On some slightly different datasets:
tt1<-structure(list(subj = c(1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5,
6, 6, 6, 7, 7, 7, 8, 8, 8), response = structure(c(2L, 2L, 1L,
2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L,
1L, 2L, 1L), .Label = c("buy", "sample"), class = "factor"),
product = c(1, 2, 3, 2, 2, 3, 2, 1, 1, 4, 4, 2, 2, 4, 5,
5, 4, 3, 4, 5, 4, 2)), .Names = c("subj", "response", "product"
), class = "data.frame", row.names = c(NA, 22L))
tt2<- structure(list(subj = c(1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5,
6, 6, 6, 7, 7, 7, 8, 8, 8), response = structure(c(2L, 2L, 1L,
2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L,
1L, 2L, 2L), .Label = c("buy", "sample"), class = "factor"),
product = c(1, 2, 3, 2, 2, 3, 2, 1, 1, 4, 1, 4, 5, 1, 4,
2, 3, 3, 2, 5, 3, 4)), .Names = c("subj", "response", "product"
), class = "data.frame", row.names = c(NA, 22L))
tt3<- structure(list(subj = c(1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5,
6, 6, 6, 7, 7, 7, 8, 8, 8), response = structure(c(2L, 2L, 1L,
2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L,
1L, 1L, 2L), .Label = c("buy", "sample"), class = "factor"),
product = c(1, 2, 3, 2, 2, 3, 2, 1, 1, 4, 1, 1, 3, 5, 2,
2, 2, 2, 4, 3, 2, 5)), .Names = c("subj", "response", "product"
), class = "data.frame", row.names = c(NA, 22L))
#Tried David's solution:
tt1$rown <- rownames(tt1)
as.numeric ( apply(tt1, 1, function(x) {
x['product'] %in% tt1[ rownames(tt1) < x['rown'] & tt1$response == "buy",
"product"] } ) )
#gave inconsistent results especially since the first 10 rows were from `tt`
# [1] 0 1 1 1 1 1 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1
#similarly for `tt2` and `tt3`.
##Created this function. It seems to work in the tested cases, though it is
not tested extensively.
fun1<- function(dat,colName,newColumn){
indx<- which(dat[,colName]=="buy")
dat[,newColumn]<-0
dat[unlist(lapply(seq_along(indx),function(i){
x1<- if(i==length(indx)){
seq(indx[i],nrow(dat))
}
else if((indx[i+1]-indx[i])==1){
indx[i]
}
else {
seq(indx[i]+1,indx[i+1]-1)
}
x2<- dat[unique(c(indx[i:1],x1)),]
x3<- subset(x2,response=="sample")
x4<- subset(x2,response=="buy")
if(nrow(x3)!=0) {
row.names(x3)[x3$product%in% x4$product]
}
})),newColumn]<-1
dat
}
fun1(tt,"response","newCol")
# subj response product rown newCol
#1 1 sample 1 1 0
#2 1 sample 2 2 0
#3 1 buy 3 3 0
#4 2 sample 2 4 0
#5 2 buy 2 5 0
#6 3 sample 3 6 1
#7 3 sample 2 7 1
#8 3 buy 1 8 0
#9 4 sample 1 9 1
#10 4 buy 4 10 0
fun1(tt1,"response","newCol")
# subj response product newCol
#1 1 sample 1 0
#2 1 sample 2 0
#3 1 buy 3 0
#4 2 sample 2 0
#5 2 buy 2 0
#6 3 sample 3 1
#7 3 sample 2 1
#8 3 buy 1 0
#9 4 sample 1 1
#10 4 buy 4 0
#11 5 buy 4 0
#12 5 sample 2 1
#13 5 buy 2 0
#14 6 buy 4 0
#15 6 sample 5 0
#16 6 sample 5 0
#17 7 sample 4 1
#18 7 buy 3 0
#19 7 buy 4 0
#20 8 buy 5 0
#21 8 sample 4 1
#22 8 buy 2 0
#Also
fun1(tt2,"response","newCol")
fun1(tt3,"response","newCol")
A.K.
