R help - Sep 2020 - truth[(truth[, 1]=="G3" & truth[, 2]=="G2") | (truth[, 1]=="G2" & truth[, 2]=="G3"), 3]<-1

helloout<-read.csv("outbr.csv")truth<-out[,seq(1,2)]for example
:
?If row1= G1 and row2=G2 , and row 1 = G2 and row 2= G1,make G3=1?# note G1 and
G2 are values from 1 to 2000?#if this happend add to thrid column in truth 1
otherwise add 0 as in statment follow
truth<-cbind(as.character(truth[,1]),as.character(truth[,2])? ? ? ? ? ?
?,as.data.frame(rep(0,,dim(out)[1])));#here just G2 and G3, i want make loop to
cam[are all values from G1 to G2000?
truth[(truth[,1]=="G3" & truth[,2]=="G2") |
(truth[,1]=="G2" &
truth[,2]=="G3"),3]<-1?###############################3(Simply they
regulate the other. If element A is in the first group , and it is related to
element B in the second group , and element B also in? in the first group , and
it is related to element A(the same element? in the first group)? in the second
group , we write 1 and otherwise 0.
this the distination result:
I want this result
G1 G2? G3?D? ? B? ? 1?B? ?D? ? ?1?A? ? D? ? 0?B? ? A? ? 1B? ? C? ? 0A ? B? ? 1


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I am afraid this is completely unreadable because you posted in HTML ad 
this is a plain text list. Best to resend it having set your mailer to 
send plain text as HTML gets mangled here.

Michael

On 06/09/2020 10:58, Hesham A. AL-bukhaiti via R-help
wrote:
> helloout<-read.csv("outbr.csv")truth<-out[,seq(1,2)]for
example :
>  ?If row1= G1 and row2=G2 , and row 1 = G2 and row 2= G1,make G3=1?# note
G1 and G2 are values from 1 to 2000?#if this happend add to thrid column in
truth 1 otherwise add 0 as in statment follow
> truth<-cbind(as.character(truth[,1]),as.character(truth[,2])? ? ? ? ? ?
?,as.data.frame(rep(0,,dim(out)[1])));#here just G2 and G3, i want make loop to
cam[are all values from G1 to G2000
> truth[(truth[,1]=="G3" & truth[,2]=="G2") |
(truth[,1]=="G2" &
truth[,2]=="G3"),3]<-1?###############################3(Simply they
regulate the other. If element A is in the first group , and it is related to
element B in the second group , and element B also in? in the first group , and
it is related to element A(the same element? in the first group)? in the second
group , we write 1 and otherwise 0.
> this the distination result:
> I want this result
> G1 G2? G3?D? ? B? ? 1?B? ?D? ? ?1?A? ? D? ? 0?B? ? A? ? 1B? ? C? ? 0A ? B?
? 1
> 
> 
> 	[[alternative HTML version deleted]]
> 
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-- 
Michael
http://www.dewey.myzen.co.uk/home.html