 1 6 5 2 1.5 6 2 2.5 7 2 2.5 8 2 2.5 > > > > add rank column regarding id coulmn where rank for the highest value would be 1, others rank would be the (value/ value of heighest)/ number of rows that took the same value > > thus the data frame should be > > id value Rank1 1 4 0.332 1 4 0.333 1 6 0.54 1 6 0.55 2 1.5 0.6 6 2 2.5 0.337 2 2.5 0.338 2 2.5 0.33 > > > > how to reach this resultthanks in advanceRagia > > [[alternative HTML version deleted]] > > > > -- > Sarah Goslee > http://www.functionaldiversity.org-------------- next part -------------- An embedded and charset-unspecified text was scrubbed... Name: measures.txt URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20150905/18b6e044/attachment.txt>
Please use dput() to provide data, rather than expecting people to open random attachments. Besides, there are multiple options for getting data into R, and we need to know exactly what you did. dput() is faster and easier. What have you tried? Did you look at aggregate() as I suggested? Sarah On Sat, Sep 5, 2015 at 10:44 AM, Ragia Ibrahim <ragia11 at hotmail.com> wrote:> thanks for replying, I attached the data frame > for source "i" I want to sum the values and get the max value then add a > new column called rank . That new column cell value for each source i and > for specific id would be (value/max value) * count of rows that have the > same criteria "same i and same id" > > many thanks > Ragia > >> Date: Fri, 4 Sep 2015 10:19:35 -0400 >> Subject: Re: [R] groups Rank >> From: sarah.goslee at gmail.com >> To: ragia11 at hotmail.com >> CC: r-help at r-project.org >> >> Hi Ragia, >> >> I can't make out your data or desired result, but it sounds like >> aggregate() might get you started. If you need more help, please >> repost your data using dput() and do NOT post in HTML so that we can >> see what your desired result looks like. >> >> Sarah >> >> On Fri, Sep 4, 2015 at 10:12 AM, Ragia Ibrahim <ragia11 at hotmail.com> >> wrote: >> > Dear Group,kinldy, I have the following data frame df >> > id value1 1 4 2 1 4 3 1 6 4 1 6 5 2 1.5 6 2 2.5 7 2 2.5 8 2 2.5 >> > >> > add rank column regarding id coulmn where rank for the highest value >> > would be 1, others rank would be the (value/ value of heighest)/ number of >> > rows that took the same value >> > thus the data frame should be >> > id value Rank1 1 4 0.332 1 4 0.333 1 6 0.54 1 6 0.55 2 1.5 0.6 6 2 2.5 >> > 0.337 2 2.5 0.338 2 2.5 0.33 >> > >> > how to reach this resultthanks in advanceRagia >> > [[alternative HTML version deleted]] >> >
apology for re sending the Email, I changed the format to plain text as I have
been advised
the?data ?is as follow??
thanks Sarah,?
I used pdut, and here is the data as written on R..I attached the dput result
structure(list(Measure_id = c(1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3,?
3, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 2,?
3, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3), i = c(5, 5,?
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,?
5, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,?
7, 7, 7, 7), j = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2,?
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,?
3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5), id = c(1, 2, 3, 4, 5,?
6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8,?
9, 10, 11, 12), value = c(2, 1.5, 0, 0, 1, 0.5, 0, 0, 0, 0, 0.5,?
2, 2, 1.5, 0, 1, 2, 0, 0.5, 1.44269504088896, 0, 0, 0, 0, 1,?
1.5, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0,?
0, 0, 0)), .Names = c("Measure_id", "i", "j",
"id", "value"), row.names = c(NA,?
48L), class = "data.frame")
the?data ?is as follow??:
? ?Measure_id i j id value
1 ? ? ? ? ? 1 5 1 ?1 ? 2.0
2 ? ? ? ? ? 1 5 2 ?1 ? 2.0
3 ? ? ? ? ? 1 5 1 ?2 ? 1.5
4 ? ? ? ? ? 1 5 2 ?2 ? 1.5
5 ? ? ? ? ? 1 5 1 ?3 ? 0.0
6 ? ? ? ? ? 1 5 2 ?3 ? 0.0
7 ? ? ? ? ? 1 5 1 ?4 ? 0.0
8 ? ? ? ? ? 1 5 2 ?4 ? 1.0
9 ? ? ? ? ? 1 5 1 ?5 ? 1.0
10 ? ? ? ? ?1 5 2 ?5 ? 2.0
.. ? ? ? ?... . . .. ? ...
I want to add a probability column, ?the prob column depends on id grouped by
for each i
the rank will be current (value / max value ) for the same id for specific i, it
would be
?Measure_id i j id value ? ?prob
1 ? ? ? ? ? 1 5 1 ?1 ? 2.0 ? ? ? ? ?2/2 ?
2 ? ? ? ? ? 1 5 2 ?1 ? 2.0 ? ? ? ? ?2/2 ?
