[R] Replace missing value within group with non-missing value
Leask, Graham
g.leask at aston.ac.uk
Sat Apr 6 18:16:16 CEST 2013
Hi Rui,
Data as follows
structure(list(dn = c(4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), obs = c(1, 1,
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4,
4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8,
8, 8, 8, 8, 9, 9), choice = c(0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0), br = c(1,
2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4,
5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1,
2, 3, 4, 5, 6, 1, 2), mth = c(NA, NA, NA, NA, NA, 487, NA, NA,
488, NA, NA, NA, NA, NA, NA, NA, NA, 488, NA, NA, 489, NA, NA,
NA, NA, NA, NA, NA, NA, 489, NA, NA, NA, NA, NA, 489, NA, NA,
NA, NA, NA, 490, NA, NA, NA, NA, NA, 491, NA, NA)), .Names = c("dn",
"obs", "choice", "br", "mth"), row.names = c("1", "2", "3", "4",
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15",
"16", "17", "18", "19", "20", "21", "22", "23", "24", "25", "26",
"27", "28", "29", "30", "31", "32", "33", "34", "35", "36", "37",
"38", "39", "40", "41", "42", "43", "44", "45", "46", "47", "48",
"49", "50"), class = "data.frame")
Best wishes
Graham
-----Original Message-----
From: Rui Barradas [mailto:ruipbarradas at sapo.pt]
Sent: 06 April 2013 16:32
To: Leask, Graham
Cc: r-help at r-project.org
Subject: Re: [R] Replace missing value within group with non-missing value
Hello,
Can't you post a data example? If your dataset is named 'dat' use
dput(head(dat, 50)) # paste the output of this in a post
Rui Barradas
Em 06-04-2013 15:34, Leask, Graham escreveu:
> Hi Rui,
>
> Thank you for your suggestion which is very much appreciated. Unfortunately running this code produces the following error.
>
> error in '$<-.data.frame' ('*tmp*', "mth", value = NA_real_) :
> replacement has 1 rows, data has 0
>
> I'm sure there must be an elegant solution to this problem?
>
> Best wishes
>
>
>
> Graham
>
> On 6 Apr 2013, at 12:15, "Rui Barradas" <ruipbarradas at sapo.pt> wrote:
>
>> Hello,
>>
>> That's not a very good way of posting your data, preferably paste the output of ?dput in a post.
>> Some thing along the lines of the following might do what you want.
>> It seems that the groups are established by 'dn' and 'obs' numbers.
>> If so, try
>>
>>
>> # Make up some data
>> dat <- data.frame(dn = 4, obs = rep(1:5, each = 6), mth = NA)
>> dat$mth[6] <- 487 dat$mth[9] <- 488 dat$mth[18] <- 488 dat$mth[21] <-
>> 489 dat$mth[30] <- 489
>>
>>
>> sp <- split(dat, list(dat$dn, dat$obs))
>> names(sp) <- NULL
>> tmp <- lapply(sp, function(x){
>> idx <- which(!is.na(x$mth))[1]
>> x$mth <- x$mth[idx]
>> x
>> })
>> do.call(rbind, tmp)
>>
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>>
>> Em 06-04-2013 11:33, Leask, Graham escreveu:
>>> Dear List members
>>>
>>> I have a large dataset organised in choice groups see sample below
>>>
>>> +-------------------------------------------------------------------------------------------------+
>>> | dn obs choice acid br date cdate situat~n mth year set |
>>> |-------------------------------------------------------------------------------------------------|
>>> 1. | 4 1 0 LOSEC 1 . . . . 1 |
>>> 2. | 4 1 0 NEXIUM 2 . . . . 1 |
>>> 3. | 4 1 0 PARIET 3 . . . . 1 |
>>> 4. | 4 1 0 PROTIUM 4 . . . . 1 |
>>> 5. | 4 1 0 ZANTAC 5 . . . . 1 |
>>> |-------------------------------------------------------------------------------------------------|
>>> 6. | 4 1 1 ZOTON 6 23aug2000 01:00:00 23aug2000 NS 487 2000 1 |
>>> 7. | 4 2 0 LOSEC 1 . . . . 2 |
>>> 8. | 4 2 0 NEXIUM 2 . . . . 2 |
>>> 9. | 4 2 1 PARIET 3 25sep2000 01:00:00 25sep2000 L 488 2000 2 |
>>> 10. | 4 2 0 PROTIUM 4 . . . . 2 |
>>> |-------------------------------------------------------------------------------------------------|
>>> 11. | 4 2 0 ZANTAC 5 . . . . 2 |
>>> 12. | 4 2 0 ZOTON 6 . . . . 2 |
>>> 13. | 4 3 0 LOSEC 1 . . . . 3 |
>>> 14. | 4 3 0 NEXIUM 2 . . . . 3 |
>>> 15. | 4 3 0 PARIET 3 . . . . 3 |
>>> |-------------------------------------------------------------------------------------------------|
>>> 16. | 4 3 0 PROTIUM 4 . . . . 3 |
>>> 17. | 4 3 0 ZANTAC 5 . . . . 3 |
>>> 18. | 4 3 1 ZOTON 6 20sep2000 00:00:00 20sep2000 R 488 2000 3 |
>>> 19. | 4 4 0 LOSEC 1 . . . . 4 |
>>> 20. | 4 4 0 NEXIUM 2 . . . . 4 |
>>> |-------------------------------------------------------------------------------------------------|
>>> 21. | 4 4 1 PARIET 3 27oct2000 00:00:00 27oct2000 NL 489 2000 4 |
>>> 22. | 4 4 0 PROTIUM 4 . . . . 4 |
>>> 23. | 4 4 0 ZANTAC 5 . . . . 4 |
>>> 24. | 4 4 0 ZOTON 6 . . . . 4 |
>>> 25. | 4 5 0 LOSEC 1 . . . . 5 |
>>> |-------------------------------------------------------------------------------------------------|
>>> 26. | 4 5 0 NEXIUM 2 . . . . 5 |
>>> 27. | 4 5 0 PARIET 3 . . . . 5 |
>>> 28. | 4 5 0 PROTIUM 4 . . . . 5 |
>>> 29. | 4 5 0 ZANTAC 5 . . . . 5 |
>>> 30. | 4 5 1 ZOTON 6 23oct2000 03:00:00 23oct2000 NS 489 2000 5 |
>>>
>>> I wish to fill in the missing values in each choice set - delineated by dn (Doctor) obs (Observation number) and choices (1 to 6).
>>> For each choice set one choice is chosen which contains full time
>>> information for that choice set ie in set 1 choice 6 was chosen and shows the month 487. The other 5 choices show mth as missing. I want to fill these with the correct mth.
>>>
>>> I am sure there must be an elegant way to do this in R?
>>>
>>>
>>> Best wishes
>>>
>>>
>>>
>>> Graham
>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list
>>> 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.
>>
>
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