[R] Solve system of non linear equations using nasted loops

[Bert Gunter]

I haven't gone through your code carefully, but I believe this can be done
in a tiny fraction of the time you are taking by eschewing loops. See
?expand.grid to get started. In general,nested loops should be avoided if
possible.

I  also suggest you spend some time with a good R tutorial to learn how to
write good R code instead of recapitulating C (or whatever) in R. All
languages have their own coding paradigms(R's is functional) and you will
do better in the long run if you spend the time to learn R's.

Cheers,
Bert



On May 10, 2017 1:20 PM, "Santiago Burone" <santiagoburone at icloud.com>
wrote:

Hello,

I'm new at R and I would like to use it in order to solve a system of non
linear equations. I have the code that works but im not able to save the
results.



My system has three equations and i would like to solve this using three
nested loops, becouse i need solutions for all the values combinations.





The code im using is this:



library(nleqslv)
#x3 es gamma
#x2 es alpha
#x1 es beta
MA <- c(50000, 43600, 40000, 38800, 37600, 34400, 31600, 27200, 24400,
20000)
MI <- c(10000, 21800, 20000, 19400, 18800, 17200, 15800, 13600, 12200,
10000)
ME <- c(30000, 32700, 30000, 29100, 28200, 25800, 23700, 20400, 18300,
15000)
DE <- c(0.384900179, 0.19245009, 0.19245009, 0.19245009, 0.19245009,
0.19245009, 0.19245009, 0.19245009, 0.19245009, 0.19245009)
for (i in 1:9) {
for (j in 1:9){
for (k in 1:9){
df.names <- paste("SociedadB",1:10,sep="")

fun <- function(x) {
f <- numeric(length(x)) # read as:
f[1] <- 1*x[3] - (log(ME[1+i]/ME[1])+(1*x[1]+1*
x[2])*(2)*(log((MA[1]-ME[1])/(MA[i+1]-ME[i+1]))))/(log(0.
19245009/0.384900179))
f[2] <- 1*x[3] - (log(MI[1+j]/MI[1])-(1*x[2])*(
2)*(log((MA[1+j]-MI[1+j])/(MA[1]-MI[1]))))/(log(0.19245009/0.384900179))
f[3] <- 1*x[3] - (log(MA[1+k]/MA[1])-(1*x[1])*(
2)*(log((MI[1+k]-MA[1+k])/(MI[1]-MA[1]))))/(log(0.19245009/0.384900179))
f
}
startx <- c(1,1,1)
answers<-as.data.frame(nleqslv(startx,fun))
d.frame <- (answers$x)
assign(df.names[k+20], d.frame)
despejes<- data.frame(SociedadB2, SociedadB3, SociedadB4, SociedadB5,
SociedadB6, SociedadB7, SociedadB8, SociedadB9, SociedadB10)

}
}
}

despejes<- data.frame(SociedadB2, SociedadB3, SociedadB4, SociedadB5,
SociedadB6, SociedadB7, SociedadB8, SociedadB9, SociedadB10)



If im not wrong i should have 1000 different solutions but the data frame
only save 10 results. I tried copying the last part of the code between the
last two brackets (changing the name of the data frame despejes for
despejes 1, 2 and 3 but i only get three data frames with exactly the same
results. I dont know how could i save all the results of every combination.
I mean, i need to solve for i=1, j=1 and the 9 diferent values of k and
save those results, then i=1, j=2 and the 9 values of k..



I know this might be a easy question but i have not been able to find a
solution in the last days.



Thanks in advance if you can help me.


Santiago.
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