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Package nleqslv Error: Length of fn result length of x
Using R's nleqslv to solve system of equations with vector valuesMigrating from using S-Plus Design package to R RMS packageWhy is nleqslv function returning a NULL output?How do you construct a flexible function input when creating a function in R?RStudio crashing with RcppGSL functionHandling very small numbers in ratio and how to keep exponential valuegillnets/gillnetfit error: Non-numeric argument… invalid type (list) for variable…How to pass nonlinear objective functions into ROI package in R?Question about pol() function in the ASReml-R packageRShiny - Warning: Error in : Aesthetics must be either length 1 or the same as the data (48): y
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I am solving a nonlinear equation using the package nleqslv, but I keep getting the error: Length of fn result <> length of x!
I can't spot the place where vector length can be a problem. Anyone knows what mistakes did I made in my code?
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(x) F_C0(x) - x[3]
nleqslv( c(40, 1, 17.35), eqn)
I am trying to solve for x[2] and I have the input of x[1] and x[3]. It looks like I should get the numeric solution of x[2].
r
asked Mar 25 at 3:28
SocRortySocRorty
32
add a comment |
I am solving a nonlinear equation using the package nleqslv, but I keep getting the error: Length of fn result <> length of x!
I can't spot the place where vector length can be a problem. Anyone knows what mistakes did I made in my code?
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(x) F_C0(x) - x[3]
nleqslv( c(40, 1, 17.35), eqn)
I am trying to solve for x[2] and I have the input of x[1] and x[3]. It looks like I should get the numeric solution of x[2].
r
asked Mar 25 at 3:28
SocRortySocRorty
32
pnormis vectorised in its first function argumentq. SinceF_C0accepts avectorxwhich you then pass on topnorm,pnormreturns a vector (which you then operate on with scalars) and so does the whole functionF_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.
– Maurits Evers
Mar 25 at 3:55
add a comment |
I am solving a nonlinear equation using the package nleqslv, but I keep getting the error: Length of fn result <> length of x!
I can't spot the place where vector length can be a problem. Anyone knows what mistakes did I made in my code?
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(x) F_C0(x) - x[3]
nleqslv( c(40, 1, 17.35), eqn)
I am trying to solve for x[2] and I have the input of x[1] and x[3]. It looks like I should get the numeric solution of x[2].
r
asked Mar 25 at 3:28
SocRortySocRorty
32
I am solving a nonlinear equation using the package nleqslv, but I keep getting the error: Length of fn result <> length of x!
I can't spot the place where vector length can be a problem. Anyone knows what mistakes did I made in my code?
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(x) F_C0(x) - x[3]
nleqslv( c(40, 1, 17.35), eqn)
I am trying to solve for x[2] and I have the input of x[1] and x[3]. It looks like I should get the numeric solution of x[2].
r
r
asked Mar 25 at 3:28
SocRortySocRorty
32
asked Mar 25 at 3:28
SocRortySocRorty
32
edited Mar 25 at 12:41
SocRorty
asked Mar 25 at 3:28
SocRortySocRorty
32
asked Mar 25 at 3:28
SocRortySocRorty
32
asked Mar 25 at 3:28
SocRortySocRorty
32
32
pnormis vectorised in its first function argumentq. SinceF_C0accepts avectorxwhich you then pass on topnorm,pnormreturns a vector (which you then operate on with scalars) and so does the whole functionF_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.
– Maurits Evers
Mar 25 at 3:55
add a comment |
pnormis vectorised in its first function argumentq. SinceF_C0accepts avectorxwhich you then pass on topnorm,pnormreturns a vector (which you then operate on with scalars) and so does the whole functionF_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.
