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3 messages · Ragia Ibrahim, PIKAL Petr

Original
Ragia Ibrahim
# 
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
PIKAL Petr
# 
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
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1 day later
Ragia Ibrahim
# 
many thanks, It WORKS.

But

what If I want to add a condition that considered?Measure_id , if it is '1' ? rank reverse the probability and if it is ' 2 ' rank is ordered like probability??

Replying is highly appreciated?
Ragia
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