Problem assigning probabilities in sample function - r

I have generated a bunch of variables through the following :
max_no=10
list2env(setNames(as.list(c(0.2, 0.25,0.15, 0.1, 0.1, 0.05, 0.03, 0.06, 0.03, 0.02, 0.01)), paste0("proportion", 0:max_no)), envir = .GlobalEnv)
These generate objects like "proportion0", "proportion1",..., "proportion10" with values 0.2, 0.25.... etc.
I want to plug these proportion values as a vector of probabilities into the sample function like the following:
sample(seq.int(0, max_no, 1), size=10000, replace=TRUE, prob=c(paste0("proportion", 0:max_no)))
But I get the error message: Error in sample.int(length(x), size, replace, prob) : NA in probability vector
What would be a simple way of feeding in the probabilities?

I am not sure what is the use-case for this but you can use mget and unlist
sample(seq.int(0, max_no, 1), size=10000, replace=TRUE,
prob=unlist(mget(c(paste0("proportion", 0:max_no)))))
Why not pass probability directly instead of storing each number in a different variable?
sample(seq.int(0, max_no, 1), size=10000, replace=TRUE,
prob=c(0.2, 0.25,0.15, 0.1, 0.1, 0.05, 0.03, 0.06, 0.03, 0.02, 0.01))

Related

bnlearn Error: Wrong number of conditional probability distributions

I am learning to work with bnlearn and I keep running into the following error in the last line of my code below:
Error in custom.fit(dag, cpt) : wrong number of conditional probability distributions
What am I doing wrong?
modelstring(dag)= "[s][r][nblw|r][nblg|nblw][mlw|s:r][f|s:r:mlw][mlg|mlw:f]
[mlgr|mlg:nblg]"
###View DAG Specifics
dag
arcs(dag)
nodes(dag)
# Create Levels
State <- c("State0", "State1")
##Create probability distributions given; these are all 2d b/c they have 1 or 2 nodes
cptS <- matrix(c(0.6, 0.4), ncol=2, dimnames=list(NULL, State))
cptR <- matrix(c(0.7, 0.3), ncol=2, dimnames=list(NULL, State))
cptNBLW <- matrix(c(0.95, 0.05, 0.05, 0.95), ncol=2, dimnames=list(NULL, "r"= State))
cptNBLG <- matrix(c(0.9, 0.099999999999999998, 0.2, 0.8), ncol=2, dimnames=list(NULL,
"nblw"=State))
cptMLG <- matrix(c(0.95, 0.05, 0.4, 0.6, 0.2, 0.8, 0.05, 0.95),ncol=2,nrow = 2,
dimnames=list("mlw"= State, "f"=State))
cptMLGR <- matrix(c(0.6,0.4,0.95,0.05,0.2,0.8,0.55,0.45),ncol=2,nrow = 2,
dimnames=list("mlg"= State, "nblg"=State))
cptMLW <-matrix(c(0.95, 0.05, 0.1, 0.9, 0.2, 0.8, 0.01, 0.99), ncol=2,nrow = 2,byrow = TRUE,
dimnames=list("r"= State, "s"=State))
# Build 3-d matrices( becuase you have 3 nodes, you can't use the matrix function; you
have to build it from scratch)
cptF <- c(0.05, 0.95, 0.4, 0.6, 0.9, 0.1, 0.99, 0.01, 0.9, 0.1, 0.95, 0.05, 0.95, 0.05, 0.99,
0.01)
dim(cptF) <- c(2, 2, 2, 2)
dimnames(cptF) <- list("s"=State, "r"=State, "mlw"=State)
###Create CPT Table
cpt <- list(s = cptS, r = cptR, mlw = cptMLW,nblw= cptNBLW,
mlg= cptMLG, nblg= cptNBLG, mlgr= cptMLGR)
# Construct BN network with Conditional Probability Table
S.net <- custom.fit(dag,cpt)
Reference: https://rpubs.com/sarataheri/bnlearnCGM
You have several errors in your CPT definitions. Primarily, you need to make sure that:
the number of probabilities supplied are equal to the product of the number of states in the child and parent nodes,
that the number of dimensions of the matrix/array is equal to the number of parent nodes plus one, for the child node,
the child node should be given in the first dimension when the node dimension is greater than one.
the names given in the dimnames arguments (e.g. the names in dimnames=list(ThisName = ...)) should match the names that were defined in the DAG, in your case with modelstring and in my answer with model2network. (So my earlier suggestion of using dimnames=list(cptNBLW = ...) should be dimnames=list(nblw = ...) to match how node nblw was declared in the model string)
You also did not add node f into your cpt list.
Below is your code with comments where things have been changed. (I have commented out the offending lines and added ones straight after)
library(bnlearn)
dag <- model2network("[s][r][nblw|r][nblg|nblw][mlw|s:r][mlg|mlw:f][mlgr|mlg:nblg][f|s:r:mlw]")
State <- c("State0", "State1")
cptS <- matrix(c(0.6, 0.4), ncol=2, dimnames=list(NULL, State))
cptR <- matrix(c(0.7, 0.3), ncol=2, dimnames=list(NULL, State))
# add child node into first slot of dimnames
cptNBLW <- matrix(c(0.95, 0.05, 0.05, 0.95), ncol=2, dimnames=list(nblw=State, "r"= State))
cptNBLG <- matrix(c(0.9, 0.099999999999999998, 0.2, 0.8), ncol=2, dimnames=list(nblg=State,"nblw"=State))
# Use a 3d array and not matrix, and add child node into dimnames
# cptMLG <- matrix(c(0.95, 0.05, 0.4, 0.6, 0.2, 0.8, 0.05, 0.95),ncol=2,nrow = 2, dimnames=list("mlw"= State, "f"=State))
cptMLG <- array(c(0.95, 0.05, 0.4, 0.6, 0.2, 0.8, 0.05, 0.95),dim=c(2,2,2), dimnames=list(mlg = State, "mlw"= State, "f"=State))
# cptMLGR <- matrix(c(0.6,0.4,0.95,0.05,0.2,0.8,0.55,0.45),ncol=2,nrow = 2, dimnames=list("mlg"= State, "nblg"=State))
cptMLGR <- array(c(0.6,0.4,0.95,0.05,0.2,0.8,0.55,0.45), dim=c(2,2,2), dimnames=list(mlgr=State, "mlg"= State, "nblg"=State))
# cptMLW <-matrix(c(0.95, 0.05, 0.1, 0.9, 0.2, 0.8, 0.01, 0.99), ncol=2,nrow = 2,byrow = TRUE, dimnames=list("r"= State, "s"=State))
cptMLW <-array(c(0.95, 0.05, 0.1, 0.9, 0.2, 0.8, 0.01, 0.99), dim=c(2,2,2), dimnames=list(mlw=State, "r"= State, "s"=State))
# add child into first slot of dimnames
cptF <- c(0.05, 0.95, 0.4, 0.6, 0.9, 0.1, 0.99, 0.01, 0.9, 0.1, 0.95, 0.05, 0.95, 0.05, 0.99, 0.01)
dim(cptF) <- c(2, 2, 2, 2)
dimnames(cptF) <- list("f" = State, "s"=State, "r"=State, "mlw"=State)
# add missing node f into list
cpt <- list(s = cptS, r = cptR, mlw = cptMLW,nblw= cptNBLW, mlg= cptMLG, nblg= cptNBLG, mlgr= cptMLGR, f=cptF)
# Construct BN network with Conditional Probability Table
S.net <- custom.fit(dag, dist=cpt)