3 ? ? ? ? ? 1 5 1 ?2 ? 1.5 ? ? ? ? ?1.5/1.5 ??
4 ? ? ? ? ? 1 5 2 ?2 ? 1.5 ? ? ? ? ?1.5/1.5?
5 ? ? ? ? ? 1 5 1 ?3 ? 0.0 ? ? ? ? ?0
6 ? ? ? ? ? 1 5 2 ?3 ? 0.0 ? ? ? ? ?0
7 ? ? ? ? ? 1 5 1 ?4 ? 0.0 ? ? ? ? ?0/1 ?
8 ? ? ? ? ? 1 5 2 ?4 ? 1.0 ? ? ? ? ?1/1 ?
9 ? ? ? ? ? 1 5 1 ?5 ? 1.0 ? ? ? ? ?1/2
10 ? ? ? ? ?1 5 2 ?5 ? 2.0 ? ? ? ? ?2/3
.. ? ? ? ?... . . .. ? ...
then I want to add a rank column that rank regarding probability, if the
probability equal they took the same rank for the
same id belongs to the same i, otherwize lower probability took higher rank for
examole if we have three values for i=7 and for the three values the id is 1 and
the probability is ( .2,.4,.5) the rank should be 3,2,1
I looked at aggregate and dplyr...should I use for loop and subset each i and id
rows do calculations and then group them again ??
is there easier way?
replying ?highly appreciated>
>
>
>> Date: Sun, 6 Sep 2015 19:02:02 -0400
>> Subject: Re: [R] groups Rank
>> From: sarah.goslee at gmail.com
>> To: ragia11 at hotmail.com
>> CC: r-help at r-project.org
>>
>> Please use dput() to provide data, rather than expecting people to
>> open random attachments. Besides, there are multiple options for
>> getting data into R, and we need to know exactly what you did. dput()
>> is faster and easier.
>>
>> What have you tried? Did you look at aggregate() as I suggested?
>>
>> Sarah
>>
>> On Sat, Sep 5, 2015 at 10:44 AM, Ragia Ibrahim <ragia11 at
hotmail.com> wrote:
>>> thanks for replying, I attached the data frame
>>> for source "i" I want to sum the values and get the max
value then add a
>>> new column called rank . That new column cell value for each source
i and
>>> for specific id would be (value/max value) * count of rows that
have the
>>> same criteria "same i and same id"
>>>
>>> many thanks
>>> Ragia
>>>
>>>> Date: Fri, 4 Sep 2015 10:19:35 -0400
>>>> Subject: Re: [R] groups Rank
>>>> From: sarah.goslee at gmail.com
>>>> To: ragia11 at hotmail.com
>>>> CC: r-help at r-project.org
>>>>
>>>> Hi Ragia,
>>>>
>>>> I can't make out your data or desired result, but it sounds
like
>>>> aggregate() might get you started. If you need more help,
please
>>>> repost your data using dput() and do NOT post in HTML so that
we can
>>>> see what your desired result looks like.
>>>>
>>>> Sarah
>>>>
>>>> On Fri, Sep 4, 2015 at 10:12 AM, Ragia Ibrahim <ragia11 at
hotmail.com>
>>>> wrote:
>>>>> Dear Group,kinldy, I have the following data frame df
>>>>> id value1 1 4 2 1 4 3 1 6 4 1 6 5 2 1.5 6 2 2.5 7 2 2.5 8 2
2.5
>>>>>
>>>>> add rank column regarding id coulmn where rank for the
highest value
>>>>> would be 1, others rank would be the (value/ value of
heighest)/
> number of
>>>>> rows that took the same value
>>>>> thus the data frame should be
>>>>> id value Rank1 1 4 0.332 1 4 0.333 1 6 0.54 1 6 0.55 2 1.5
0.6 6 2 2.5
>>>>> 0.337 2 2.5 0.338 2 2.5 0.33
>>>>>
>>>>> how to reach this resultthanks in advanceRagia
>>>>> [[alternative HTML version deleted]]
>>>>>
Hi OK, thanks for sending dput result. I am still not sure what exactly you want. Using ?ave you can get result of x/max(x) dat$prob <- ave(dat$value, paste(dat$id, dat$i), FUN= function(x) x/max(x)) however in case max(x) is zero the result is NA You can change it to zero dat$prob[is.nan(dat$prob)] <- 0 and compute rank value by similar process. dat$rankvalue <- ave(dat$prob, paste(dat$id, dat$i), FUN = rank) But I am not sure if this is the desired result. Cheers Petr> -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ragia > Ibrahim > Sent: Monday, September 07, 2015 10:12 AM > To: Sarah Goslee; r-help at r-project.org > Subject: Re: [R] groups Rank > > apology for re sending the Email, I changed the format to plain text as > I have been advised the data is as follow > > thanks Sarah, > I used pdut, and here is the data as written on R..I attached the dput > result structure(list(Measure_id = c(1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, > 3, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, > 3, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3), i = c(5, 5, 5, 5, 5, 5, 5, 5, > 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 7, > 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7), j = c(1, 1, 1, 1, 1, > 