similar to: Problem with minimization that I failed to understand

?s 19:36 de 27/03/2025, Daniel Lobo escreveu: > My code is to minimize the objective function > > therefore, shouldnt I expect that > > StartingValue = c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, > 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02) > Fn(q1$par) < Fn(StartingValue) > ## FALSE > > Below is the corrected code that can

My code is to minimize the objective function therefore, shouldnt I expect that StartingValue = c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02) Fn(q1$par) < Fn(StartingValue) ## FALSE Below is the corrected code that can be reproduced: MyDat = structure(list(c(50L, 0L, 0L, 50L, 75L, 100L, 50L, 0L, 50L, 0L, 25L,

?s 18:35 de 27/03/2025, Daniel Lobo escreveu: > Hi, > > I have below minimization problem > > > MyDat = structure(list(c(50L, 0L, 0L, 50L, 75L, 100L, 50L, 0L, 50L, 0L, > 25L, 50L, 50L, 75L, 75L, 75L, 0L, 75L, 75L, 75L, 0L, 25L, 75L, > 75L, 0L, 75L, 100L, 0L, 25L, 100L), c(75L, 0L, 0L, 50L, 100L, > 50L, 75L, 75L, 100L, 25L, 0L, 25L, 100L, 0L, 50L, 0L, 25L, 25L, >

I haven't run your code, but since Kendall correlation is based on ranks, your Fn is probably locally constant with jumps when the ranks change. That's a really hard kind of function to maximize, and the algorithm used by fmincon is not appropriate to do it. Sorry, but I don't know if there is an R function that can do constrained discrete maximization. Duncan Murdoch On

Hi Duncan, Thanks for your comment, I agree with that. But, how it can be justified that an Optimizer gives a result which is inferior to the starting value? At most, resulting value can remain at the same level, isnt it? On Fri, 28 Mar 2025 at 14:34, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > I haven't run your code, but since Kendall correlation is based on >

?s 13:59 de 28/03/2025, Daniel Lobo escreveu: > Hi Duncan, > > Thanks for your comment, I agree with that. > > But, how it can be justified that an Optimizer gives a result which is > inferior to the starting value? At most, resulting value can remain at the > same level, isnt it? > > On Fri, 28 Mar 2025 at 14:34, Duncan Murdoch <murdoch.duncan at gmail.com>

Hi R-Help, data at bottom I've been struggling with a problem where I need to order based on 1) the Frequency "Freq" and 2) keeping each group of 3 of the same type together "Var2" but I want across all groups it to go "high to low" based on the earn factor. Thank you! structure(list(Var1 = structure(c(1L, 3L, 2L, 1L, 2L, 3L, 1L, 3L, 2L, 3L, 1L, 2L, 3L, 1L,

I would like to preform a t.test to each of the measured variables (sand.silt etc.) with a mean and sd for each of the treatments (up or down), and out put this as a table.... I am having a hard time starting- maybe it is to close to lunch. Any suggestions would be greatly appreciated. Stephen Sefick x <- (structure(list(sample. = structure(c(1L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 2L, 3L,

Dear R users, A have a dataframe (matrix) with two collumns (plot, and diameter (d)). I want all diameters values for different combination of plots. For example I want all d values for all posible combination, 100C2 (all d values for plot 1 with all d values in the plot 2.......with all d values from plot 1 with all d values from plot 100, ...... with all d values from plot 99 with all d values

Hi, I'm trying to understand some details about an example maintened in [1]. According that link, I have total.data as a data set (am I right?). But I don't understand how is built that table. I saved the dataset in a file, with dput(), and had something like this: structure(list(df.all = structure(list(V1 = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L,

what am I doing wrong? chron(as.character(f), format=c(dates="%m/%d/%y", times="%h:%m")) f <- structure(c(51L, 60L, 66L, 87L, 90L, 115L, 23L, 35L, 37L, 6L, 12L, 55L, 84L, 96L, 109L, 17L, 29L, 41L, 3L, 74L, 94L, 102L, 30L, 8L, 46L, 69L, 107L, 15L, 25L, 39L, 1L, 71L, 95L, 19L, 56L, 62L, 76L, 85L, 99L, 111L, 42L, 4L, 52L, 61L, 67L, 91L, 13L, 24L, 36L, 38L, 7L, 81L, 82L, 57L,

Hi, I need some help with running a recursive function. I like to run funlp2 recursively. When I try to run recursive function in another function named "calclp" I get this "Error: any(!dat2$norm_sd) >= 1 is not TRUE". I have never built a recursive function before so having trouble executing it in this case. I would appreciate any help or guidance to resolve this issue.

You seem to have a typo at this expression (and some others like it) Namely, you write any(!dat2$norm_sd) >= 1 when you possibly meant to write !( any(dat2$norm_sd) >= 1 ) i.e. I think your ! seems to be in the wrong place. HTH, Eric On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] < nilesh.dighe at monsanto.com> wrote: > Hi, I need some help with running a

My own typo ... whoops ... !( any(dat2$norm_sd >= 1 )) On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger <ericjberger at gmail.com> wrote: > You seem to have a typo at this expression (and some others like it) > > Namely, you write > > any(!dat2$norm_sd) >= 1 > > when you possibly meant to write > > !( any(dat2$norm_sd) >= 1 ) > > i.e. I think your !

Eric: Thanks for taking time to look into my problem. Despite of making the change you suggested, I am still getting the same error. I am wondering if the logic I am using in the stopifnot and if functions is a problem. I like the recursive function to stop whenever the norm_sd column has zero values that are above or equal to 1. Below is the calclp function after the changes you suggested.

Hi, I accidently left out few lines of code from the calclp function. Updated function is pasted below. I am still getting the same error ?Error: !(any(data1$norm_sd >= 1)) is not TRUE? I would appreciate any help. Nilesh dput(calclp) function (dataset) { dat1 <- funlp1(dataset) recursive_funlp <- function(dataset = dat1, func = funlp2) { dat2 <- dataset %>%

The message is coming from your stopifnot() condition being met. On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] < nilesh.dighe at monsanto.com> wrote: > Hi, I accidently left out few lines of code from the calclp function. > Updated function is pasted below. > > I am still getting the same error ?Error: !(any(data1$norm_sd >= 1)) is > not TRUE? > > >

Hi all: I am trying to use the ddply function to estimate the mean of 'Total','Fry','Smolt' and 'Fry.Eq' columns without success. I have the dput of my dataset below. I wonder if someone can give me a hand with this function. # dput(winter) winter <-structure(list(IDDate = structure(c(37L, 48L, 59L, 62L, 63L, 64L, 65L, 66L, 67L, 38L, 39L, 40L, 41L, 42L, 43L,

Eric: I will try and see if I can figure out the issue by debugging as you suggested. I don?t know why my code after stopifnot is not getting executed where I like the code to run the funlp2 function when the if statement is TRUE but when it is false, I like it to keep running until the stopifnot condition is met. When the stopifnot condition is met, I like to get the output from if statement

If you are trying to understand why the "stopifnot" condition is met you can replace it by something like: if ( any(dat2$norm_sd >= 1) ) browser() This will put you in a debugging session where you can examine your variables, e.g. > dat$norm_sd HTH, Eric On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger <ericjberger at gmail.com> wrote: > The message is coming from