# [R] choose the lines2

arun smartpink111 at yahoo.com
Tue Jun 4 17:37:32 CEST 2013

```Hi,
May be this helps:
dat.bru<- dat1[!is.na(dat1\$evnmt_brutal),]

fun2<- function(dat){
lst1<- split(dat,dat\$patient_id)
lst2<- lapply(lst1,function(x) x[cumsum(x\$evnmt_brutal==0)>0,])
lst3<- lapply(lst2,function(x) x[!(all(x\$evnmt_brutal==1)|all(x\$evnmt_brutal==0)),])
lst4<- lst3[lapply(lst3,nrow)!=0]
lst5<- lapply(seq_along(lst4),function(i){
do.call(rbind,lapply(which(lst4[[i]]\$evnmt_brutal==1),function(x) {
x1<-c(x-2,x-1,x)
x2<-x1[!any(x1==0)]
x3<-lst4[[i]][x2,]
x4<-x3[!is.na(match(paste(x3\$evnmt_brutal,collapse=""),"001")),]
x4[!any(duplicated(x4\$number))]
}
))
})
lst6<-lst5[lapply(lst5,nrow)!=0]
names(lst6)<- unlist(lapply(lst6,function(x) unique(x\$patient_id)))
Mean0bet_01<- do.call(rbind,lapply(lst6,function(x) mean(x[seq(nrow(x))%%3==2,"basdai_d"])))
lst7<-list(lst6,Mean0bet_01)
lst7
}

fun2(dat.bru)

##output from first 2 patients

#[[1]]
#[[1]]\$`2`
#   X patient_id number responsed_at  t basdai_d evnmt_brutal
#13 13          2     12   2011-07-05 12     -1.0            0
#14 14          2     13   2011-08-07 13      0.9            0
#15 15          2     14   2011-09-11 14     -0.8            1
#
#[[1]]\$`5`
#    X patient_id number responsed_at t basdai_d evnmt_brutal
#52 52          5      8   2011-01-11 7     -2.8            0
#53 53          5      9   2011-02-13 8      0.0            0
#54 54          5     10   2011-03-19 9     -1.2            1
#

#[[2]]
# [,1]
#2  0.9
#5  0.0

A.K.

________________________________
From: GUANGUAN LUO <guanguanluo at gmail.com>
To: arun <smartpink111 at yahoo.com>
Sent: Tuesday, June 4, 2013 3:54 AM
Subject: choose the lines2

Hello, Arun,
now it is nearly the same problem.
I want to know if I want to choose three period : one line with evnmt_brutal ==0 , one line with evnmt_brutal==0 and one line with evnmt_brutal==1, then I want to choose the second and third period for each patient, so that i can calculate the average of scores of basdai when evnmt_brutal==0 (in condition that the precedent line with evnmt_brutal==0) and evnmt_brutal==1.
In writing with the phrase "if", if i have two conditions, i don't know how can i write that.

This is when i want to choose the period with evnmt_brutal ==0 et evnmt_brutal==1 for each patient, you have written this code. If i want to add one condition that before the line evnmt_brutal==0, evnmt_brutal of that line equal to 0 too.
The result i want to get is just the two last lines.

Do you know how can i realize that?

dat.bru<- dat1[!is.na(dat1\$evnmt_brutal),]

fun1<- function(dat){
lst1<- split(dat,dat\$patient_id)
lst2<- lapply(lst1,function(x) x[cumsum(x\$evnmt_brutal==0)>0,])
lst3<- lapply(lst2,function(x) x[!(all(x\$evnmt_brutal==1)|all(x\$evnmt_brutal==0)),])
lst4<-lapply(lst3,function(x) {vect.brutal=c()
for(line in which(x\$evnmt_brutal==1)){
if(x\$evnmt_brutal[line-1]==0){
vect.brutal=c(vect.brutal,line)
}
}
vect.brutal1<- sort(c(vect.brutal,vect.brutal-1))
x[vect.brutal1,]
}
)
res<- do.call(rbind,lst4)
row.names(res)<- 1:nrow(res)
res
}

#    X patient_id number responsed_at  t basdai_d evnmt_brutal
#1  14          2     13   2011-08-07 13    0.900            0
#2  15          2     14   2011-09-11 14   -0.800            1
#3  22          3      2   2010-06-29  1   -0.800            0
#4  23          3      3   2010-08-05  2    0.000            1
#5  24          3      4   2010-09-05  3    1.200            0
#6  25          3      5   2010-10-13  4    1.925            1
#7  26          3      6   2010-11-15  5   -2.525            0
#8  27          3      7   2010-12-18  6   -0.200            1
#9  53          5      9   2011-02-13  8    0.000            0
#10 54          5     10   2011-03-19  9   -1.200            1

```