Thank you Arun, that was exactly what I was looking for.
This is the second time you helped me in this mailing list. Thanks!
And also thanks to all of the others of you whou helped me. It is amazing
how many possibilities there are to solve my problem (which isn't a prob now
anymore).
Have a gre
On Sep 17, 2012, at 6:07 AM, Tagmarie wrote:
> Thank you Michael, that worked perfectly!
>
> Now I wonder, if it is possible to break my data further apart and put it
> together again.
> Assume I include a column for an ID in the data frame like this:
>
> dattrial2<-data.frame(a=c(1,NA,NA,NA
4 Ernie 0
A.K.
- Original Message -
From: Tagmarie
To: r-help@r-project.org
Cc:
Sent: Monday, September 17, 2012 9:07 AM
Subject: Re: [R] count NAs per week
Thank you Michael, that worked perfectly!
Now I wonder, if it is possible to break my data further apart and put it
together
Thank you Michael, that worked perfectly!
Now I wonder, if it is possible to break my data further apart and put it
together again.
Assume I include a column for an ID in the data frame like this:
dattrial2<-data.frame(a=c(1,NA,NA,NA,2,3), Week=c(3,3,3,4,4,4),
AnimalID=c("Ernie","Bert", "Ernie
Hi,
Try this:
dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4))
aggregate(dattrial$a,list(dattrial$Week),function(x) sum(is.na(x)))
# Group.1 x
#1 3 1
#2 4 0
#or
ddply(dattrial,.(Week),summarize, sum(is.na(a)))
# Week ..1
#1 3 1
#2 4 0
#or
list1<-split(dat
> with(dattrial, table(A_is_missing=is.na(a), Week))
Week
A_is_missing 3 4
FALSE 2 3
TRUE 1 0
> with(dattrial, table(A_is_missing=is.na(a), Week))["TRUE", ] # extract the
> missing="TRUE" row, note quotes
3 4
1 0
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
On Mon, Sep 17, 2012 at 11:03 AM, Tagmarie wrote:
> Even though I work with R since a year or so I still struggle with simple
> problems. I hope someone can help me with this. Been trying for days and am
> a little frustrated now.
>
> I have a data frame somewhat like the one bellow:
>
> dattrial<
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