– Maurits Evers
Mar 25 at 3:55
pnorm is vectorised in its first function argument q. Since F_C0 accepts a vector x which you then pass on to pnorm, pnorm returns a vector (which you then operate on with scalars) and so does the whole function F_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.– Maurits Evers
Mar 25 at 3:55
pnorm is vectorised in its first function argument q. Since F_C0 accepts a vector x which you then pass on to pnorm, pnorm returns a vector (which you then operate on with scalars) and so does the whole function F_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.– Maurits Evers
Mar 25 at 3:55
add a comment |
1 Answer
1
active
oldest
votes
Your example is not reproducible since you have not shown all your code: library(nleqslv) is missing. Please show all your code.
As the first comment on your question stated you are providing a vector to eqn but functions d_plus, d_minus and thus F_C0 return a scalar.
That implies that the length of the function result is not the same as the length of the input.
From your explanation you want to solve for x[2]. So the function presented to nleqslv must take a scalar as input and return a scalar.
This can be achieved as follows:
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(xpar) x <- c(40,xpar,17.35);F_C0(x) - x[3]
Insert the scalar argument of eqn, which is your x[2], into a vector x where the first and third entries are what you provided as starting values in your code .
Then running this
xstart <- 1
nleqslv( xstart, eqn)
results in this
$x
[1] 0.6815036
$fvec
[1] 6.18563e-11
$termcd
[1] 1
$message
[1] "Function criterion near zero"
$scalex
[1] 1
$nfcnt
[1] 5
$njcnt
[1] 1
$iter
[1] 5
Read the documentation of nleqslv to see what these items mean.
As you can see nleqslv found a solution to your problem.
answered Mar 25 at 6:34
BhasBhas
1,466198
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argumentfnofnleqslvsays so.
– Bhas
Mar 25 at 13:45
add a comment |
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Your example is not reproducible since you have not shown all your code: library(nleqslv) is missing. Please show all your code.
As the first comment on your question stated you are providing a vector to eqn but functions d_plus, d_minus and thus F_C0 return a scalar.
That implies that the length of the function result is not the same as the length of the input.
From your explanation you want to solve for x[2]. So the function presented to nleqslv must take a scalar as input and return a scalar.
This can be achieved as follows:
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(xpar) x <- c(40,xpar,17.35);F_C0(x) - x[3]
Insert the scalar argument of eqn, which is your x[2], into a vector x where the first and third entries are what you provided as starting values in your code .
Then running this
xstart <- 1
nleqslv( xstart, eqn)
results in this
$x
[1] 0.6815036
$fvec
[1] 6.18563e-11
$termcd
[1] 1
$message
[1] "Function criterion near zero"
$scalex
[1] 1
$nfcnt
[1] 5
$njcnt
[1] 1
$iter
[1] 5
Read the documentation of nleqslv to see what these items mean.
As you can see nleqslv found a solution to your problem.
answered Mar 25 at 6:34
BhasBhas
1,466198
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argumentfnofnleqslvsays so.
– Bhas
Mar 25 at 13:45
add a comment |
Your example is not reproducible since you have not shown all your code: library(nleqslv) is missing. Please show all your code.
As the first comment on your question stated you are providing a vector to eqn but functions d_plus, d_minus and thus F_C0 return a scalar.
That implies that the length of the function result is not the same as the length of the input.
From your explanation you want to solve for x[2]. So the function presented to nleqslv must take a scalar as input and return a scalar.
This can be achieved as follows:
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(xpar) x <- c(40,xpar,17.35);F_C0(x) - x[3]
Insert the scalar argument of eqn, which is your x[2], into a vector x where the first and third entries are what you provided as starting values in your code .
Then running this
xstart <- 1
nleqslv( xstart, eqn)
results in this
$x
[1] 0.6815036
$fvec
[1] 6.18563e-11
$termcd
[1] 1
$message
[1] "Function criterion near zero"
$scalex
[1] 1
$nfcnt
[1] 5
$njcnt
[1] 1
$iter
[1] 5
Read the documentation of nleqslv to see what these items mean.
As you can see nleqslv found a solution to your problem.
answered Mar 25 at 6:34
BhasBhas
1,466198
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argumentfnofnleqslvsays so.
