721
> #5 2 900
> #6 2 1010
> #7 2 1120
>
>
> A.K.
>
>
>
>
>
>
> - Original Message -
> From: William Dunlap
> To: arun ; Christopher Desjardins <
> cddesjard...@gmail.com>
&g
n ; Christopher Desjardins
Cc: R help
Sent: Saturday, November 3, 2012 9:21 PM
Subject: RE: [R] Replacing NAs in long format
Or, even simpler,
> flag <- with(dat2, ave(schyear<=5 & year==0, idr, FUN=any))
> data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 TRUE
hristopher Desjardins
> Cc: R help
> Subject: Re: [R] Replacing NAs in long format
>
> ave() or split<-() can make that easier to write, although it
> may take some time to internalize the idiom. E.g.,
>
> > flag <- rep(NA, nrow(dat2)) # add as.integer if you prefer 1,
ilto:r-help-boun...@r-project.org] On
> Behalf
> Of arun
> Sent: Saturday, November 03, 2012 5:01 PM
> To: Christopher Desjardins
> Cc: R help
> Subject: Re: [R] Replacing NAs in long format
>
> Hi,
> May be this helps:
> dat2<-read.table(text="
> i
0
#6 2 10 1 0
#7 2 11 2 0
A.K.
- Original Message -
From: Christopher Desjardins
To: jim holtman
Cc: r-help@r-project.org
Sent: Saturday, November 3, 2012 7:09 PM
Subject: Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could re
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear < = 5
and year = 0 for a given idr. For example
> dat
idr schyear year
1 4 -1
1 50
1
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman wrote:
> > x <- read.table(text = "idr schyear year
> + 1 80
> + 1 91
> + 1 10 NA
> + 2 4 NA
> + 2 5 -1
> + 2 60
> + 2 7
> x <- read.table(text = "idr schyear year
+ 1 80
+ 1 91
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 60
+ 2 71
+ 2 82
+ 2 93
+ 2 104
+ 2 11 NA
+ 2 126
+ 3 4 NA
+ 3 5 -2
+
Hello,
Try the following. I've called your data.frames 'dat' and 'dat2'
# First your datasets, see ?dput
dput(dat)
structure(list(idr = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), schyear = c(8L, 9L,
10L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 4L, 5L, 6L, 7
Hi,
I have the following data:
> data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3
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