Number of items to replace is not a multiple of replacement length? - r

Question

Not sure what is wrong here.
Building the model along with the example
LBsPD <- c()
for (i in 1:5000) {
FishCaught <- sample(x=c(7,4,2), size=1, prob = c(.1,.6,.3),replace = TRUE)
YellowPercent <- sample(x=c(0,.25,.35), size=1, prob = c(.25,.5,.25),replace = TRUE)
BluePercent <- 1-YellowPercent
BlueLBs <- rnorm(n=365, mean=35, sd=18)
YellowLBs <- rnorm(n=365, mean=30, sd=18)
LBsPerDay <- FishCaught * ((BluePercent * BlueLBs[BlueLBs > 20]) + (YellowPercent * YellowLBs[YellowLBs > 20]))
LBsPD[i] <- LBsPerDay
}
Keep getting the 50+ errors "Number of items to replace is not a multiple of replacement length" But in the example it is the same.

Answer 1

Here's a drawn-out explanation that #AndrewChisholm started.
I'll start by (1) setting the random seed, so you can repeat this in your console, and (2) stepping through the for loop's first iteration.
set.seed(42) # since this is a random process
i <- 1 # first pass in the loop
FishCaught <- sample(x=c(7,4,2), size=1, prob = c(.1,.6,.3),replace = TRUE)
YellowPercent <- sample(x=c(0,.25,.35), size=1, prob = c(.25,.5,.25),replace = TRUE)
BluePercent <- 1-YellowPercent
BlueLBs <- rnorm(n=365, mean=35, sd=18)
Now, let's look at the components of your next expression:
# FishCaught * ((BluePercent * BlueLBs[BlueLBs > 20]) +
# (YellowPercent * YellowLBs[YellowLBs > 20]))
str(BlueLBs[BlueLBs > 20])
# num [1:291] 24.8 41.5 46.4 42.3 33.1 ...
str(YellowLBs[YellowLBs > 20])
# num [1:255] 64.1 22.5 36.3 59.3 31.6 ...
It doesn't matter now that BluePercent is 1 and YellowPercent is 0, since 0*somevec is still the length of the vector, so you are effectively trying to add vectors of different sizes. What does this mean to you?
c(1,3,5,7,9) + c(1,1000,1)
# Warning in c(1, 3, 5, 7, 9) + c(1, 1000, 1) :
# longer object length is not a multiple of shorter object length
# [1] 2 1003 6 8 1009
The bigger problem here is that R does not consider this a problem: it warns you that it is suspect, but it happily "recycles" the values for you. So this is not what is causing your error.
LBsPerDay <- FishCaught * ((BluePercent * BlueLBs[BlueLBs > 20]) + (YellowPercent * YellowLBs[YellowLBs > 20]))
# Warning in (BluePercent * BlueLBs[BlueLBs > 20]) + (YellowPercent * YellowLBs[YellowLBs > :
# longer object length is not a multiple of shorter object length
str(LBsPerDay)
# num [1:291] 174 291 325 296 232 ...
This is not a syntax error, but you should treat it as a "data is now corrupt" error, because you really don't know for certain what numbers were added/multiplied to other numbers (see my previous example of with c(1,1000,1) to see that if we think things should be aligned, then those results are going to result in some incorrect logical conclusions from this process).
Here's the real problem:
LBsPD[i] <- LBsPerDay
# Warning in LBsPD[i] <- LBsPerDay :
# number of items to replace is not a multiple of replacement length
First, some clarification:
This is a Warning, not an Error. The only way I get an error with that is if I had previous set options(warn=2) (which, btw, is not a bad idea here). Warnings can often be ignored if you expect them, but in this case you should really pay attention to it and treat it as an error.
LBsPerDay is length 291, but you try trying to cram 291 numbers into one position in the vector LBsPD[i]. That is, the length of the LHS using [i] is always going to be the length of i, which is 1; whereas the length of the RHS is (in this case) 291.
Options:
I'm inferring that your BlueLBs[BlueLBs > 20] might be a filter so that fish caught below 20 (pounds? kilos? grams?) will not be "caught". In that case, let's just replace those under 20 with 0 ... but please check me on this logic ... a blue/yellow that is below 20 will be changed to 0, effectively "not caught":
LBsPerDay <- FishCaught * ((BluePercent * replace(BlueLBs, BlueLBs <= 20, 0)) + (YellowPercent * replace(YellowLBs, YellowLBs <= 20, 0)))
str(LBsPerDay)
# num [1:365] 174 291 325 296 232 ...
(No warning, no error.)
If you intend LBsPD to contain all of the weights for each iteration in your simulation, then start with LBsPD <- list(), in which case you'll eventually use
LBsPD <- list()
for (i in 1:5000) {
# ...
LBsPD[[i]] <- LBsPerDay
}
where after 3 (of 5000) iterations, your LBsPD looks like:
str(LBsPD)
# List of 3
# $ : num [1:365] 174 291 325 296 232 ...
# $ : num [1:365] 160.7 97 161.5 145 99.6 ...
# $ : num [1:365] 30.3 121.4 111.7 210.8 139.7 ...
BTW, you might notice that both BlueLBs and YellowLBs have negatives ... not sure if that's a problem, but negative pounds seems suspect. (Because "normal distributions" are by definition unbounded, many things labeled as "normally distributed" are typically not asymptotically compliant. For hasty simulations like this, I often revert to a triangular distribution, which may be normal-enough for some applications, and certainly never gives negative or extremely-positive weights.)