P.S. Below is OP's clarification regarding the conditional statement in a
private message:
I am sorry i didnt question it very clearly, let me change the
conditional statement, I hope you can understand. i will explain by
example
as you can see, almost every number is duplicated, but only in row 6th,7th,and
9th the value on column is 1.
on row4th, the value is duplicated( 2 already occurred on 2nd row),but
since the value is considered as duplicated only if the value is
duplicated where the response is 'buy' than the value on column, on
row4th still zero.
On row 6th, where the value product column is 3. 3 is already occurred
in 3rd row where the value on response is 'buy', so the value on column
should be 1
I hope it can understand the conditional statement.
----- Original Message -----
From: David Winsemius <dwinsem...@comcast.net>
To: David Winsemius <dwinsem...@comcast.net>
Cc: R-help@r-project.org; Uwe Ligges <lig...@statistik.tu-dortmund.de>
Sent: Friday, July 26, 2013 5:16 PM
Subject: Re: [R] Duplicated function with conditional statement
On Jul 26, 2013, at 2:06 PM, David Winsemius wrote:
>
> On Jul 26, 2013, at 11:51 AM, Uwe Ligges wrote:
>
>>
>>
>> On 25.07.2013 21:05, vanessa van der vaart wrote:
>>> Hi everybody,,
>>> I have a question about R function duplicated(). I have spent days try to
>>> figure this out,but I cant find any solution yet. I hope somebody can help
>>> me..
>>> this is my data:
>>>
>>> subj=c(1,1,1,2,2,3,3,3,4,4)
>>> response=c('sample','sample','buy','sample','buy','sample','
>>> sample','buy','sample','buy')
>>> product=c(1,2,3,2,2,3,2,1,1,4)
>>> tt=data.frame(subj, response, product)
>>>
>>> the data look like this:
>>>
>>> subj response product
>>> 1 1 sample 1
>>> 2 1 sample 2
>>> 3 1 buy 3
>>> 4 2 sample 2
>>> 5 2 buy 2
>>> 6 3 sample 3
>>> 7 3 sample 2
>>> 8 3 buy 1
>>> 9 4 sample 1
>>> 10 4 buy 4
>>>
>>> I want to create new column based on the value on response and product
>>> column. if the value on product is duplicated, then the value on new column
>>> is 1, otherwise is 0.
>>
>>
>> According to your description:
>>
>
> Agree that the description did not match the output. I tried to match the
> output using a rule that could be expressed as:
>
> if( a "buy"- associated "product" value precedes the current "product"
> value){1}else{0}
>
So this delivers the specified output:
tt$rown <- rownames(tt)
as.numeric ( apply(tt, 1, function(x) {
x['product'] %in% tt[ rownames(tt) < x['rown'] & tt$response == "buy",
"product"] } ) )
# [1] 0 0 0 0 0 1 1 0 1 0
> --
> David.
>
>> tt$newcolumn <- as.integer(duplicated(tt$product) & tt$response=="buy")
>>
>> which is different from what you show us below, where I cannot derive any
>> systematic rule from.
>>
>> Uwe Ligges
>>
>>> but I want to add conditional statement that the value on product column
>>> will only be considered as duplicated if the value on response column is
>>> 'buy'.
>>> for illustration, the table should look like this:
>>>
>>> subj response product newcolumn
>>> 1 1 sample 1 0
>>> 2 1 sample 2 0
>>> 3 1 buy 3 0
>>> 4 2 sample 2 0
>>> 5 2 buy 2 0
>>> 6 3 sample 3 1
>>> 7 3 sample 2 1
>>> 8 3 buy 1 0
>>> 9 4 sample 1 1
>>> 10 4 buy 4 0
>>>
>>>
>>> can somebody help me?
>>> any help will be appreciated.
>>> I am new in this mailing list, so forgive me in advance, If I did not ask
>>> the question appropriately.
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help@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.
>>>
>>
>> ______________________________________________
>> R-help@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.
>
> David Winsemius
> Alameda, CA, USA
>
> ______________________________________________
> R-help@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.
David Winsemius
Alameda, CA, USA
______________________________________________
R-help@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.
______________________________________________
R-help@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.
