How to prevent two NPCs from damaging each other with their AoE spells in Azerothcore? - azerothcore

I am running a local server of Azerothcore with rev. 3716ddf3e4e1 2021-05-25 18:05:13 +0200 (master branch).
I have created two custom NPCs and added them to my DB with the following code:
INSERT INTO `creature_template` (`entry`, `difficulty_entry_1`, `difficulty_entry_2`, `difficulty_entry_3`, `KillCredit1`, `KillCredit2`, `modelid1`, `modelid2`, `modelid3`, `modelid4`, `name`, `subname`, `gossip_menu_id`, `minlevel`, `maxlevel`, `exp`, `faction`, `npcflag`, `speed_walk`, `speed_run`, `scale`, `rank`, `dmgschool`, `DamageModifier`, `BaseAttackTime`, `RangeAttackTime`, `BaseVariance`, `RangeVariance`, `unit_class`, `unit_flags`, `unit_flags2`, `dynamicflags`, `family`, `trainer_type`, `trainer_spell`, `trainer_class`, `trainer_race`, `type`, `type_flags`, `lootid`, `pickpocketloot`, `skinloot`, `PetSpellDataId`, `VehicleId`, `mingold`, `maxgold`, `AIName`, `MovementType`, `InhabitType`, `HoverHeight`, `HealthModifier`, `ManaModifier`, `ArmorModifier`, `RacialLeader`, `movementId`, `RegenHealth`, `mechanic_immune_mask`, `spell_school_immune_mask`, `flags_extra`, `ScriptName`, `VerifiedBuild`) VALUES
(1112001, 0, 0, 0, 0, 0, 3456, 0, 0, 0, 'Mob1', '', 0, 43, 43, 0, 63, 0, 1, 1.14286, 3, 3, 0, 30, 2000, 2000, 1, 1, 1, 32832, 2048, 0, 0, 0, 0, 0, 0, 7, 4, 0, 0, 0, 0, 0, 50000, 60000, 'SmartAI', 1, 3, 1, 100, 1, 1, 0, 0, 1, 0, 0, 256, '', 12340),
(1112003, 0, 0, 0, 0, 0, 21443, 0, 0, 0, 'Mob2', '', 0, 42, 42, 0, 63, 0, 1, 1.14286, 1, 1, 0, 10, 2000, 2000, 1, 1, 1, 0, 2048, 0, 0, 0, 0, 0, 0, 6, 0, 4860, 0, 0, 0, 0, 297, 393, 'SmartAI', 1, 3, 1, 60, 1, 1, 0, 0, 1, 650854271, 0, 0, '', 12340);
Once i let Mob1 cast spell Death and Decay, Mob2 will take damage from it once they walk across it. How can i prevent the mobs from damaging each other, but only do damage to players instead?


Have you tried to set the two NPC to an allied faction that is also an enemy faction for the player ?

Related

What is the most typographically safe way to write a linear program in R?

