thr3ads.net - R help - [R] Solve system of non linear equations using nasted loops [May 2017]

If this information is useful, please help other people find it:
Share via:

Santiago Burone

2017-May-10 18:20 UTC

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

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. 
	[[alternative HTML version deleted]]

David Winsemius

2017-May-10 20:55 UTC

head link

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

> On May 10, 2017, at 11:20 AM, 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){
I got an error on the first pass through the loop.

> df.names <- paste("SociedadB",1:10,sep="")
# seems kind of excessive to repeat that process 9*9*9 times
# Maybe this should be done outside the loop?
> 
> 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 
> } 
# So you need 9*9*9 different functions? 

> startx <- c(1,1,1) 
> answers<-as.data.frame(nleqslv(startx,fun)) 
> d.frame <- (answers$x)
> assign(df.names[k+20], d.frame)
# I don't think that df.names[k+20] would be anything other than NA since it
only has 10 elements
> despejes<- data.frame(SociedadB2, SociedadB3, SociedadB4, SociedadB5,
SociedadB6, SociedadB7, SociedadB8, SociedadB9, SociedadB10)
> 
> }
> }
> }
> 
> despejes<- data.frame(SociedadB2, SociedadB3, SociedadB4, SociedadB5,
SociedadB6, SociedadB7, SociedadB8, SociedadB9, SociedadB10)
> 
The error message is:

Error in data.frame(SociedadB2, SociedadB3, SociedadB4, SociedadB5, SociedadB6, 
:
  object 'SociedadB2' not found

That's because column names are not first class data-objects in R.

> 
> 
> 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..
> 
If you need to save separate results for each combination of i,j,k , then you
need at a minimum an array (possibly with a further 4th dimension if you are
storing numerical vectors)  or possibly a list if these items are more
complicated than an atomic vector.> 
> 
> 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. 
> 	[[alternative HTML version deleted]]
Rhelp is a plain text list although you don't seem to have run into trouble
on that account. You should however determine how your mail client can be setup
to just send pain text to this mailing list.

> 
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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

Bert Gunter

2017-May-10 22:51 UTC

head link

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

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.
        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.

	[[alternative HTML version deleted]]

Berend Hasselman

2017-May-11 08:12 UTC

head link

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

Your code looks needlessly complicated and inefficient in my view.
Even if you don't need or use nested for loops it can still be simplified.

1: First define  some new constants to avoid the  the M? stuff in your function

<code>
pME <- ME[2:9]/ME[1]
pMI <- MI[2:9]/MI[1]
pMA <- MA[2:9]/MA[1]
pMAE <- (MA[1]-ME[1])/(MA[2:9]-ME[2:9])
pMAI <- (MA[2:9]-MI[2:9])/(MA[1]-MI[1])
pMIA <- (MI[2:9]-MA[2:9])/(MI[1]-MA[1])
</code>

2. Redefine your function and avoid silly and unnecessary  use of 1* and 2* and
use a constant for log(.../...)
The vector pXX is a vector consisting of elements from the vector pME etc.

<code>
fun <- function(x,pXX) {
    f <- numeric(length(x)) # read as:
    pME <- pXX[1]
    pMI <- pXX[2]
    pMA <- pXX[3]
    pMAE <- pXX[4]
    pMAI <- pXX[5]
    pMIA <- pXX[6]
    const <- log(0.19245009/0.384900179)
    f[1] <- x[3] - (log(pME)+(x[1]+x[2])*2*(log(pMAE)))/const
    f[2] <- x[3] - (log(pMI)-(x[2])*2*(log(pMAI)))/const
    f[3] <- x[3] - (log(pMA)-(x[1])*2*(log(pMIA)))/const
    f
}
</code>

This can be simplified further by doing the log(pXX) beforehand. 

3. I'm not going to have a go at using expand.grid. Too lazy and more
involved to get everything correct.
So use a nested loop as follows:

- prepare a result matrix of correct dimensions with7 columns: one for nleaslv
termination code, 3 for the result x-vales and another 3 to store i,j,k.
- storing the termination code is a must to see if convergence was obtained.
- and before invoking nleqslv test if fun(startx) contains NA values. If it does
fill the appropriate row of Result with NA.
- use an index counter (indx) to store the results in the rows of Result

<code>
Result <- matrix(0,nrow=9*9*9,ncol=7)

indx <- 1
for (i in 1:9) {
    for (j in 1:9) {
        for (k in 1:9) {
            pXXijk <- c(pME[i],pMI[j],pMA[k],pMAE[i],pMAI[j],pMIA[k])
            f.startx <- fun(startx,pXXijk)
            if(anyNA(f.startx)) {
                Result[indx,1:4] <- NA
            } else {
                z <- nleqslv(startx,fun,pXX=pXXijk)
                Result[indx,1:4] <- c(z$termcd,z$x)
            }
            Result[indx,5:7] <- c(i,j,k)
            indx <- indx+1
        }
    }
}
</code>

And now you can inspect Result, which contains 729 rows (and not 1000) and
continue to make  a data.frame with column names.
The rest is up to you.

Berend

On 10 May 2017, at 20:20, Santiago Burone <santiago
burone 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. 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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 - May 2017 - Solve system of non linear equations using nasted loops

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

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

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

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