Output selected variables to global environment R function

I have function which is an extension of an earlier question here
Function to calculate median by column to an R dataframe that is done regularly to multiple dataframes
my function below
library(outliers)
MscoreMax <- 3
scores_na <- function(x, ...) {
not_na <- !is.na(x)
scores <- rep(NA, length(x))
scores[not_na] <- outliers::scores(na.omit(x), ...)
scores
}
mediansFunction <- function(x){
labmedians <- sapply(x[-1], median)
median_of_median <- median(labmedians)
grand_median <- median(as.matrix(x[-1]))
labMscore <- as.vector(round(abs(scores_na(labmedians, "mad")), digits = 2)) #calculate mscore by lab
labMscoreIndex <- which(labMscore > MscoreMax) #get the position in the vector that exceeds Mscoremax
x[-1][labMscoreIndex] <- NA # discharge values above threshold by making NA
return(x)
}
the function has the desired outcome of converting my Mscore values above the threshold to NA. However, I would like to send
labmedians
grand_median
labMscore
As their own variables to the global environment from within the function, but not as a list of items as 3 variables. Can i do this or is better to create a second function which is slightly different that sends the variables to the global environment as a function then use list2env outside the function afterwards to extract the variables as seperate items?
my df below
structure(list(Determination_No = 1:6, `2` = c(0.08, 0.08, 0.08,
0.08, 0.08, 0.08), `3` = c(0.08, 0.07, 0.07, 0.08, 0.07, 0.07
), `4` = c(0.07, 0.08, 0.08, 0.08, 0.07, 0.08), `5` = c(0.08,
0.08, 0.08, 0.08, 0.09, 0.09), `7` = c(0.09, 0.09, 0.11, 0.1,
0.1, 0.1), `8` = c(0.086, 0.087, 0.086, 0.09, 0.083, 0.079),
`10` = c(0.049748274, 0.049748274, 0.066331032, 0.066331032,
0.066331032, 0.049748274), `12` = c(0.086, 0.078, 0.078,
0.077, 0.077, 0.068)), class = "data.frame", row.names = c(NA,
-6L))
It is not recommended to write to global environment from inside the function. If you want to create multiple objects in the global environment return a named list from the function and use list2env.
mediansFunction <- function(x){
labmedians <- sapply(x[-1], median)
median_of_median <- median(labmedians)
grand_median <- median(as.matrix(x[-1]))
labMscore <- as.vector(round(abs(scores_na(labmedians, "mad")), digits = 2)) #calculate mscore by lab
labMscoreIndex <- which(labMscore > MscoreMax) #get the position in the vector that exceeds Mscoremax
x[-1][labMscoreIndex] <- NA # discharge values above threshold by making NA
dplyr::lst(data = x, labmedians, grand_median, labMscore)
}
result <- mediansFunction(df)
list2env(result, .GlobalEnv)
Now you have variables data, labmedians, grand_median and labMscore in the global environment.