In R, linear programming using the lpsolve library requires an extensive amount of manually typed values for the needed matrices and vectors for the objective function coefficients, left-hand side coefficients, constraints, etc. Messing up even one value or one comma will make the script error or worse, the program will find a solution but it will be the wrong one due to incorrect setup. For large, real-world problems like network flow, simply typing the program itself is time prohibitive. What am I missing about R's capabilities in this space? Or, is there an alternate tool better fit for the job? Open source preferred due to budget.
Here is an example of the type of code needed in R for a relatively simple optimization problem:
# fleet size optimization, with an added computation of total miles driven
library(lpSolve)
# Objective function coefficients
ObjCoeff<-c(1300, 690, 421.5, 531, 690, 427.50, 277.50, 421.5, 427.50, 303, 531, 277.50, 303,
460, 281, 354, 460, 285, 185, 281, 285, 202, 354, 185, 202, 0)
# Constraint matrix
Amatrix<-matrix(c(0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, -1, 0, 0, -1, 0, 0, -1, 0, 0, 1, 1, 1, -1, 0, 0, -1, 0, 0, -1, 0, 0,0,
0, -1,0, 0, 1, 1, 1, 0, -1, 0, 0,-1, 0,-1, 0, 0, 1, 1, 1, 0, -1, 0, 0, -1, 0,0,
0, 0,-1, 0, 0,-1, 0, 1, 1, 1, 0, 0,-1, 0,-1, 0, 0, -1, 0, 1, 1, 1, 0, 0,-1,0,
0, 0, 0, -1, 0, 0,-1, 0, 0,-1, 1, 1, 1, 0, 0,-1, 0, 0,-1, 0, 0,-1, 1, 1, 1,0,
-1300, 460, 281,354,460,285,185,281,285,202,354,185,202,460,281,354, 460, 285,185, 281, 285,202, 354, 185, 202, 0,
0, 460, 281,354,460,285,185,281,285,202,354,185,202,460,281,354, 460, 285,185, 281, 285,202, 354, 185, 202,-1), nrow=18, byrow=TRUE)
# Right hand side constraint vector
Bvector<-c(10, 10, 10, 20, 10, 10, 10, 10, 10, 10, 10, 10, 0, 0, 0, 0, 0,0 )
# Constraint inequality direction vector
constrainttype<- c(">=", ">=",">=", ">=",">=", ">=",">=", ">=",">=", ">=",">=", ">=","=", "=" , "=" , "=" ,"<=", "=" )
# Solve the specified integer program by setting all.int=TRUE
optimum<-lp(direction="min", objective.in=ObjCoeff, const.mat = Amatrix, const.dir = constrainttype, const.rhs = Bvector, all.int =TRUE)
# Print constraint matrix to verify it was specified correctly
print(optimum$constraints)
# Check to see if the solver reached optimality (0 means yes)
print(optimum$status)
# Print values of each variable in the optimal solution
# note that they are all integer valued
print(optimum$solution)
# Print the optimal objective function value
print(optimum$objval)

R has its strengths for sure, but formulation of non-trivial LP's is not one of them.
There are several other open source frameworks that are much more expressive. If you are a python user, you can pick from many including:
pyomo, gekko, cvxpy, pulp, or-tools and others I'm sure.
I'm a fan of pyomo for model building, but it requires installation of a separate solver, which isn't too difficult. There are several open-source free solvers that are excellent and in common use with different capabilities such as cbc, glpk, ipopt and others, and of course many excellent licensed solvers.
If you want to start with an "all in one" the pulp framework includes a solver with the build--I forget which one.
There are many examples on this site for all the pkgs above if you search by tag.

