Loops

Hi
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of bett kimutai
Sent: Wednesday, June 26, 2013 3:17 AM
To: Law, Jason; r-help at r-project.org
Subject: Re: [R] Loops

Thanks for you response. The issue is that I need to run a thousand
simulations for each pipe (20000) and then get an average of the values
that are below the condition I have set. When I ran the code, I only
get just a thousand values meaning it gives me values for a single
How do you get result with syntax error code? 

Anyway did you give a chance to help provided by Jason? If I understand your code, you have three "k" values based on them you compute three "l" values with a differnt constant each time. Based on this you can get
k<-c(1.15, .504, .43) 
(7.16-0.44+0.12-0.016)
[1] 6.824
(8.01-1.5+0.35+0.45)
[1] 7.31
9.55-2.45+0.40+0.65
[1] 8.15
m<-c(6.824,7.31,8.15)
l=exp((1/k)*(m))
l
[1] 3.776293e+02 1.990643e+06 1.703709e+08

then you want to get z vector in 3 steps

z <- (log(1/p)*l[1])^k[1]
z[z<=684] <-  (log(1/p)*l[2])^k[2]
z[z<=684] <-  (log(1/p)*l[3])^k[3]

you maybe want to subset also p

and you need top repeat this 1000 times probably in loop to populate resulting matrix.

Anyyway some example of data and working code would be helpful.

Regards
Petr
pipe. How can I rectify my loop to run for all the pipes and then
create a table to hold these values. Sorry for the syntax errors,
I?guess I will get them with time.

thanks again..

________________________________
 From: "Law, Jason" <Jason.Law at portlandoregon.gov>

ect.org>
Sent: Tuesday, June 25, 2013 3:32 PM
Subject: RE: [R] Loops

Not sure what you're trying to do, but it looks like most of what
you're attempting to do in the code can be done just using vectors
rather than loops, at least the inner loop.? For example:

k <- 1.15
l <- exp((1 / k) * (7.16 - 0.44 + 0.12 - 0.016))
z <- (log(1 / p) * l)^k

See ifelse for how to do the if tests on a vector.? In addition, much
of the code in your loops doesn't vary with the loop indices and can be
moved outside your loop (e.g., setting k = 1.15).? If you really
want/need to use loops, you'll have to initialize the vectors/matrices
within your loop with some value:

z <- numeric(1000)

Finally, you have some plain syntax errors:

p[i]=[i+1].

That's not valid R code; '[' is the extraction operator, see
help('[').? I'm not sure what you're trying to do there.? Perhaps:

p[i] <- p[i + 1]

HTH,

Jason Law

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of bett kimutai
Sent: Tuesday, June 25, 2013 1:32 PM
To: r-help at r-project.org
Subject: [R] Loops

Dear All,

I? have spent most of my time trying to figure out how to simulate the
number of breaks in a pipe using Monte Carlo simulation.
i have 20,000 individual pipes that i have to run, and for each pipe i
have to run 1000 times while checking some conditions and therefore, i
have to use a nested loop.
what i would like to have as a final result is a matrix table with with
all the individual pipe elements and the simulated runs here is the
loop that i tried to create x=20000 y=matrix(x, z)
p=runif(1000)
for(j in 1:20000) {
for(i in 1:1000) {
?? ??? k=1.15
??? l=exp((1/k)*(7.16-0.44+0.12-0.016))

??? z[i]=(log(1/p[i])*l)^k

??? if (z[i] <=684)
??? {
??? ??? k1=0.504
??? ??? l1=exp((1/k)*(8.01-1.5+0.35+0.45))
??? ??? z1[i]=(log(1/p[i])*l1)^k1
??? if (z1[i] <=684)
??? {
??? ??? ??? ??? k2=0.43
??? ??? l2=exp((1/k2)*(9.55-2.45+0.40+0.65))
??? ??? z2[i]=(log(1/p[i])*l2)^k2
??? ??????????? p[i]=[i+1]
??? ??? ??? ??? break()
??? ??? ??? }
??? ??? }
??? }
x[ j ]=[ j+1 ]
}
the last column of the table,? in addition to the simulated runs, i
would like to have the summary of the means (for z<=684) of individual
row as this means will give me the number of breaks.
i will really appreciate if anyone who can help me figure out how to go
about this. pardon me, I am new to R and programming.

Thank you in Advance,

Eliab
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Loops

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