Efficient way to calculate average MAPE and MSE in R

I have a real data and predicted data and I want to calculate overall MAPE and MSE. The data are time series, with each column representing data for different weeks. I predict value for each of the 52 weeks for each of the items as shown below. What would be the best possible calculate overall Error in R.
real = matrix(
c("item1", "item2", "item3", "item4", .5, .7, 0.40, 0.6, 0.3, 0.29, 0.7, 0.09, 0.42, 0.032, 0.3, 0.37),
nrow=4,
ncol=4)
colnames(real) <- c("item", "week1", "week2", "week3")
predicted = matrix(
c("item1", "item2", "item3", "item4", .55, .67, 0.40, 0.69, 0.13, 0.9, 0.47, 0.19, 0.22, 0.033, 0.4, 0.37),
nrow=4,
ncol=4)
colnames(predicted) <- c("item", "week1", "week2", "week3")
How do you get the predicted values in the first place? The model you use to get the predicted values is probably based on minimising some function of prediction errors (usually MSE). Therefore, if you calculate your predicted values, the residuals and some metrics on MSE and MAPE have been calculated somewhere along the line in fitting the model. You can probably retrieve them directly.
If the predicted values happened to be thrown into your lap and you have nothing to do with fitting the model, then you calculate MSE and MAPE as per below:
You have only one record per week for every item. So for every item, you can only calculate one prediction error per week. Depending on your application, you can choose to calculate the MSE and MAPE per item or per week.
This is what your data looks like:
real <- matrix(
c(.5, .7, 0.40, 0.6, 0.3, 0.29, 0.7, 0.09, 0.42, 0.032, 0.3, 0.37),
nrow = 4, ncol = 3)
colnames(real) <- c("week1", "week2", "week3")
predicted <- matrix(
c(.55, .67, 0.40, 0.69, 0.13, 0.9, 0.47, 0.19, 0.22, 0.033, 0.4, 0.37),
nrow = 4, ncol = 3)
colnames(predicted) <- c("week1", "week2", "week3")
Calculate the (percentage/squared) errors for every entry:
pred_error <- real - predicted
pct_error <- pred_error/real
squared_error <- pred_error^2
Calculate MSE, MAPE:
# For per-item prediction errors
apply(squared_error, MARGIN = 1, mean) # MSE
apply(abs(pct_error), MARGIN = 1, mean) # MAPE
# For per-week prediction errors
apply(squared_error, MARGIN = 0, mean) # MSE
apply(abs(pct_error), MARGIN = 0, mean) # MAPE