Errors with distance-decay using betapart and ddecay packages

My goal is to create a distance-decay curve for species data vs geographic distance. However, I am running into errors. For the betapart package, this may be due to the lack of columns relative to the number of rows. Is there a way to get past this? If not, is there another method for creating a distance-decay curve (and plotting it)? I also tried the ddecay package but ran into errors there too. Any help is much appreciated. Data is in structure form below.
# BETAPART -------------------------------------------------
library(betapart)
spat.dist<-dist(coords)
dissim.BCI<-beta.pair.abund(spec)$beta.bray.bal
plot(spat.dist, dissim.BCI, ylim=c(0,1), xlim=c(0, max(spat.dist)))
BCI.decay.exp<-decay.model(dissim.BCI, spat.dist, y.type="dissim", model.type="exp", perm=100)
#========================================================================================================
I also tried a few other packages --------------------------
# ddecay package -------------------------------------------
devtools::install_github("chihlinwei/ddecay")
the issue with this method is that it requires the use of a gradient however, I would like to avoid that if possible but I do not see a way around this. Also they do not include their example data in the package.
dd <- beta.decay(gradient=spat.dist, counts=decostand(spec, method="pa"),
coords=coords, nboots=1000,
dis.fun = "beta.pair", index.family = "sorensen", dis = 1, like.pairs=T)
x <- vegdist(coords, method = "euclidean")
y <- 1 - dist(decostand(spec, method="pa"), index.family = "sorensen")[[1]]
plot(x, y)
lines(dd$Predictions[, "x"], dd$Predictions[,"mean"], col="red", lwd=2)
#========================================================================================================
# DATA -----------------------------------------------------
spec <- structure(list(Ccol = c(0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Acol = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0), NYcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0), Mcol = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0), AAcol = c(14, 0, 14, 3, 11, 1, 0, 2, 0,
3, 0, 4, 0, 1, 8, 2, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 7),
Ncol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1), ATBcol = c(0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 3), CVcol = c(0, 0, 0, 0, 0, 0, 1, 20,
0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 7, 0, 2, 0, 0,
0, 6), AZNcol = c(0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GBcol = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0), KHAcol = c(0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0), AFcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0), AFPcol = c(0,
0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1), TIAcol = c(4, 1, 0, 2, 6, 0,
1, 1, 0, 2, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0), AUcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), AScol = c(0,
4, 0, 2, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 5, 0, 0), NSAcol = c(0, 0, 0, 0, 0, 0,
0, 0, 0, 7, 0, 0, 3, 0, 0, 0, 4, 0, 2, 0, 1, 0, 9, 5, 1,
0, 0, 2, 0), WZcol = c(0, 0, 0, 0, 0, 0, 1, 0, 0, 10, 4,
0, 0, 0, 0, 0, 0, 1, 5, 0, 0, 0, 17, 4, 0, 0, 0, 0, 0), AJcol = c(0,
3, 6, 0, 0, 1, 0, 4, 0, 0, 0, 0, 39, 12, 0, 0, 0, 0, 0, 0,
0, 4, 5, 1, 12, 13, 16, 0, 5), EADcol = c(4, 1, 2, 1, 2,
0, 0, 0, 0, 4, 0, 2, 1, 1, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0,
0, 0, 0, 0, 1), CAcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0), Pcol = c(0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 60, 0, 0,
13, 0, 8, 1, 0, 0, 0, 0, 0), ASDcol = c(3, 5, 6, 17, 3, 5,
26, 2, 0, 17, 3, 10, 6, 3, 2, 4, 0, 0, 5, 25, 0, 0, 0, 2,
2, 9, 0, 2, 8), RMAcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
OUcol = c(0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), KAcol = c(0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12,
0, 0, 0, 0, 0, 8, 1), PACcol = c(0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 2, 0, 37, 0, 24,
1, 0, 0), LAAcol = c(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), GAcol = c(1,
0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0,
0, 0, 3, 0, 0, 0, 2, 0, 0), AAcol = c(1, 0, 1, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0), EVAcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0), EAcol = c(0,
0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0), AKcol = c(0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0), Acol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 1, 0), QAcol = c(0,
0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0), YAcol = c(11, 24, 21, 63, 44,
95, 12, 43, 0, 5, 26, 22, 25, 48, 86, 2, 0, 0, 13, 0, 0,
2, 0, 0, 60, 6, 7, 0, 45), BANcol = c(0, 0, 0, 3, 0, 0, 0,
0, 0, 0, 0, 0, 24, 0, 6, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0,
9, 17, 17), VCcol = c(0, 38, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Vcol = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0), Ocol = c(0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0), AVcol = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1), JXcol = c(0,
3, 3, 0, 0, 0, 0, 0, 8, 0, 0, 10, 3, 0, 0, 5, 0, 0, 0, 1,
0, 0, 0, 2, 4, 1, 0, 0, 0)), class = "data.frame", row.names = c(NA,
-29L))
coords <- structure(list(Lat.x = c(34.43363, 34.36784, 34.32587, 34.19891,
34.24217, 34.24863, 34.18137, 34.16838, 34.10961, 34.08329, 34.40571,
34.39591, 34.39292, 34.37466, 34.28948, 34.26146, 34.04687, 34.0409,
34.068339, 34.34679, 34.17161, 34.23308, 34.21544, 34.14922,
34.27539, 34.2323, 34.19057, 34.07042, 34.06289), Lon.x = c(-94.94494,
-94.92512, -94.94429, -94.84497, -94.8573, -94.85641, -94.887,
-94.91322, -94.92913, -94.93276, -95.02622, -95.04382, -94.96295,
-94.83733, -94.81071, -94.79161, -95.03968, -95.0608, -95.086986,
-95.03345, -95.23862, -95.25619, -95.1041, -95.02286, -95.02672,
-95.02626, -95.02941, -95.01746, -94.98786)), class = "data.frame", row.names = c(NA,
-29L))