– Bhas
Mar 25 at 13:45
add a comment |
Your example is not reproducible since you have not shown all your code: library(nleqslv) is missing. Please show all your code.
As the first comment on your question stated you are providing a vector to eqn but functions d_plus, d_minus and thus F_C0 return a scalar.
That implies that the length of the function result is not the same as the length of the input.
From your explanation you want to solve for x[2]. So the function presented to nleqslv must take a scalar as input and return a scalar.
This can be achieved as follows:
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(xpar) x <- c(40,xpar,17.35);F_C0(x) - x[3]
Insert the scalar argument of eqn, which is your x[2], into a vector x where the first and third entries are what you provided as starting values in your code .
Then running this
xstart <- 1
nleqslv( xstart, eqn)
results in this
$x
[1] 0.6815036
$fvec
[1] 6.18563e-11
$termcd
[1] 1
$message
[1] "Function criterion near zero"
$scalex
[1] 1
$nfcnt
[1] 5
$njcnt
[1] 1
$iter
[1] 5
Read the documentation of nleqslv to see what these items mean.
As you can see nleqslv found a solution to your problem.
answered Mar 25 at 6:34
BhasBhas
1,466198
Your example is not reproducible since you have not shown all your code: library(nleqslv) is missing. Please show all your code.
As the first comment on your question stated you are providing a vector to eqn but functions d_plus, d_minus and thus F_C0 return a scalar.
That implies that the length of the function result is not the same as the length of the input.
From your explanation you want to solve for x[2]. So the function presented to nleqslv must take a scalar as input and return a scalar.
This can be achieved as follows:
library(nleqslv)
d_plus <- function(x) (log(55.75/x[1])+(0.026 + x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
d_minus <- function(x) (log(55.75/x[1])+(0.026 - x[2]^2 / 2) * 0.25) / (x[2]*0.5) + 0 * x[3]
F_C0 <- function(x) 55.75 * pnorm(d_plus(x)) - x[1] * exp(-0.026 * 0.25) * pnorm(d_minus(x)) + 0 * x[3]
eqn <- function(xpar) x <- c(40,xpar,17.35);F_C0(x) - x[3]
Insert the scalar argument of eqn, which is your x[2], into a vector x where the first and third entries are what you provided as starting values in your code .
Then running this
xstart <- 1
nleqslv( xstart, eqn)
results in this
$x
[1] 0.6815036
$fvec
[1] 6.18563e-11
$termcd
[1] 1
$message
[1] "Function criterion near zero"
$scalex
[1] 1
$nfcnt
[1] 5
$njcnt
[1] 1
$iter
[1] 5
Read the documentation of nleqslv to see what these items mean.
As you can see nleqslv found a solution to your problem.
answered Mar 25 at 6:34
BhasBhas
1,466198
answered Mar 25 at 6:34
BhasBhas
1,466198
answered Mar 25 at 6:34
BhasBhas
1,466198
answered Mar 25 at 6:34
BhasBhas
1,466198
1,466198
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argumentfnofnleqslvsays so.
– Bhas
Mar 25 at 13:45
add a comment |
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argumentfnofnleqslvsays so.
– Bhas
Mar 25 at 13:45
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Thank you for the explanation! It looks like I always have to match the number of real equations with the number of real unknowns.
– SocRorty
Mar 25 at 12:46
Yes and the documentation for argument
fn of nleqslv says so.– Bhas
Mar 25 at 13:45
Yes and the documentation for argument
fn of nleqslv says so.– Bhas
Mar 25 at 13:45
add a comment |
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pnormis vectorised in its first function argumentq. SinceF_C0accepts avectorxwhich you then pass on topnorm,pnormreturns a vector (which you then operate on with scalars) and so does the whole functionF_C0. Could that be the reason for the error? Otherwise it would help if you were to explicitly write down the non-linear equation with some explanations.– Maurits Evers
Mar 25 at 3:55