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, > 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5), id = c(1, 2, > 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, > 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, > 11, 12), value = c(2, 1.5, 0, 0, 1, 0.5, 0, 0, 0, 0, 0.5, 2, 2, 1.5, 0, > 1, 2, 0, 0.5, 1.44269504088896, 0, 0, 0, 0, 1, 1.5, 0, 0, 1, 0, 0, 0, > 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0)), .Names > c("Measure_id", "i", "j", "id", "value"), row.names = c(NA, 48L), class > = "data.frame") > > the data is as follow : > > Measure_id i j id value > 1 1 5 1 1 2.0 > 2 1 5 2 1 2.0 > 3 1 5 1 2 1.5 > 4 1 5 2 2 1.5 > 5 1 5 1 3 0.0 > 6 1 5 2 3 0.0 > 7 1 5 1 4 0.0 > 8 1 5 2 4 1.0 > 9 1 5 1 5 1.0 > 10 1 5 2 5 2.0 > .. ... . . .. ... > I want to add a probability column, the prob column depends on id > grouped by for each i the rank will be current (value / max value ) for > the same id for specific i, it would be > > Measure_id i j id value prob > 1 1 5 1 1 2.0 2/2 > 2 1 5 2 1 2.0 2/2 > 3 1 5 1 2 1.5 1.5/1.5 > 4 1 5 2 2 1.5 1.5/1.5 > 5 1 5 1 3 0.0 0 > 6 1 5 2 3 0.0 0 > 7 1 5 1 4 0.0 0/1 > 8 1 5 2 4 1.0 1/1 > 9 1 5 1 5 1.0 1/2 > 10 1 5 2 5 2.0 2/3 > .. ... . . .. ... > > then I want to add a rank column that rank regarding probability, if > the probability equal they took the same rank for the same id belongs > to the same i, otherwize lower probability took higher rank for examole > if we have three values for i=7 and for the three values the id is 1 > and the probability is ( .2,.4,.5) the rank should be 3,2,1 > > > > I looked at aggregate and dplyr...should I use for loop and subset each > i and id rows do calculations and then group them again ?? > is there easier way? > > replying highly appreciated > > > > > > > >> Date: Sun, 6 Sep 2015 19:02:02 -0400 > >> Subject: Re: [R] groups Rank > >> From: sarah.goslee at gmail.com > >> To: ragia11 at hotmail.com > >> CC: r-help at r-project.org > >> > >> Please use dput() to provide data, rather than expecting people to > >> open random attachments. Besides, there are multiple options for > >> getting data into R, and we need to know exactly what you did. > dput() > >> is faster and easier. > >> > >> What have you tried? Did you look at aggregate() as I suggested? > >> > >> Sarah > >> > >> On Sat, Sep 5, 2015 at 10:44 AM, Ragia Ibrahim <ragia11 at hotmail.com> > wrote: > >>> thanks for replying, I attached the data frame for source "i" I > want > >>> to sum the values and get the max value then add a new column > called > >>> rank . That new column cell value for each source i and for > specific > >>> id would be (value/max value) * count of rows that have the same > >>> criteria "same i and same id" > >>> > >>> many thanks > >>> Ragia > >>> > >>>> Date: Fri, 4 Sep 2015 10:19:35 -0400 > >>>> Subject: Re: [R] groups Rank > >>>> From: sarah.goslee at gmail.com > >>>> To: ragia11 at hotmail.com > >>>> CC: r-help at r-project.org > >>>> > >>>> Hi Ragia, > >>>> > >>>> I can't make out your data or desired result, but it sounds like > >>>> aggregate() might get you started. If you need more help, please > >>>> repost your data using dput() and do NOT post in HTML so that we > >>>> can see what your desired result looks like. > >>>> > >>>> Sarah > >>>> > >>>> On Fri, Sep 4, 2015 at 10:12 AM, Ragia Ibrahim > >>>> <ragia11 at hotmail.com> > >>>> wrote: > >>>>> Dear Group,kinldy, I have the following data frame df id value1 1 > >>>>> 4 2 1 4 3 1 6 4 1 6 5 2 1.5 6 2 2.5 7 2 2.5 8 2 2.5 > >>>>> > >>>>> add rank column regarding id coulmn where rank for the highest > >>>>> value would be 1, others rank would be the (value/ value of > >>>>> heighest)/ > > number of > >>>>> rows that took the same value > >>>>> thus the data frame should be > >>>>> id value Rank1 1 4 0.332 1 4 0.333 1 6 0.54 1 6 0.55 2 1.5 0.6 6 > 2 > >>>>> 2.5 > >>>>> 0.337 2 2.5 0.338 2 2.5 0.33 > >>>>> > >>>>> how to reach this resultthanks in advanceRagia [[alternative HTML > >>>>> version deleted]] > >>>>> > > ______________________________________________ > R-help at 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.________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. 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