You can get more answers, if you tell what was the problem. For instance, which functions failed and what was the error message. I had a look at betapart::decay.model(), where I could get this error message:
Error in eval(family$initialize) :
cannot find valid starting values: please specify some
I cut the long story short: you cannot use this function with your data because you have dissimilarities of 1 in your data, dissimilarities are turned into similarities with 1-dissimilarity and this makes these values zero similarities (that is, these pairs of sampling unit have nothing in common, they share no species). Function decay.model uses glm with gaussian family with log-link, and log-link requires that you give the starting values, if you have zeros in the y-variate.
I think that you have four alternatives:
You do not use the method as it does not suit your data.
You modify the decay.model function so that you can specify the starting values, like the error message suggested. This means that you add mustart to the function call so that it reads, e.g., glm(y ~ x, family=gaussian(link="log"), mustart=pmax(y, 0.01)). This replaces zeros with 0.01 as starting values.
You change maximum distances from 1 to something smaller, for instance, 0.99: dissim.BCI[dissim.BCI==1] <- 0.99. However, this changes the data, and also changes the results from those you get with alternative 2 (which only changes starting values, but data are unmodified). However, the effect is not very large and any Bayesian would claim that dissimilarity 1 is just a frequentist folly (you just haven't seen the case that is in common with these sampling units).
You change the maximum distance to missing values. This will change data more than alternative 3. It removes maximum dissimilarities and these no longer influence the decay curve. The effect is the same as censoring greatest dissimilarities. The results change more than in alternative 3.

The output from LPSolver is not the optimal result

Since I need to create an automatic shift planning tool, R lpSolve is the package I plan to use. However, I can't get the optimal output by the code shown as below. The output(hourly supply by accumulating all shifts available) can't fulfill the demand. e.g. The demand for a hour is 46 but the supply is only 25 which means 21 unit of demand can't be satisfied.
Background:
The objective
To minimize the difference between total supply and total demand.
To minimize the difference between supply and demand for each hour.
Constraint
Shift constraint - each hour might have several shifts available to be assigned.
Max capacity - I have the cap of supply. the sum of total shift cant exceed the cap for each hour. (46 is the cap in this example).
In the constr matrix, 24 rows represent 24 hours starting from 7 am and 9 columns refers to no. of shifts I have.
The constr.val refers to hourly demand.
Please let me know if there is anything unclear. thanks!
library(lpSolve)
obj.fun<-c(1,1,1,1,1,1,1,1,1)
constr<-matrix(c(
1, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 1, 0, 0, 0, 0, 0,
1, 0, 0, 1, 0, 1, 0, 0, 0,
1, 0, 0, 1, 0, 1, 0, 1, 0,
1, 0, 0, 1, 0, 1, 0, 1, 0,
1, 0, 0, 1, 0, 1, 0, 1, 0,
0, 1, 0, 1, 0, 1, 0, 1, 0,
0, 1, 0, 1, 0, 1, 0, 1, 0,
0, 1, 0, 1, 0, 1, 0, 1, 0,
0, 1, 0, 0, 1, 1, 0, 1, 0,
0, 1, 0, 0, 1, 0, 1, 1, 0,
0, 1, 0, 0, 1, 0, 1, 0, 1,
0, 1, 0, 0, 1, 0, 1, 0, 1,
0, 1, 0, 0, 1, 0, 1, 0, 1,
0, 0, 1, 0, 1, 0, 1, 0, 1,
0, 0, 1, 0, 1, 0, 1, 0, 1,
0, 0, 1, 0, 1, 0, 1, 0, 1,
0, 0, 1, 0, 0, 0, 1, 0, 1,
0, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0), nrow = 24, byrow = TRUE)
constr.dir <- rep("<=", 24)
constr.val <- c(24, 20, 21, 22, 26, 34, 40, 44,
46, 46, 46, 46, 46, 46, 46, 46,
46, 46, 46, 46, 46, 46, 41, 27)
day.shift <- lp("max", obj.fun, constr, constr.dir,
constr.val, compute.sens = TRUE)
day.shift$solution

