R help - Mar 2019 - data frame solution

Hello All,

wonder if you have thoughts on a clever solution for this code:



df? ? ? ?<- data.frame(a = c(6,1), b = c(1000,1200), c =c(-1,3))?

#the caveat here is that the number of rows for df can be anything from 1 row to
in the hundreds. I kept it to 2 to have minimal reproducible

t<-seq(-5,24,0.1) #min(t) will always be <=df$c[1], which is the value
that is always going to equal to min(df$c)

times1 <- c(rbind(df$c[1],df$c[1]+df$a[1]),max(t)) #length of times1 will
always be 3, see times2 is of length 4

input1? ?<- c(rbind(df$b[1]/df$a[1],rep(0,length(df$b[1]))),0)?#length of
input1 will always be 3, see input2 is of length 4

out1
<-data.frame(t,ifelse(t>=times1[1]&t<times1[2],input1[1],ifelse(t>=times1[2]&t<times1[3],input1[2],input1[3])))

times2 <- c(times1[1],rbind(df$c[2],df$c[2]+df$a[2]),max(t)) #note 1st value
of times2, (or for times3, times4,.. if more rows in df) is times1[1], which
will always be < times2[1]?(or times3[1], times4[1],.. if more rows in df)?

input2? ?<- c(0,rbind(df$b[2]/df$a[2],rep(0,length(df$b[2]))),0)?#note 1st
value of input2??(or for input3, input4,.. if more rows in df)? is 0, which will
always be 0 (or?for all input2-n other then for input1 as that will be as
above?,.. if more rows in df)

out2
<-data.frame(t,ifelse(t>=times2[1]&t<times2[2],input2[1],ifelse(t>=times2[2]&t<times2[3],input2[2],ifelse(t>=times2[3]&t<times2[4],input2[3],input2[3]))))

result<-data.frame(t,out1[,2]+out2[,2])

so if I did it all manually then for row 3 in df I would calculate out3 and so
on... Would like to be able to do this with a clean function solution that
allows for different row numbers in df...

as always your help is appreciated?

thanks? ? ? ? ? ? ? ? ?

Andras?

Statements like c(rbind(x, xx+yy), max(t)) and rep(0,length(df$b[1]))  don't
make any sense. You're example will be easier to understand if you show us
the nrow(df) ==3
case. Thanks


Grant Izmirlian, Ph.D.
Mathematical Statistician
izmirlig at mail.nih.gov

Delivery Address:
9609 Medical Center Dr, RM 5E130
Rockville MD 20850

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BG 9609 RM 5E130 MSC 9789
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________________________________
From: Andras Farkas <motyocska at yahoo.com>
Sent: Tuesday, March 19, 2019 7:06 AM
To: R-help Mailing List
Subject: [R] data frame solution

Hello All,

wonder if you have thoughts on a clever solution for this code:



df       <- data.frame(a = c(6,1), b = c(1000,1200), c =c(-1,3))

#the caveat here is that the number of rows for df can be anything from 1 row to
in the hundreds. I kept it to 2 to have minimal reproducible

t<-seq(-5,24,0.1) #min(t) will always be <=df$c[1], which is the value
that is always going to equal to min(df$c)

times1 <- c(rbind(df$c[1],df$c[1]+df$a[1]),max(t)) #length of times1 will
always be 3, see times2 is of length 4

input1   <- c(rbind(df$b[1]/df$a[1],rep(0,length(df$b[1]))),0) #length of
input1 will always be 3, see input2 is of length 4

out1
<-data.frame(t,ifelse(t>=times1[1]&t<times1[2],input1[1],ifelse(t>=times1[2]&t<times1[3],input1[2],input1[3])))

times2 <- c(times1[1],rbind(df$c[2],df$c[2]+df$a[2]),max(t)) #note 1st value
of times2, (or for times3, times4,.. if more rows in df) is times1[1], which
will always be < times2[1] (or times3[1], times4[1],.. if more rows in df)

input2   <- c(0,rbind(df$b[2]/df$a[2],rep(0,length(df$b[2]))),0) #note 1st
value of input2  (or for input3, input4,.. if more rows in df)  is 0, which will
always be 0 (or for all input2-n other then for input1 as that will be as above
,.. if more rows in df)

out2
<-data.frame(t,ifelse(t>=times2[1]&t<times2[2],input2[1],ifelse(t>=times2[2]&t<times2[3],input2[2],ifelse(t>=times2[3]&t<times2[4],input2[3],input2[3]))))

result<-data.frame(t,out1[,2]+out2[,2])

so if I did it all manually then for row 3 in df I would calculate out3 and so
on... Would like to be able to do this with a clean function solution that
allows for different row numbers in df...

as always your help is appreciated

thanks

Andras



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