Trying to fit f distribution to a vector

Would anyone know why the following code fails to execute fitdist with error "the function mle failed to estimate the parameters, with the error code 100".
I have encountered this error in the past when working with the normal distribution; the solution in that case was increasing the variance of the vector (by multiplying it by say 100), but that does not help on this case. Please note all elements in the vector are positive. Thank you.
library(fitdistrplus)
VH <- c(0.36, 0.3, 0.36, 0.47, 0, 0.05, 0.4, 0, 0, 0.15, 0.89, 0.03, 0.45, 0.21, 0, 0.18, 0.04, 0.53, 0, 0.68, 0.06, 0.09, 0.58, 0.03, 0.23, 0.27, 0, 0.12, 0.12, 0, 0.32, 0.07, 0.04, 0.07, 0.39, 0, 0.25, 0.28, 0.42, 0.55, 0.04, 0.07, 0.18, 0.17, 0.06, 0.39, 0.65, 0.15, 0.1, 0.32, 0.52, 0.55, 0.71, 0.93, 0, 0.36)
f <- fitdist(na.exclude(VH),"f", start =list(df1=1, df2=2))
The error you get here is actually somewhat informative:
simpleError in optim(par = vstart, fn = fnobj, fix.arg = fix.arg, obs = data, ddistnam = ddistname, hessian = TRUE, method = meth, lower = lower, upper = upper, ...): function cannot be evaluated at initial parameters
Error in fitdist(na.exclude(VH), "f", start = list(df1 = 1, df2 = 2)) :
the function mle failed to estimate the parameters,
with the error code 100
That means something went wrong right away, not in the middle of the optimization process.
Taking a guess, I looked and saw that there was a zero value in your data (so your statement that all the elements are positive is not technically correct -- they're all non-negative ...). The F distribution has an infinite value at 0: df(0,1,2) is Inf.
If I exclude the zero value, I get an answer ...
f <- fitdist(na.exclude(VH[VH>0]),"f", start =list(df1=1, df2=2))
... the estimated value for the second shape parameter is very large (approx. 6e6, with a big uncertainty), but seems to fit OK ...
par(las=1); hist(VH,freq=FALSE,col="gray")
curve(df(x,1.37,6.45e6),add=TRUE)

Interpolate within points in a vector

Vector V1 contains 56 observations for X, and vector BS contains a bootstrapped sample of V1 of length 100000. I would like to interpolate linearly within points in BS to fill in any missing values. For example, V1 contains no 0.27 values, and hence neither does BS. But BS would contain a few 0.28 and 0.26. I would like the interpolation to create a few 0.27 values and add those to BS. And so on for any missing values within the two extremes in the vector.
V1 <- c(0.18, 0.2, 0.24, 0.35, -0.22, -0.17, 0.28, -0.28, -0.14, 0.03, 0.87, -0.2, 0.06, -0.1, -0.72, 0.18, 0.01, 0.31, -0.36, 0.61, -0.16, -0.07, -0.13, 0.01, -0.09, 0.26, -0.14, 0.08, -0.62, -0.2, 0.3, -0.21, -0.11, 0.05, 0.06, -0.28, -0.27, 0.17, 0.42, -0.05, -0.15, 0.05, -0.07, -0.22, -0.34, 0.16, 0.34, 0.1, -0.12, 0.24, 0.45, 0.37, 0.61, 0.9, -0.25, 0.02)
BS <- sample(V1, 100000, replace=TRUE)
The approxfun functions do not help as are for interpolating within data sets. Have found a few questions/answers covering interpolating within different data sets, but not within one data set. Thank you for your help.
EDIT: please note I do not want to fit a normal distribution (or any other) to create those points.
You can use approx() (or approxfun()) to do this by treating BS as the y-coordinate and using sequential x-coordinates:
set.seed(1L); BS <- sample(V1,1e5L,T);
res <- approx(seq_along(BS),BS,n=length(BS)*2L-1L)$y;
The specification of n here is important. It ensures that exactly one interpolated value will be produced halfway between each adjacent pair of input values.
Here's a plot of an excerpt of the result, centered around the first occurrence of an adjacent pair of 0.26 and 0.28:
i <- which(BS[-length(BS)]==0.26 & BS[-1L]==0.28)[1L];
j <- i*2L-1L;
xlim <- c(j-6L,j+8L);
ylim <- c(-1,1);
xticks <- seq(xlim[1L],xlim[2L]);
yticks <- seq(ylim[1L],ylim[2L],0.05);
plot(NA,xlim=xlim,ylim=ylim,xlab='res index',ylab='y',axes=F,xaxs='i',yaxs='i');
abline(v=xticks,col='lightgrey');
abline(h=yticks,col='lightgrey');
axis(1L,xticks,cex.axis=0.7);
axis(2L,yticks,sprintf('%.02f',round(yticks,2L)),las=1L,cex.axis=0.7);
x <- seq(xlim[1L],xlim[2L],2L); y <- BS[seq(i-3L,len=8L)];
points(x,y,pch=16L,col='red',xpd=NA);
x <- seq(xlim[1L],xlim[2L]); y <- res[x];
points(x,y,pch=4L,cex=1.2,col='blue',xpd=NA);
text(x+0.24,y+0.03,y,cex=0.7,xpd=NA);
legend(xlim[1L]+1.5,0.87,c('input value','interpolated'),col=c('red','blue'),pch=c(16L,4L));

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