Why is auto.arima() giving me the Error: "'by' argument is much too small"

I am working on predicting intra-day sales for a retailer. We want to know if we can predict sales through the rest of the day, based off sales within that day. I'm working with roughly 3 years of data in a time series, which has given me roughly 26,000 rows of data.
I've never worked with a time series this large so my approach might be off. Or auto.arima() may not have been made to handle data this large.
I've tried limiting my data down to even 300 rows and had marginal success, but have not found anything that works with my larger data set. auto.arima() doesn't even have a by = argument from what I can find.
my_ts <- structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 3,
3, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 7, 4, 1,
1, 0, 3, 1, 0, 8, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4,
1, 9, 1, 6, 5, 1, 0, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 3, 1, 0, 3, 5, 2, 3, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 3, 1, 0, 0, 6, 0, 6, 0, 1, 2, 3, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 2, 4, 6, 5, 0, 1, 0, 2, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 2, 0, 2, 0, 0, 0, 1, 1,
3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 3, 0, 1, 0,
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1438779600, 1438783200, 1438786800), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), class = c("zooreg", "zoo"), frequency = 24)
fit1 <-auto.arima(my_ts,seasonal = TRUE)
I was hoping to get a model through arima, but I'm only getting the error:
"Error in seq.default(head(tt, 1), tail(tt, 1), deltat) :
'by' argument is much too small"

Stacked barplot with partitioned main and sub x label

I'm new to R. I want to create a stacked barplot based on the following data. I want to plot these data into a stacked barplot that describes percentage of disease score for each combination of genotype and race in x axis, e.g. 76R-race 1, rmc-race 1. As well how to simplify the x axis, by splitting each race to have 76R and rmc combination instead of labelling each combination, i.e. instead of labelling each bar with 76R-race 1, rmc-race 1, etc, how to label race 1 as main axis and have 76R and rmc as sub-axis, and so on.
disease.nov <- data.frame(disease.score = c(0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 6, 7, 5, 5, 4, 4, 6, 5, 8, 5, 5, 5, 6, 4, 7, 5, 8, 6, 1, 2, 2, 4, 5, 8, 5, 6, 7, 4, 4, 4, 2, 3, 5, 6, 7, 7, 5, 2, 6, 6, 6, 4, 5, 8, 7, 5, 2, 5, 6, 3, 7, 4, 7, 7, 8, 6, 8, 8, 7, 9, 9, 7, 4, 9, 9, 5, 3, 8, 8, 6, 5, 7, 7, 8, 6, 6, 5, 7, 7, 8, 8, 8, 8, 7, 6, 6, 8, 4, 7, 7, 8, 6, 7, 7, 8, 6, 5, 6, 7, 7, 4, 6, 8, 7, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
genotype=gl(2,52,624, labels=c("76R","rmc")),
race=gl(6,104,624, labels=c("race 1","race 2","race 3","WAC 7673","WAC 7591","control")))

I'm not sure if this is what you're looking for:
library(ggplot2)
library(scales)
ggplot(disease.nov, aes(x=genotype, fill=factor(disease.score))) +
geom_bar(position="fill") +
facet_wrap(~ race) +
scale_y_continuous(labels = percent_format())
Which produces:

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