On Jun 25, 2014, at 1:49 PM, David Winsemius wrote:
>
> On Jun 24, 2014, at 11:18 PM, Abhinaba Roy wrote:
>
>> Hi David,
>>
>> I was thinking something like this:
>>
>> ID Disease
>> 1 A
>> 2 B
>> 3 A
>> 1C
>> 2D
>> 5A
>> 4B
>> 3D
>> 2A
>> ....
>>
>>
On Jun 24, 2014, at 11:18 PM, Abhinaba Roy wrote:
> Hi David,
>
> I was thinking something like this:
>
> ID Disease
> 1 A
> 2 B
> 3 A
> 1C
> 2D
> 5A
> 4B
> 3D
> 2A
> ....
>
> How can this be done?
do.call(rbind, lapply( 1:20, function(pt) {
Also, you can do:
library(dplyr)
dat%>%group_by(ID)%>%filter(length(unique(Disease))>1)%>%arrange(Disease,ID)
A.K.
On Wednesday, June 25, 2014 3:45 AM, arun wrote:
Forgot about:
library(reshape2)
On , arun wrote:
Hi,
Check if this works:
set.seed(495)
dat <- data.frame(ID=sample
Hi,
Check if this works:
set.seed(495)
dat <- data.frame(ID=sample(1:10,20,replace=TRUE),
Disease=sample(LETTERS[1:6], 20, replace=TRUE) )
subset(melt(table(dat)[rowSums(!!table(dat))>1,]), !!value,select=1:2)
ID Disease
1 2 A
3 4 A
4 6 A
6 10 A
8 3
# build off of david's suggestion
x <-
data.frame(
patient= 1:20 ,
disease =
sapply(
pmin( 2 + rpois( 20 , 2 ) , 6 ) ,
function( n ) paste0( sample( c('A','B','C','D','E','F'),
n), collapse="+" )
)
)
# break the diseas
Hi David,
I was thinking something like this:
ID Disease
1 A
2 B
3 A
1C
2D
5A
4B
3D
2A
....
How can this be done?
On Wed, Jun 25, 2014 at 11:34 AM, David Winsemius
wrote:
>
> On Jun 24, 2014, at 10:14 PM, Abhinaba Roy wrote:
>
> > Dear R helpers,
> >
On Jun 24, 2014, at 10:14 PM, Abhinaba Roy wrote:
> Dear R helpers,
>
> I want to generate data for say 1000 patients (i.e., 1000 unique IDs)
> having suffered from various diseases in the past (say diseases
> A,B,C,D,E,F). The only condition imposed is that each patient should've
> suffered fro
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